Order of operations with fractions and exponents

WEBVTT
Kind: captions
Language: en

00:00:00.360 --> 00:00:01.210
- [Instructor] Pause this video

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and see if you can
evaluate this expression

00:00:03.230 --> 00:00:05.460
before we do it together.

00:00:05.460 --> 00:00:07.850
All right, now let's
work on this together.

00:00:07.850 --> 00:00:10.720
And we see that we have a lot
of different operations here.

00:00:10.720 --> 00:00:12.720
We have exponents. We have multiplication.

00:00:12.720 --> 00:00:14.500
We have addition. We have division.

00:00:14.500 --> 00:00:16.280
We have parentheses.

00:00:16.280 --> 00:00:19.020
And so to interpret this properly,

00:00:19.020 --> 00:00:23.430
we just have to remind ourselves
of the order of operations.

00:00:23.430 --> 00:00:26.700
So you start with parentheses,
then go to exponents,

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then multiplication and division,

00:00:28.050 --> 00:00:30.470
then addition and subtraction.

00:00:30.470 --> 00:00:33.780
So we see that we're going
to, whatever is over here,

00:00:33.780 --> 00:00:35.510
we're eventually going to square it.

00:00:35.510 --> 00:00:38.510
That's the only place that
we have the parentheses.

00:00:38.510 --> 00:00:39.930
But how are we going to evaluate

00:00:39.930 --> 00:00:42.890
what's inside of these parentheses?

00:00:42.890 --> 00:00:45.660
So let's, then, think about, all right,

00:00:45.660 --> 00:00:49.810
we have an exponent here
that we can evaluate.

00:00:49.810 --> 00:00:53.490
We know that 2 squared is
the same thing as 2 times 2,

00:00:53.490 --> 00:00:55.730
which is the same thing as 4.

00:00:55.730 --> 00:00:57.790
No more exponents to evaluate.

00:00:57.790 --> 00:01:00.470
So then we go to
multiplication and division.

00:01:00.470 --> 00:01:04.820
So we know by how this
fraction sign is written

00:01:04.820 --> 00:01:06.640
that we need to evaluate the numerator

00:01:06.640 --> 00:01:09.380
and then divide it by
the entire denominator

00:01:09.380 --> 00:01:10.830
right over here.

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Now in this numerator, we
have to remind ourselves

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that we do this multiplication
before we do this addition.

00:01:16.350 --> 00:01:18.650
We don't just go left to right.

00:01:18.650 --> 00:01:20.780
So we know that it's 1 plus,

00:01:20.780 --> 00:01:23.560
and I could put parentheses
here to really emphasize

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that we do the multiplication first.

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So before this gets too messy,

00:01:26.880 --> 00:01:28.590
let me just rewrite everything.

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I'm going to do this
multiplication up here first,

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and actually in the denominator

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I'm going to do this
multiplication first as well.

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So this is all going to
simplify to 1 over 14

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or 1/14 times,

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Now this numerator here is going to be

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1 plus 4 times 3.

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4 times 3 is 12.

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All of that is going to be over

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7 plus 2 times 3,

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which is of course equal to 6.

00:02:00.240 --> 00:02:04.650
And then I am going to
have our plus 1 here,

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and then I square everything.

00:02:07.370 --> 00:02:11.270
Well now we can evaluate this numerator

00:02:11.270 --> 00:02:13.900
and this denominator.

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Find another color to do it in.

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This numerator, 1 plus 12,

00:02:18.710 --> 00:02:21.770
is going to be equal to 13.

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And 7 plus 6, interestingly,
is also equal to 13.

00:02:26.410 --> 00:02:28.740
So we 1/14 or 1 divided by 14

00:02:28.740 --> 00:02:30.930
times this whole thing squared,

00:02:30.930 --> 00:02:34.980
and inside you have 13
divided by 13 plus 1.

00:02:34.980 --> 00:02:38.090
Well, we know we need to do
division before we do addition.

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So we will want to evaluate this part

00:02:41.600 --> 00:02:43.940
before we do the addition.

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What is 13 divided by 13?

00:02:46.540 --> 00:02:48.610
Well that's just going to be equal to 1.

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So I can rewrite this as 1 over 14

00:02:52.840 --> 00:02:57.383
times 1 plus 1,

00:02:58.400 --> 00:02:59.880
all of that squared.

00:02:59.880 --> 00:03:03.640
And now we'll want to do this
parentheses. So let's do that.

00:03:03.640 --> 00:03:08.423
1 plus 1 is going to be
equal, of course, to 2.

00:03:09.400 --> 00:03:11.580
And then we're going to do the exponents.

00:03:11.580 --> 00:03:16.100
2 squared is, of course, equal to 4.

00:03:16.100 --> 00:03:19.870
And then we're going to
multiply 1 over 14 times 4.

00:03:19.870 --> 00:03:21.900
Now you could interpret
this, and they're equivalent.

00:03:21.900 --> 00:03:23.200
You could say, hey, this is the same thing

00:03:23.200 --> 00:03:26.040
as multiplying 1/14 times 4.

00:03:26.040 --> 00:03:27.680
Or you could say this is the same thing

00:03:27.680 --> 00:03:31.460
as multiplying 1 times 4 divided by 14.

00:03:31.460 --> 00:03:35.490
1 times 4 divided by 14.

00:03:35.490 --> 00:03:36.590
Either way you look at it,

00:03:36.590 --> 00:03:40.880
you're going to get 4
over 14, and we're done.

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If you want, you could rewrite this

00:03:42.440 --> 00:03:46.090
by dividing both the numerator
and the denominator by 2,

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and you could get 2 over 7.

00:03:48.730 --> 00:03:51.260
But that's how we can evaluate

00:03:51.260 --> 00:03:53.110
this pretty complex expression

00:03:53.110 --> 00:03:56.853
just step-by-step looking at
what we can simplify first.


Order of operations introduction

WEBVTT
Kind: captions
Language: en

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- Every few months,

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you'll see an expression like
this go viral on social media

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because it looks simple

00:00:06.610 --> 00:00:09.490
but depending on how people
interpret this expression,

00:00:09.490 --> 00:00:11.800
they often get different answers.

00:00:11.800 --> 00:00:14.240
So first, why don't you pause this video

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and think about what you think
three plus two times five

00:00:18.550 --> 00:00:19.913
is equal to.

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All right, now let's work
through this together.

00:00:24.260 --> 00:00:26.560
Now some of you might
have interpreted this

00:00:26.560 --> 00:00:29.140
as we should just go from left to right.

00:00:29.140 --> 00:00:30.230
So you might've said, all right,

00:00:30.230 --> 00:00:32.280
let me first add three plus two.

00:00:32.280 --> 00:00:33.660
So I can do that part first,

00:00:33.660 --> 00:00:35.680
which is of course equal to five.

00:00:35.680 --> 00:00:39.850
And then I could multiply that times five.

00:00:39.850 --> 00:00:44.630
And that is going to get me to 25.

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Now others of you would have said no,

00:00:47.170 --> 00:00:49.410
that makes no sense at all.

00:00:49.410 --> 00:00:53.400
We know that we should do
some operations before others.

00:00:53.400 --> 00:00:57.130
For example, multiplication
should come before addition.

00:00:57.130 --> 00:00:58.990
And so if you follow that,

00:00:58.990 --> 00:01:03.490
you would do the two times
five first to get 10.

00:01:03.490 --> 00:01:06.830
And so this would become three plus 10,

00:01:06.830 --> 00:01:11.210
which is of course,
going to be equal to 13.

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Who do you think would be right?

00:01:14.020 --> 00:01:16.130
Well, it turns out that
the second way of doing it

00:01:16.130 --> 00:01:17.340
is the correct way

00:01:17.340 --> 00:01:19.100
and that's because we have something known

00:01:19.100 --> 00:01:21.920
as order of operations.

00:01:21.920 --> 00:01:25.290
And the order of operations
is the convention

00:01:25.290 --> 00:01:27.750
that mathematicians have decided to use

00:01:27.750 --> 00:01:31.127
in order for us to have one
way to interpret an expression

00:01:31.127 --> 00:01:33.810
like this and the order of operations

00:01:33.810 --> 00:01:35.760
are to do parentheses first.

00:01:35.760 --> 00:01:36.700
So, for example,

00:01:36.700 --> 00:01:40.700
if you really wanted to add
the three and the two first,

00:01:40.700 --> 00:01:43.180
you should put parentheses around it

00:01:43.180 --> 00:01:46.500
to say, hey, that's what
you've got to do first.

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But then after parentheses, do exponents,

00:01:49.360 --> 00:01:52.500
which is really, you could view
as repeated multiplication,

00:01:52.500 --> 00:01:54.900
then do multiplication and division.

00:01:54.900 --> 00:01:58.050
Then do addition and subtraction.

00:01:58.050 --> 00:02:01.910
Now, some people might just
memorize this as PEMDAS,

00:02:01.910 --> 00:02:05.340
or PEDMAS or something like that.

00:02:05.340 --> 00:02:06.290
And you can do that,

00:02:06.290 --> 00:02:08.220
but there's a rationale to this.

00:02:08.220 --> 00:02:09.600
Parentheses are just,

00:02:09.600 --> 00:02:11.520
the person who writing
the expression saying

00:02:11.520 --> 00:02:13.740
do this for sure first.

00:02:13.740 --> 00:02:14.710
But then if you think about it,

00:02:14.710 --> 00:02:17.490
exponents are repeated multiplication

00:02:17.490 --> 00:02:19.950
and multiplication is repeated addition.

00:02:19.950 --> 00:02:23.200
So you're doing the most
repeated things first,

00:02:23.200 --> 00:02:25.620
then next repeated things
like multiplication.

00:02:25.620 --> 00:02:27.910
And then you go straight to the addition.

00:02:27.910 --> 00:02:29.930
The reason why multiplication and division

00:02:29.930 --> 00:02:31.030
are on the same line

00:02:31.030 --> 00:02:33.480
and addition and
subtraction on the same line

00:02:33.480 --> 00:02:36.690
is the convention there
is just go left to right.

00:02:36.690 --> 00:02:40.420
So now that we are armed
with our order of operations,

00:02:40.420 --> 00:02:44.280
let's tackle another
slightly hairier expression.

00:02:44.280 --> 00:02:47.270
And I use the term hairy for
a little bit more complex.

00:02:47.270 --> 00:02:51.560
So two times 10, minus eight,

00:02:51.560 --> 00:02:56.170
divided by four, plus one.

00:02:56.170 --> 00:02:58.610
Pause the video and see
how you could evaluate this

00:02:58.610 --> 00:03:01.023
now that you know about
order of operations.

00:03:02.560 --> 00:03:04.470
All right, now let's do this together.

00:03:04.470 --> 00:03:06.480
We don't see any parentheses here.

00:03:06.480 --> 00:03:08.220
We don't see any exponents here

00:03:08.220 --> 00:03:12.170
but we do see some
multiplication and division here.

00:03:12.170 --> 00:03:14.360
So we could see it over here.

00:03:14.360 --> 00:03:16.070
This two times 10.

00:03:16.070 --> 00:03:19.470
And we also have this
eight divided by four.

00:03:19.470 --> 00:03:21.600
So that's what we're
going to want to do first.

00:03:21.600 --> 00:03:25.530
So that two times 10, that
is going to be equal to 20.

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The eight divided by four,
is of course, equal to two.

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And then we have 20 minus two plus one.

00:03:35.050 --> 00:03:38.420
And this is now going to get
us, if we go left to right,

00:03:38.420 --> 00:03:41.580
which we should do when we're
just adding and subtracting,

00:03:41.580 --> 00:03:46.580
is we get 20 minus two is 18, plus one,

00:03:46.630 --> 00:03:49.860
which is going to be equal to 19.

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And we're done.

00:03:51.270 --> 00:03:52.103
Now, you might say, well,

00:03:52.103 --> 00:03:54.980
what if we wanted someone to
add the four and the one first?

00:03:54.980 --> 00:03:56.170
Well, one way to do that

00:03:56.170 --> 00:03:59.300
is you could have added a
parentheses there to force it

00:03:59.300 --> 00:04:02.130
or you could have even written
your division differently.

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For example, you could use a fraction sign

00:04:05.170 --> 00:04:06.210
to make things clearer.

00:04:06.210 --> 00:04:08.620
If we wanted to add the
four plus one first,

00:04:08.620 --> 00:04:12.890
instead of writing eight divided
by four plus one like that,

00:04:12.890 --> 00:04:17.890
you could write eight over four plus one.

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Now of course, this is
fundamentally different

00:04:21.240 --> 00:04:22.560
than this right over here.

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'Cause here we would do the
eight divided by four first,

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then add the plus one,

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while here, you would
add the four plus one

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and then divide eight by that.


Quantitative electrolysis

WEBVTT
Kind: captions
Language: en

00:00:00.350 --> 00:00:02.600
- [Instructor] We already know
that in an electrolytic cell,

00:00:02.600 --> 00:00:04.260
current or movement of electrons

00:00:04.260 --> 00:00:06.990
is used to drive a redox reaction.

00:00:06.990 --> 00:00:10.700
And if we look at a generic
reduction half-reaction,

00:00:10.700 --> 00:00:12.880
the stoichiometry of the half reaction

00:00:12.880 --> 00:00:15.580
shows how many electrons are needed

00:00:15.580 --> 00:00:18.610
to reduce a generic metal ion.

00:00:18.610 --> 00:00:22.780
For example, if we say that
our generic metal ion is M2+,

00:00:22.780 --> 00:00:27.560
it takes two electrons to
turn M2+ into the solid metal.

00:00:27.560 --> 00:00:32.560
So one mole of M2+ ions
requires two moles of electrons

00:00:33.110 --> 00:00:35.800
to form one mole of the metal.

00:00:35.800 --> 00:00:39.240
So we could use this mole
ratio of one mole of metal

00:00:39.240 --> 00:00:42.350
to two moles of electrons to figure out,

00:00:42.350 --> 00:00:44.840
say, how much of a metal is produced

00:00:44.840 --> 00:00:47.130
in an electrolytic reaction.

00:00:47.130 --> 00:00:48.480
So this is an example

00:00:48.480 --> 00:00:51.840
of a quantitative electrolysis problem.

00:00:51.840 --> 00:00:55.630
Moles of electrons are related
to the amount of charge

00:00:55.630 --> 00:00:58.540
that passes through the electrolytic cell

00:00:58.540 --> 00:01:02.687
because one mole of electrons
carries 96,485 coulombs.

00:01:04.890 --> 00:01:06.660
So we could set this up as a constant

00:01:06.660 --> 00:01:08.190
called Faraday's constant.

00:01:08.190 --> 00:01:10.390
So Faraday's constant is symbolized by F,

00:01:10.390 --> 00:01:15.390
and it's equal to 96,485
coulombs per mole of electrons.

00:01:16.460 --> 00:01:20.550
And charge is related to current
and time by this equation.

00:01:20.550 --> 00:01:23.440
So I is equal to Q over t;

00:01:23.440 --> 00:01:27.720
where I is the current; Q,
capital Q is the charge;

00:01:27.720 --> 00:01:29.240
and t is the time.

00:01:29.240 --> 00:01:33.200
Sometimes, you see a
lower case Q for charge.

00:01:33.200 --> 00:01:35.740
Charge is measured in coulombs,

00:01:35.740 --> 00:01:38.580
and time is in seconds.

00:01:38.580 --> 00:01:42.380
And one coulomb per second
is equal to one ampere,

00:01:42.380 --> 00:01:45.870
so current is measured in amperes.

00:01:45.870 --> 00:01:48.090
Next, let's look at a
conversion chart that shows

00:01:48.090 --> 00:01:51.500
how to approach quantitative
electrolysis problems.

00:01:51.500 --> 00:01:54.570
And let's start over here
with this box on the far left,

00:01:54.570 --> 00:01:57.900
which says that charge is
equal to current times time.

00:01:57.900 --> 00:01:59.970
This comes from our equation for current,

00:01:59.970 --> 00:02:02.000
I is equal to Q over t.

00:02:02.000 --> 00:02:04.550
So solving for charge, solving for Q,

00:02:04.550 --> 00:02:06.920
Q is equal to I times t

00:02:06.920 --> 00:02:10.190
or charge is equal to current times time.

00:02:10.190 --> 00:02:13.440
Charge is related to moles of electrons

00:02:13.440 --> 00:02:15.770
by Faraday's constant.

00:02:15.770 --> 00:02:19.000
So we can convert back
and forth between the two

00:02:19.000 --> 00:02:21.260
using Faraday's constant.

00:02:21.260 --> 00:02:24.690
And moles of electrons
are related to the moles

00:02:24.690 --> 00:02:27.320
of the substance that we're
interested in the problem

00:02:27.320 --> 00:02:29.500
by the balanced half-reaction.

00:02:29.500 --> 00:02:31.330
So the balanced half-reaction allows us

00:02:31.330 --> 00:02:34.310
to convert back and forth
between moles of electrons

00:02:34.310 --> 00:02:36.130
and moles of substance.

00:02:36.130 --> 00:02:37.380
And from stoichiometry,

00:02:37.380 --> 00:02:39.370
we know that the moles of a substance

00:02:39.370 --> 00:02:42.840
is related to the grams of the
substance by the molar mass.

00:02:42.840 --> 00:02:45.710
So the molar mass allows us
to convert back and forth

00:02:45.710 --> 00:02:47.580
between these quantities.

00:02:47.580 --> 00:02:49.640
So it helps to look at a conversion chart

00:02:49.640 --> 00:02:51.400
when doing these sorts of problems.

00:02:51.400 --> 00:02:53.910
However, if you don't remember the chart,

00:02:53.910 --> 00:02:56.860
you can always use
dimensional analysis and units

00:02:56.860 --> 00:02:59.460
to figure out the answer to the question.

00:02:59.460 --> 00:03:01.190
Next, let's use our conversion chart

00:03:01.190 --> 00:03:04.530
to figure out a quantitative
electrolysis problem.

00:03:04.530 --> 00:03:07.740
Let's say we have an aqueous
solution of silver nitrate

00:03:07.740 --> 00:03:12.740
and a constant current of
2.40 amperes is applied

00:03:12.940 --> 00:03:16.020
for 1,225 seconds.

00:03:16.020 --> 00:03:19.550
And our goal is to figure
out the mass of silver

00:03:19.550 --> 00:03:20.870
that has formed.

00:03:20.870 --> 00:03:24.290
As a quick reminder, an aqueous
solution of silver nitrate

00:03:24.290 --> 00:03:29.290
contains Ag+ ions and
nitrate anions in solution.

00:03:29.340 --> 00:03:34.150
So the applied current is
used to reduce the Ag+ ions

00:03:34.150 --> 00:03:36.570
to form solid silver.

00:03:36.570 --> 00:03:38.690
And since our goal is
to figure out the mass

00:03:38.690 --> 00:03:40.700
of the solid silver that's formed,

00:03:40.700 --> 00:03:44.130
we're trying to find grams of substance.

00:03:44.130 --> 00:03:49.130
And in the problem we're given
the current and the time,

00:03:49.230 --> 00:03:51.870
so we're starting over
here in the first box

00:03:51.870 --> 00:03:54.230
with the current and the time.

00:03:54.230 --> 00:03:56.450
So let's think about
the steps that we need

00:03:56.450 --> 00:03:58.410
to solve this problem.

00:03:58.410 --> 00:04:01.750
The first step is to find the charge,

00:04:01.750 --> 00:04:04.580
which we can get from
the current and the time.

00:04:04.580 --> 00:04:07.550
The second step is to convert charge

00:04:07.550 --> 00:04:10.440
into moles of electrons.

00:04:10.440 --> 00:04:11.900
From moles of electrons,

00:04:11.900 --> 00:04:16.900
we can convert that into the
moles of silver in this case.

00:04:17.070 --> 00:04:19.820
And in the fourth step for this problem,

00:04:19.820 --> 00:04:24.600
we can convert moles of
silver into grams of silver.

00:04:24.600 --> 00:04:28.240
So step one is to find the charge.

00:04:28.240 --> 00:04:32.440
And Q, the charge, is equal
to the current times the time.

00:04:32.440 --> 00:04:35.760
So the current is equal to 2.40 amperes.

00:04:35.760 --> 00:04:36.593
And remember,

00:04:36.593 --> 00:04:39.880
one ampere is equal to
one coulomb per second.

00:04:39.880 --> 00:04:44.610
So we plug in 2.40 coulombs per second.

00:04:44.610 --> 00:04:48.180
And the time is 1,225 seconds,

00:04:48.180 --> 00:04:51.260
so we plug that in, seconds cancels out,

00:04:51.260 --> 00:04:56.020
and we get that Q is
equal to 2,940 coulombs.

00:04:56.020 --> 00:05:00.440
So that's the charge that's
transferred in that time period.

00:05:00.440 --> 00:05:01.700
Now that we have the charge,

00:05:01.700 --> 00:05:02.810
in the second step,

00:05:02.810 --> 00:05:05.120
our goal is to find
the moles of electrons.

00:05:05.120 --> 00:05:07.730
And charge is related
to moles of electrons

00:05:07.730 --> 00:05:09.800
by Faraday's constant.

00:05:09.800 --> 00:05:13.340
So dividing the charge
by Faraday's constant

00:05:13.340 --> 00:05:15.330
causes coulombs to cancel out,

00:05:15.330 --> 00:05:20.330
and we get 0.0305 moles of electrons.

00:05:20.330 --> 00:05:22.280
And now that we have moles of electrons,

00:05:22.280 --> 00:05:26.230
in step three, we can
find the moles of silver.

00:05:26.230 --> 00:05:29.340
Looking at the balanced
equation for our half reaction

00:05:29.340 --> 00:05:32.540
showing the reduction of silver ions,

00:05:32.540 --> 00:05:37.540
one mole of silver ions
requires one mole of electrons

00:05:38.060 --> 00:05:41.240
to form one mole of solid silver.

00:05:41.240 --> 00:05:44.660
So our mole ratio of silver
to moles of electrons

00:05:44.660 --> 00:05:46.410
is one-to-one,

00:05:46.410 --> 00:05:48.800
which gives us this conversion factor

00:05:48.800 --> 00:05:52.810
of one mole of silver per
one mole of electrons.

00:05:52.810 --> 00:05:55.480
So if we multiply our moles of electrons

00:05:55.480 --> 00:05:59.240
by our conversion factor,
moles of electrons cancels out,

00:05:59.240 --> 00:06:03.630
and that gives us 0.0305 moles of silver.

00:06:03.630 --> 00:06:06.340
In this case, our mole
ratio was one-to-one,

00:06:06.340 --> 00:06:09.410
but if we were reducing
a different metal ion,

00:06:09.410 --> 00:06:11.910
our mole ratio might not be one-to-one.

00:06:11.910 --> 00:06:14.500
So we have to be extra careful on the step

00:06:14.500 --> 00:06:17.270
to make sure we get the right mole ratio.

00:06:17.270 --> 00:06:19.080
And since we have moles of silver,

00:06:19.080 --> 00:06:21.020
in our fourth and final step,

00:06:21.020 --> 00:06:24.980
we can convert moles of
silver into grams of silver.

00:06:24.980 --> 00:06:28.350
So multiplying moles of silver
by the molar mass of silver,

00:06:28.350 --> 00:06:31.750
which is 107.87 grams per mole,

00:06:31.750 --> 00:06:36.750
moles cancel and this gives
us 3.29 grams of silver.

00:06:37.190 --> 00:06:40.940
It's also possible to do
all four steps at once

00:06:40.940 --> 00:06:43.700
in a dimensional analysis approach.

00:06:43.700 --> 00:06:45.090
So looking at our units,

00:06:45.090 --> 00:06:50.090
if we multiply 1,225 seconds
by 2.40 coulombs per second,

00:06:51.460 --> 00:06:54.500
seconds cancels out and gives us coulombs.

00:06:54.500 --> 00:06:57.300
Multiplying that by one mole of electrons

00:06:57.300 --> 00:07:02.220
per 96,485 coulombs, cancels out coulombs.

00:07:02.220 --> 00:07:05.600
Next, we have our conversion
factor of one mole of silver

00:07:05.600 --> 00:07:07.640
per one mole of electrons,

00:07:07.640 --> 00:07:10.040
so moles of electrons cancels out.

00:07:10.040 --> 00:07:12.030
And multiplying that by the molar mass

00:07:12.030 --> 00:07:16.150
of 107.87 grams per one mole,

00:07:16.150 --> 00:07:18.640
moles of silver will cancel out

00:07:18.640 --> 00:07:22.410
and give us a final answer of 3.29 grams.

00:07:22.410 --> 00:07:25.953
So we get the same answer, no
matter which approach we take.


The Nernst equation

WEBVTT
Kind: captions
Language: en

00:00:00.320 --> 00:00:01.153
- [Instructor] We already know

00:00:01.153 --> 00:00:02.810
how to calculate cell potential

00:00:02.810 --> 00:00:04.160
when the reactants and products

00:00:04.160 --> 00:00:06.020
are in their standard states.

00:00:06.020 --> 00:00:08.650
However, what if that's not the case?

00:00:08.650 --> 00:00:11.180
We can find cell potential
when reactants and products

00:00:11.180 --> 00:00:12.680
are not in their standard states

00:00:12.680 --> 00:00:15.870
by using the Nernst equation,
which is shown here.

00:00:15.870 --> 00:00:17.660
And we're gonna call that cell potential

00:00:17.660 --> 00:00:20.580
the instantaneous cell potential.

00:00:20.580 --> 00:00:23.410
E cell is the instantaneous
cell potential,

00:00:23.410 --> 00:00:27.310
the cell potential or voltage
at a specific moment in time.

00:00:27.310 --> 00:00:30.920
E naught of the cell is the
standard cell potential.

00:00:30.920 --> 00:00:32.330
This is the cell potential

00:00:32.330 --> 00:00:33.900
when the reactants and products

00:00:33.900 --> 00:00:35.960
are in their standard states.

00:00:35.960 --> 00:00:38.520
R is the ideal gas constant.

00:00:38.520 --> 00:00:40.500
T is the temperature.

00:00:40.500 --> 00:00:43.590
N is the number of electrons transferred

00:00:43.590 --> 00:00:45.650
in the redox reaction.

00:00:45.650 --> 00:00:47.790
F is Faraday's constant.

00:00:47.790 --> 00:00:51.200
And Q is the reaction quotient.

00:00:51.200 --> 00:00:54.760
So, this is one way to
write the Nernst equation.

00:00:54.760 --> 00:00:57.750
And let's compare this
form of the Nernst equation

00:00:57.750 --> 00:01:00.210
to a simplified form.

00:01:00.210 --> 00:01:03.090
In the simplified form
of the Nernst equation,

00:01:03.090 --> 00:01:05.730
E of the cell is equal
to E naught of the cell,

00:01:05.730 --> 00:01:09.730
minus 0.0592 volts, divided by n,

00:01:09.730 --> 00:01:13.060
times the log of the reaction quotient Q.

00:01:13.060 --> 00:01:15.760
So, going back to the first
form of the Nernst equation,

00:01:15.760 --> 00:01:18.580
if we assume the temperature
is 25 degrees Celsius,

00:01:18.580 --> 00:01:20.190
or 298 Kelvin,

00:01:20.190 --> 00:01:22.620
we multiply that by
the ideal gas constant,

00:01:22.620 --> 00:01:24.520
divide that by Faraday's constant,

00:01:24.520 --> 00:01:27.650
and for changing natural log to log,

00:01:27.650 --> 00:01:31.340
then we end up with this
form for the bottom equation,

00:01:31.340 --> 00:01:35.800
0.0592 volts divided
by n times the log of Q

00:01:35.800 --> 00:01:39.430
for this second term
of the Nernst equation.

00:01:39.430 --> 00:01:42.620
You can use the Nernst equation
to quantitatively calculate

00:01:42.620 --> 00:01:46.030
the exact voltage at a
specific moment in time.

00:01:46.030 --> 00:01:48.380
However, we're gonna try to
think about the Nernst equation

00:01:48.380 --> 00:01:51.000
from a qualitative point of view.

00:01:51.000 --> 00:01:52.870
Let's use the Nernst equation to think

00:01:52.870 --> 00:01:56.160
about the instantaneous cell potential

00:01:56.160 --> 00:01:58.200
for a zinc copper cell.

00:01:58.200 --> 00:02:01.170
And we're gonna look at a
few different situations.

00:02:01.170 --> 00:02:02.550
In a zinc copper cell,

00:02:02.550 --> 00:02:05.830
solid zinc is oxidized
to zinc two plus ions

00:02:05.830 --> 00:02:10.310
and copper two plus ions are
reduced to form solid copper.

00:02:10.310 --> 00:02:13.060
The standard cell potential
for the zinc copper cell

00:02:13.060 --> 00:02:18.060
at 25 degrees Celsius is
equal to positive 1.10 volts.

00:02:18.090 --> 00:02:21.650
And our goal is to find the cell potential

00:02:21.650 --> 00:02:23.170
at the moment of time

00:02:23.170 --> 00:02:25.740
when the concentration
of copper two plus ions

00:02:25.740 --> 00:02:27.820
is equal to 1.0 Mueller

00:02:27.820 --> 00:02:31.410
and the concentration of zinc
two plus ions in solution

00:02:31.410 --> 00:02:34.550
is also equal to 1.0 Mueller.

00:02:34.550 --> 00:02:36.620
To find the instantaneous cell potential,

00:02:36.620 --> 00:02:41.010
we're gonna use the simplified
form of the Nernst equation.

00:02:41.010 --> 00:02:45.920
E naught of the cell is
equal to positive 1.10 volts.

00:02:45.920 --> 00:02:47.220
Next, let's think about n,

00:02:47.220 --> 00:02:50.900
the number of electrons
transferred in the redox reaction.

00:02:50.900 --> 00:02:53.060
So, looking at our equation here

00:02:53.060 --> 00:02:55.010
is a solid zinc going to zinc two plus,

00:02:55.010 --> 00:02:56.320
is a loss of two electrons,

00:02:56.320 --> 00:02:58.720
and copper two plus going to solid copper

00:02:58.720 --> 00:03:00.450
is a gain of two electrons.

00:03:00.450 --> 00:03:05.060
So, two electrons were
transferred and n is equal to two.

00:03:05.060 --> 00:03:09.310
So, I've gone ahead and made n
equal to two in our equation.

00:03:09.310 --> 00:03:12.400
Next, we need to think about
the reaction quotient Q.

00:03:12.400 --> 00:03:13.660
Q has the same form

00:03:13.660 --> 00:03:16.290
as the equilibrium constant expression.

00:03:16.290 --> 00:03:19.070
So, if we think about
the balanced equation,

00:03:19.070 --> 00:03:21.460
remember, pure solids are left out

00:03:21.460 --> 00:03:24.070
of equilibrium constant expressions.

00:03:24.070 --> 00:03:26.530
So, it would be the
concentration of products

00:03:26.530 --> 00:03:29.050
over the concentration of reactants.

00:03:29.050 --> 00:03:31.890
And therefore, Q is equal
to the concentration

00:03:31.890 --> 00:03:35.610
of zinc two plus ions,
divided by the concentration

00:03:35.610 --> 00:03:37.990
of copper two plus ions.

00:03:37.990 --> 00:03:39.610
Because our goal is to find

00:03:39.610 --> 00:03:41.720
the instantaneous cell potential

00:03:41.720 --> 00:03:44.830
when the concentrations are
both equal to one Mueller,

00:03:44.830 --> 00:03:47.560
Q is equal to one divided by one.

00:03:47.560 --> 00:03:49.600
So, Q is equal to one.

00:03:49.600 --> 00:03:53.010
And the log of one is equal to zero.

00:03:53.010 --> 00:03:55.710
Therefore this entire second term

00:03:55.710 --> 00:03:59.080
in the Nernst equation is equal to zero,

00:03:59.080 --> 00:04:01.590
which means the
instantaneous cell potential

00:04:01.590 --> 00:04:04.930
is equal to the standard cell potential.

00:04:04.930 --> 00:04:07.230
So, the instantaneous cell potential

00:04:07.230 --> 00:04:12.230
is equal to positive 1.10 volts.

00:04:12.390 --> 00:04:14.170
And this answer makes a lot of sense

00:04:14.170 --> 00:04:18.240
because standard state for solutions

00:04:18.240 --> 00:04:20.670
is a one Mueller concentration.

00:04:20.670 --> 00:04:23.160
Therefore, the
instantaneous cell potential

00:04:23.160 --> 00:04:25.840
is equal to the standard cell potential

00:04:25.840 --> 00:04:27.393
at this moment in time.

00:04:28.330 --> 00:04:31.400
Next, let's think about the
instantaneous cell potential

00:04:31.400 --> 00:04:34.130
when the concentration
of copper two plus ions

00:04:34.130 --> 00:04:36.030
is equal to 10.0 Mueller

00:04:36.030 --> 00:04:38.340
and the concentration
of zinc two plus ions

00:04:38.340 --> 00:04:40.890
is equal to 1.0 Mueller.

00:04:40.890 --> 00:04:43.880
The standard cell potential
at 25 degrees Celsius

00:04:43.880 --> 00:04:47.540
is still equal to positive 1.10 volts

00:04:47.540 --> 00:04:49.770
and n is still equal to two.

00:04:49.770 --> 00:04:52.370
So, we need to think about
the reaction quotient Q

00:04:52.370 --> 00:04:54.250
at this moment in time.

00:04:54.250 --> 00:04:56.280
The concentration of zinc two plus ions

00:04:56.280 --> 00:04:59.100
is equal to 1.0 Mueller so we plug that in

00:04:59.100 --> 00:05:01.230
and the concentration
of copper two plus ions

00:05:01.230 --> 00:05:03.610
is equal to 10.0 Mueller.

00:05:03.610 --> 00:05:08.610
So, at this moment in
time, Q is less than one.

00:05:09.140 --> 00:05:13.450
The log of a number less
than one is negative.

00:05:13.450 --> 00:05:15.807
Therefore, this second
term would be negative

00:05:15.807 --> 00:05:19.490
or less than zero.

00:05:19.490 --> 00:05:22.810
And since we are subtracting
a negative number,

00:05:22.810 --> 00:05:25.210
we would actually be adding that value

00:05:25.210 --> 00:05:28.140
to the standard cell potential,

00:05:28.140 --> 00:05:30.870
which means that the
instantaneous cell potential

00:05:30.870 --> 00:05:34.290
will be greater than the
standard cell potential.

00:05:34.290 --> 00:05:36.060
And if you went ahead and calculated

00:05:36.060 --> 00:05:39.070
the instantaneous cell potential
with the Nernst equation,

00:05:39.070 --> 00:05:42.730
you would find it's equal
to positive 1.13 volts,

00:05:42.730 --> 00:05:47.200
which is greater than positive 1.10 volts.

00:05:47.200 --> 00:05:49.890
Next, let's think about the
instantaneous cell potential

00:05:49.890 --> 00:05:52.300
when the concentration
of copper two plus ions

00:05:52.300 --> 00:05:55.280
is equal to 1.0 Mueller
and the concentration

00:05:55.280 --> 00:05:58.810
of zinc two plus ions is
equal to 10.0 Mueller.

00:05:58.810 --> 00:06:00.040
Going to the Nernst equation,

00:06:00.040 --> 00:06:03.840
the standard cell potential
is still positive 1.10 volts

00:06:03.840 --> 00:06:06.300
and n is still equal to two.

00:06:06.300 --> 00:06:09.450
So next, we think about
the reaction quotient Q.

00:06:09.450 --> 00:06:10.610
At this moment in time,

00:06:10.610 --> 00:06:12.850
the concentration of zinc two plus ions

00:06:12.850 --> 00:06:14.700
is equal to 10.0 Mueller

00:06:14.700 --> 00:06:17.050
and the concentration
of copper two plus ions

00:06:17.050 --> 00:06:19.270
is equal to 1.0 Mueller.

00:06:19.270 --> 00:06:24.250
So, at this moment in time,
Q is greater than one.

00:06:24.250 --> 00:06:28.260
The log of a number greater
than one is positive.

00:06:28.260 --> 00:06:30.680
Therefore the second term in our equation

00:06:30.680 --> 00:06:34.920
will be positive or greater than zero.

00:06:34.920 --> 00:06:38.460
And because we are
subtracting a positive number,

00:06:38.460 --> 00:06:40.790
we would be subtracting a value

00:06:40.790 --> 00:06:43.330
from the standard cell potential,

00:06:43.330 --> 00:06:46.080
which means that the
instantaneous cell potential

00:06:46.080 --> 00:06:49.580
will be less than the
standard cell potential.

00:06:49.580 --> 00:06:51.860
And if you plug the numbers
into the Nernst equation,

00:06:51.860 --> 00:06:54.840
you would find that the
instantaneous cell potential

00:06:54.840 --> 00:06:58.330
is equal to positive 1.07 volts,

00:06:58.330 --> 00:07:03.330
which is less than the standard
cell voltage of 1.10 volts.

00:07:03.480 --> 00:07:05.400
We just saw that at the moment in time

00:07:05.400 --> 00:07:07.500
when the concentration
of copper two plus ions

00:07:07.500 --> 00:07:08.930
is equal to 1.0 Mueller

00:07:08.930 --> 00:07:11.000
and the concentration
of zinc two plus ions

00:07:11.000 --> 00:07:12.910
is equal to 10.0 Mueller,

00:07:12.910 --> 00:07:14.940
the instantaneous cell potential

00:07:14.940 --> 00:07:18.220
is equal to positive 1.07 volts.

00:07:18.220 --> 00:07:20.850
Let's use this moment in
time as a starting point

00:07:20.850 --> 00:07:22.410
and let's think about what happens

00:07:22.410 --> 00:07:24.810
to the instantaneous cell potential

00:07:24.810 --> 00:07:27.340
as the reaction progresses.

00:07:27.340 --> 00:07:31.090
Remember that the instantaneous
cell potential is related

00:07:31.090 --> 00:07:32.660
to the instantaneous change

00:07:32.660 --> 00:07:35.760
in free energy, delta G, by this equation.

00:07:35.760 --> 00:07:39.420
Delta G is equal to negative nFE.

00:07:39.420 --> 00:07:41.140
Because we have a positive value

00:07:41.140 --> 00:07:43.110
for the instantaneous cell potential,

00:07:43.110 --> 00:07:45.930
if we plug in a positive
value into our equation

00:07:45.930 --> 00:07:48.060
because of the negative
sign in the equation,

00:07:48.060 --> 00:07:50.990
we would get a negative sign for delta G.

00:07:50.990 --> 00:07:52.570
And when delta G is negative,

00:07:52.570 --> 00:07:55.040
the reaction is thermodynamically favored,

00:07:55.040 --> 00:07:59.040
which means the reactants
will turn into products.

00:07:59.040 --> 00:08:00.890
And as the reactions turn into products,

00:08:00.890 --> 00:08:04.240
the concentration of zinc
two plus ions will increase

00:08:04.240 --> 00:08:08.380
and the concentration of copper
two plus ions will decrease.

00:08:08.380 --> 00:08:12.040
And think about what that does
to the reaction quotient Q.

00:08:12.040 --> 00:08:14.080
There would be an increase
in the concentration

00:08:14.080 --> 00:08:17.380
of zinc two plus ions and a
decrease in the concentration

00:08:17.380 --> 00:08:19.140
of copper two plus ions.

00:08:19.140 --> 00:08:23.690
So, Q would no longer be
equal to 10.0 over 1.0.

00:08:23.690 --> 00:08:26.360
We would see an increase in Q

00:08:26.360 --> 00:08:29.400
as the reactants turn into the products.

00:08:29.400 --> 00:08:33.730
An increased value of Q
means that the second term

00:08:33.730 --> 00:08:37.930
in the Nernst equation will
be a larger, positive number.

00:08:37.930 --> 00:08:40.800
So, if we subtract a
larger positive number

00:08:40.800 --> 00:08:42.650
than we did in the previous example

00:08:42.650 --> 00:08:44.800
from the standard cell potential,

00:08:44.800 --> 00:08:48.340
we would get a lower
instantaneous cell potential

00:08:48.340 --> 00:08:51.520
than positive 1.07 volts.

00:08:51.520 --> 00:08:53.500
So, what the Nernst equation tells us

00:08:53.500 --> 00:08:57.460
is as the reaction goes to the right,

00:08:57.460 --> 00:09:01.150
there is an increase in
Q and as Q increases,

00:09:01.150 --> 00:09:05.870
there's a decrease in the
instantaneous cell potential.

00:09:05.870 --> 00:09:08.390
So, we've just seen as the
reaction goes to the right,

00:09:08.390 --> 00:09:10.360
that's an increase in Q.

00:09:10.360 --> 00:09:11.850
And according to the Nernst equation,

00:09:11.850 --> 00:09:15.630
that's a decrease in the
instantaneous cell potential.

00:09:15.630 --> 00:09:17.700
And therefore, the
instantaneous cell potential

00:09:17.700 --> 00:09:20.530
would decrease from 1.07 volts.

00:09:20.530 --> 00:09:23.040
However, the voltage
would still be positive,

00:09:23.040 --> 00:09:26.010
which means that delta G
would still be negative,

00:09:26.010 --> 00:09:27.380
which means that the reaction

00:09:27.380 --> 00:09:29.810
is still thermodynamically favored.

00:09:29.810 --> 00:09:31.010
And so, this keeps happening.

00:09:31.010 --> 00:09:33.590
As the reaction goes to the
right, Q keeps increasing

00:09:33.590 --> 00:09:37.040
and the instantaneous cell
potential keeps decreasing.

00:09:37.040 --> 00:09:39.727
So, we had started out
at positive 1.07 volts

00:09:39.727 --> 00:09:41.470
and we saw that that had decreased

00:09:41.470 --> 00:09:44.140
and it keeps on decreasing,
keeps on decreasing.

00:09:44.140 --> 00:09:47.510
Eventually the
instantaneous cell potential

00:09:47.510 --> 00:09:49.550
will go to zero volts.

00:09:49.550 --> 00:09:51.210
And when the voltage goes to zero,

00:09:51.210 --> 00:09:53.310
if you plug in zero into our equation,

00:09:53.310 --> 00:09:55.820
delta G is equal to zero.

00:09:55.820 --> 00:09:57.700
And when delta G is equal to zero,

00:09:57.700 --> 00:10:00.020
the reaction is at equilibrium.

00:10:00.020 --> 00:10:01.740
And the reaction quotient Q

00:10:01.740 --> 00:10:05.060
is equal to the equilibrium constant K.

00:10:05.060 --> 00:10:08.540
So, if you were to plug in
the equilibrium constant K

00:10:08.540 --> 00:10:09.500
in for Q,

00:10:09.500 --> 00:10:12.710
you would find that
this entire second term

00:10:12.710 --> 00:10:15.610
would be equal to positive 1.10 volts

00:10:15.610 --> 00:10:18.370
and cancel out the
standard cell potential,

00:10:18.370 --> 00:10:22.350
giving an instantaneous
potential of zero volts.

00:10:22.350 --> 00:10:25.930
So, if we think about our
zinc copper cell as a battery,

00:10:25.930 --> 00:10:28.090
when the reaction reaches equilibrium,

00:10:28.090 --> 00:10:32.750
the voltage is equal to zero
and the battery is dead.

00:10:32.750 --> 00:10:34.780
Let's do a quick summary
of what we've learned

00:10:34.780 --> 00:10:36.090
from the Nernst equation.

00:10:36.090 --> 00:10:37.780
So, this is the simplified form,

00:10:37.780 --> 00:10:41.430
which is valid at 25 degrees Celsius.

00:10:41.430 --> 00:10:42.910
When Q is equal to one,

00:10:42.910 --> 00:10:45.430
the instantaneous cell potential

00:10:45.430 --> 00:10:48.480
is equal to the standard cell potential.

00:10:48.480 --> 00:10:50.040
When Q is greater than one,

00:10:50.040 --> 00:10:51.940
the instantaneous cell potential

00:10:51.940 --> 00:10:54.760
is less than the standard cell potential.

00:10:54.760 --> 00:10:56.200
When Q is less than one,

00:10:56.200 --> 00:10:57.820
the instantaneous cell potential

00:10:57.820 --> 00:11:00.700
is greater than the
standard cell potential.

00:11:00.700 --> 00:11:03.980
And when Q is equal to the
equilibrium constant K,

00:11:03.980 --> 00:11:07.990
the instantaneous cell potential
is equal to zero volts.

00:11:07.990 --> 00:11:12.290
When Q is equal to K and the
potential is equal to zero,

00:11:12.290 --> 00:11:14.680
the reaction is at equilibrium.

00:11:14.680 --> 00:11:16.850
However, for these first three situations,

00:11:16.850 --> 00:11:20.260
Q is not equal to K and the
instantaneous cell voltage

00:11:20.260 --> 00:11:21.940
is not equal to zero.

00:11:21.940 --> 00:11:25.683
Therefore, the reaction
is not at equilibrium.


Standard potential, free energy, and the equilibrium constant

WEBVTT
Kind: captions
Language: en

00:00:00.480 --> 00:00:02.330
- [Instructor] For a
generic redox reaction

00:00:02.330 --> 00:00:05.440
where the reactants
turn into the products,

00:00:05.440 --> 00:00:09.550
the free energy is
related to the potential

00:00:09.550 --> 00:00:11.580
for the redox reaction.

00:00:11.580 --> 00:00:13.520
The equation that relates free energy

00:00:13.520 --> 00:00:16.940
and potential is given by delta G is equal

00:00:16.940 --> 00:00:19.770
to negative nFE.

00:00:19.770 --> 00:00:23.850
Delta G is the instantaneous
difference in free energy

00:00:23.850 --> 00:00:26.500
between the reactants and the products

00:00:26.500 --> 00:00:29.290
and refers to the number of electrons

00:00:29.290 --> 00:00:32.490
that are transferred
in the redox reaction.

00:00:32.490 --> 00:00:34.860
F is Faraday's constant,

00:00:34.860 --> 00:00:39.500
which tells us the charge
carried by one mole of electrons.

00:00:39.500 --> 00:00:44.500
So one mole of electrons
carries 96,485 coulombs.

00:00:44.850 --> 00:00:48.570
E refers to the potential
of the redox reaction.

00:00:48.570 --> 00:00:52.180
You could also call this the
voltage of the redox reaction.

00:00:52.180 --> 00:00:54.190
If the reactants and products are both

00:00:54.190 --> 00:00:55.620
in their standard states,

00:00:55.620 --> 00:00:58.790
we can add the superscript
naught to delta G

00:00:58.790 --> 00:01:01.580
and the superscript naught to E.

00:01:01.580 --> 00:01:04.980
So we can use this equation
to calculate delta G naught

00:01:04.980 --> 00:01:07.560
from E naught or vice versa.

00:01:07.560 --> 00:01:10.920
We could calculate E
naught from delta G naught.

00:01:10.920 --> 00:01:13.490
And remember that delta
G naught is connected

00:01:13.490 --> 00:01:17.740
to the equilibrium constant
K by this equation.

00:01:17.740 --> 00:01:22.420
Delta G naught is equal to
negative RT natural log of K,

00:01:22.420 --> 00:01:26.930
where R is the ideal gas constant
and T is the temperature.

00:01:26.930 --> 00:01:28.780
And because delta G naught is a part

00:01:28.780 --> 00:01:30.720
of both of these two equations,

00:01:30.720 --> 00:01:35.440
these two equations connect
the standard potential E naught

00:01:35.440 --> 00:01:39.420
to the standard change in
free energy delta G naught,

00:01:39.420 --> 00:01:43.280
which is then connected to
the equilibrium constant K.

00:01:43.280 --> 00:01:45.940
So let's look at an
example of a redox reaction

00:01:45.940 --> 00:01:49.210
where we know the value for the
standard potential E naught,

00:01:49.210 --> 00:01:52.430
and first we're gonna
calculate delta G naught.

00:01:52.430 --> 00:01:53.900
And then from delta G naught,

00:01:53.900 --> 00:01:57.520
we're gonna calculate the
equilibrium constant K.

00:01:57.520 --> 00:01:59.200
So here's our redox reaction,

00:01:59.200 --> 00:02:01.380
and we're gonna calculate delta G naught

00:02:01.380 --> 00:02:05.060
at 25 degrees Celsius from E naught.

00:02:05.060 --> 00:02:07.400
So for this redox reaction,

00:02:07.400 --> 00:02:11.330
E naught is equal to positive 1.54 volts,

00:02:11.330 --> 00:02:14.200
and our goal is to
calculate delta G naught

00:02:14.200 --> 00:02:16.560
at 25 degrees Celsius.

00:02:16.560 --> 00:02:18.390
In our redox reaction,

00:02:18.390 --> 00:02:22.520
silver one plus is being
reduced to solid silver,

00:02:22.520 --> 00:02:25.280
and solid chromium is being oxidized

00:02:25.280 --> 00:02:27.290
to chromium three plus.

00:02:27.290 --> 00:02:29.650
Now that we've identified
what's being oxidized

00:02:29.650 --> 00:02:31.290
and what's being reduced,

00:02:31.290 --> 00:02:35.500
let's write the oxidation
and reduction half reactions.

00:02:35.500 --> 00:02:37.980
So for our reduction half reaction,

00:02:37.980 --> 00:02:40.840
we have silver one plus
gaining an electron

00:02:40.840 --> 00:02:42.530
to form solid silver.

00:02:42.530 --> 00:02:45.270
And for the oxidation half reaction,

00:02:45.270 --> 00:02:48.530
solid chromium is turning
into chromium three plus

00:02:48.530 --> 00:02:50.940
and losing three electrons.

00:02:50.940 --> 00:02:52.070
Next, we need to figure out

00:02:52.070 --> 00:02:54.410
how many electrons are being transferred

00:02:54.410 --> 00:02:56.400
in our redox reaction.

00:02:56.400 --> 00:02:58.090
In the reduction half reaction,

00:02:58.090 --> 00:03:01.180
the silver one plus cation
is gaining an electron,

00:03:01.180 --> 00:03:03.730
but in the oxidation half reaction,

00:03:03.730 --> 00:03:06.960
solid chromium is losing three electrons.

00:03:06.960 --> 00:03:09.730
And we need to make the
number of electrons equal.

00:03:09.730 --> 00:03:13.130
Therefore, we need to multiply
the first half reaction

00:03:13.130 --> 00:03:16.060
through by a factor of three.

00:03:16.060 --> 00:03:19.460
So now when we add together our reactants,

00:03:19.460 --> 00:03:21.230
so all the reactants on one side

00:03:21.230 --> 00:03:24.000
and all the products on the other side,

00:03:24.000 --> 00:03:26.200
the three electrons cancel out

00:03:26.200 --> 00:03:29.440
and give us our overall redox reaction.

00:03:29.440 --> 00:03:31.210
Because three electrons,

00:03:31.210 --> 00:03:33.530
three moles of electrons
were being transferred

00:03:33.530 --> 00:03:38.530
in our redox reaction,
n is equal to three.

00:03:39.440 --> 00:03:41.360
And since we know that n is equal to three

00:03:41.360 --> 00:03:46.010
and the standard potential
is equal to +1.54 volts,

00:03:46.010 --> 00:03:47.790
we're ready to plug into our equation

00:03:47.790 --> 00:03:50.010
to calculate delta G naught.

00:03:50.010 --> 00:03:53.040
So I went ahead and
rewrote our redox reaction,

00:03:53.040 --> 00:03:56.280
and our goal is to
calculate delta G naught.

00:03:56.280 --> 00:03:57.830
We already know what n is,

00:03:57.830 --> 00:03:59.530
we already know what E naught is,

00:03:59.530 --> 00:04:02.070
and F is Faraday's constant.

00:04:02.070 --> 00:04:05.080
Here's our equation with
everything plugged in.

00:04:05.080 --> 00:04:06.470
N is equal to three.

00:04:06.470 --> 00:04:09.320
There were three moles
of electrons transferred

00:04:09.320 --> 00:04:12.920
in our redox reaction
per mole of reaction.

00:04:12.920 --> 00:04:14.430
F is Faraday's constant,

00:04:14.430 --> 00:04:19.420
which is 96,485 coulombs
per mole of electrons.

00:04:19.420 --> 00:04:23.480
And the standard voltage
was equal to 1.54 volts.

00:04:23.480 --> 00:04:26.700
One volt is equal to
one joule per coulomb,

00:04:26.700 --> 00:04:28.270
so I went ahead and changed the units

00:04:28.270 --> 00:04:31.530
from volts to joules per coulomb.

00:04:31.530 --> 00:04:33.800
Next, we can cancel out some units.

00:04:33.800 --> 00:04:37.140
So moles of electrons will cancel out,

00:04:37.140 --> 00:04:39.610
and coulombs will cancel out,

00:04:39.610 --> 00:04:43.000
and that gives us joules
per mole of reaction.

00:04:43.000 --> 00:04:44.030
And after we do the math,

00:04:44.030 --> 00:04:47.170
we find that delta G naught
at 25 degrees Celsius

00:04:47.170 --> 00:04:49.210
for this redox reaction is equal

00:04:49.210 --> 00:04:53.360
to -4.46 times 10 to the 5th joules

00:04:53.360 --> 00:04:55.130
per mole of reaction.

00:04:55.130 --> 00:04:56.100
We could also write this

00:04:56.100 --> 00:04:58.770
in terms of kilojoules
per mole of reaction.

00:04:58.770 --> 00:05:00.040
So delta G naught is equal

00:05:00.040 --> 00:05:04.800
to -446 kilojoules per mole of reaction.

00:05:04.800 --> 00:05:07.150
Notice that a positive voltage

00:05:07.150 --> 00:05:10.250
and the negative in the equation means

00:05:10.250 --> 00:05:13.750
that we get a negative
sign in the final answer.

00:05:13.750 --> 00:05:15.990
And when delta G naught is negative,

00:05:15.990 --> 00:05:19.300
the reaction is thermodynamically favored,

00:05:19.300 --> 00:05:22.070
which means at 25 degrees Celsius,

00:05:22.070 --> 00:05:24.950
silver cations will
react with solid chromium

00:05:24.950 --> 00:05:29.210
to form solid silver and
chromium three plus ions.

00:05:29.210 --> 00:05:31.330
And what this example shows us is

00:05:31.330 --> 00:05:33.360
whenever the voltage is positive,

00:05:33.360 --> 00:05:36.540
that means a thermodynamically
favored reaction

00:05:36.540 --> 00:05:40.300
because delta G naught will be negative.

00:05:40.300 --> 00:05:43.030
Now that we've calculated delta
G naught for our reaction,

00:05:43.030 --> 00:05:45.650
let's calculate the equilibrium constant K

00:05:45.650 --> 00:05:47.940
at 25 degrees Celsius.

00:05:47.940 --> 00:05:50.500
So we need to plug in our
value for delta G naught,

00:05:50.500 --> 00:05:54.160
the ideal gas constant,
the temperature in Kelvin,

00:05:54.160 --> 00:05:57.640
and we'll solve for the
equilibrium constant.

00:05:57.640 --> 00:06:00.450
I've gone ahead and plugged
everything into our equation.

00:06:00.450 --> 00:06:02.480
So here's the value for delta G naught

00:06:02.480 --> 00:06:04.960
in joules per mole of reaction.

00:06:04.960 --> 00:06:08.560
The ideal gas constant
is equal to 8.314 joules

00:06:08.560 --> 00:06:10.700
per Kelvin mole of reaction.

00:06:10.700 --> 00:06:14.390
And 25 degrees Celsius,
if we add 273 to that,

00:06:14.390 --> 00:06:17.360
we get 298 Kelvin.

00:06:17.360 --> 00:06:20.500
Looking at the units, Kelvin cancels out

00:06:20.500 --> 00:06:24.630
and so does joules per mole of reaction.

00:06:24.630 --> 00:06:25.970
And solving for K,

00:06:25.970 --> 00:06:29.270
the equilibrium constant
is approximately equal

00:06:29.270 --> 00:06:32.060
to 10 to the 78th power.

00:06:32.060 --> 00:06:35.480
And since K is such an
extremely large number,

00:06:35.480 --> 00:06:38.780
that tells us the reaction
essentially goes to completion,

00:06:38.780 --> 00:06:42.020
and all the reactants turn into products.

00:06:42.020 --> 00:06:43.560
So for a reaction like this,

00:06:43.560 --> 00:06:47.410
it's not necessary to have
equilibrium arrows in here.

00:06:47.410 --> 00:06:50.970
We can just draw an
arrow going to the right.

00:06:50.970 --> 00:06:54.730
And remember, we started with
a positive standard potential

00:06:54.730 --> 00:06:56.710
for this redox reaction.

00:06:56.710 --> 00:06:59.510
So to summarize what we've
seen with these calculations,

00:06:59.510 --> 00:07:02.360
a positive value for the
standard potential leads

00:07:02.360 --> 00:07:05.000
to a negative value for delta G naught,

00:07:05.000 --> 00:07:08.350
so a thermodynamically
favored redox reaction.

00:07:08.350 --> 00:07:10.980
And a negative value
for delta G naught leads

00:07:10.980 --> 00:07:15.010
to an equilibrium constant
K that is greater than one.

00:07:15.010 --> 00:07:17.680
We've just seen what
happens to delta G naught

00:07:17.680 --> 00:07:20.060
and the equilibrium constant K

00:07:20.060 --> 00:07:23.270
when the standard potential is positive

00:07:23.270 --> 00:07:24.950
or greater than zero.

00:07:24.950 --> 00:07:28.030
Now let's think about a
generic redox reaction

00:07:28.030 --> 00:07:31.010
where the standard potential is negative,

00:07:31.010 --> 00:07:33.660
so E naught is less than zero.

00:07:33.660 --> 00:07:36.230
So let's think about the
sign for delta G naught

00:07:36.230 --> 00:07:40.760
if we plug in a negative value
for the standard potential.

00:07:40.760 --> 00:07:42.770
So if we have a negative voltage,

00:07:42.770 --> 00:07:45.890
the two negative signs in the equation

00:07:45.890 --> 00:07:49.330
will give us a positive
value for delta G naught.

00:07:49.330 --> 00:07:52.060
A positive value for delta G naught means

00:07:52.060 --> 00:07:54.246
that the forward reaction

00:07:54.246 --> 00:07:56.560
of reactants turning into products

00:07:56.560 --> 00:08:00.620
is a thermodynamically
unfavorable reaction.

00:08:00.620 --> 00:08:03.030
And if we were to plug in a positive value

00:08:03.030 --> 00:08:04.600
for delta G naught,

00:08:04.600 --> 00:08:07.300
we would find that the
equilibrium constant K

00:08:07.300 --> 00:08:09.580
would be less than one.

00:08:09.580 --> 00:08:11.200
And when K is less than one,

00:08:11.200 --> 00:08:15.000
that tells us whenever the
reaction reaches equilibrium,

00:08:15.000 --> 00:08:19.220
there will be a lot more
reactants than products.

00:08:19.220 --> 00:08:20.640
Let's look at a quick summary

00:08:20.640 --> 00:08:22.940
of what we've learned from these examples.

00:08:22.940 --> 00:08:24.600
The standard potential E naught

00:08:24.600 --> 00:08:27.440
and the standard change in
free energy delta G naught

00:08:27.440 --> 00:08:30.040
are related by the equation on the left.

00:08:30.040 --> 00:08:31.330
And delta G naught

00:08:31.330 --> 00:08:34.160
and the equilibrium constant K are related

00:08:34.160 --> 00:08:36.340
by the equation on the right.

00:08:36.340 --> 00:08:38.000
When E naught is greater than zero,

00:08:38.000 --> 00:08:40.160
delta G naught is less than zero,

00:08:40.160 --> 00:08:43.280
and the reaction is
thermodynamically favored,

00:08:43.280 --> 00:08:45.320
which means at equilibrium,

00:08:45.320 --> 00:08:47.270
there'll be more products than reactants,

00:08:47.270 --> 00:08:50.950
and the equilibrium constant
K will be greater than one.

00:08:50.950 --> 00:08:53.310
However, when E naught is less than zero,

00:08:53.310 --> 00:08:55.680
delta G naught is greater than zero,

00:08:55.680 --> 00:08:58.970
and the reaction is
thermodynamically unfavored,

00:08:58.970 --> 00:09:00.830
which means at equilibrium,

00:09:00.830 --> 00:09:03.140
there'll be more reactants than products,

00:09:03.140 --> 00:09:06.733
and the equilibrium constant
K will be less than one.


Worked example: Calculating E° using standard reduction potentials

WEBVTT
Kind: captions
Language: en

00:00:00.480 --> 00:00:02.210
- [Instructor] Let's do a worked example

00:00:02.210 --> 00:00:04.910
where we calculate the standard potential

00:00:04.910 --> 00:00:09.640
at 25 degrees Celsius for this reaction.

00:00:09.640 --> 00:00:11.020
In this redox reaction,

00:00:11.020 --> 00:00:14.950
silver cations are reduced
to form solid silver

00:00:14.950 --> 00:00:19.950
and solid chromium is
oxidized to form the Cr3+ ion.

00:00:20.160 --> 00:00:22.980
The first step is to write
down the half reactions

00:00:22.980 --> 00:00:26.060
that make up the overall redox reaction.

00:00:26.060 --> 00:00:29.360
So we said that silver
cations are reduced,

00:00:29.360 --> 00:00:31.100
therefore we need to gain the electron

00:00:31.100 --> 00:00:34.270
to turn into solid
silver and solid chromium

00:00:34.270 --> 00:00:38.610
to turn into chromium 3+ ions
must lose three electrons.

00:00:38.610 --> 00:00:41.900
Next, we need to find
the standard voltages

00:00:41.900 --> 00:00:43.800
for our two half reactions.

00:00:43.800 --> 00:00:44.633
And to do that,

00:00:44.633 --> 00:00:48.210
we could consult a standard
reduction potential table,

00:00:48.210 --> 00:00:50.790
and here's our table that
shows standard reduction

00:00:50.790 --> 00:00:53.450
potentials for some
reduction half reactions

00:00:53.450 --> 00:00:55.920
at 25 degrees Celsius.

00:00:55.920 --> 00:00:57.840
The standard reduction potentials

00:00:57.840 --> 00:01:00.500
or standard reduction voltages
for these half reactions

00:01:00.500 --> 00:01:05.064
are all compared to the
reduction of H+ ion.

00:01:05.064 --> 00:01:08.740
So two H+ plus two electrons
forming hydrogen gas

00:01:08.740 --> 00:01:12.840
has a standard reduction
potential of exactly zero volts.

00:01:12.840 --> 00:01:15.270
For our particular redox reaction,

00:01:15.270 --> 00:01:17.330
we need to know the reduction potential

00:01:17.330 --> 00:01:21.150
for the reduction of silver
cations to form solid silver.

00:01:21.150 --> 00:01:24.010
The standard reduction
potential for this half reaction

00:01:24.010 --> 00:01:27.330
is equal to positive 0.80 volts.

00:01:27.330 --> 00:01:30.180
The other half reaction that
we need to know about involves

00:01:30.180 --> 00:01:34.380
the oxidation of solid chromium
to chromium 3+ cations.

00:01:34.380 --> 00:01:37.410
But since this is a standard
reduction potential table,

00:01:37.410 --> 00:01:41.760
the half reaction is written
as a reduction half reaction.

00:01:41.760 --> 00:01:44.580
The standard reduction
potential for this half reaction

00:01:44.580 --> 00:01:48.000
is negative 0.74 volts.

00:01:48.000 --> 00:01:50.100
But since we need this
half reaction written

00:01:50.100 --> 00:01:52.300
as an oxidation half reaction,

00:01:52.300 --> 00:01:56.190
if we were to reverse this half
reaction, how it's written,

00:01:56.190 --> 00:01:59.190
we would need to change
the sign of the voltage.

00:01:59.190 --> 00:02:02.160
So the standard oxidation potential

00:02:02.160 --> 00:02:05.740
would be positive 0.74 volts.

00:02:05.740 --> 00:02:08.410
So I've gone ahead and
written in the voltages

00:02:08.410 --> 00:02:10.200
for our half reactions.

00:02:10.200 --> 00:02:13.030
The standard reduction
potential for our half reaction

00:02:13.030 --> 00:02:15.170
was positive 0.80 volts,

00:02:15.170 --> 00:02:18.450
and the standard oxidation
potential for our half reaction

00:02:18.450 --> 00:02:21.370
is positive 0.74 volts.

00:02:21.370 --> 00:02:24.620
Our next step is to make the
number of electrons equal

00:02:24.620 --> 00:02:27.930
for our two half reactions
and add them together.

00:02:27.930 --> 00:02:30.470
For our oxidation half reaction,

00:02:30.470 --> 00:02:32.210
we're losing three electrons,

00:02:32.210 --> 00:02:34.330
but for our reduction half reaction,

00:02:34.330 --> 00:02:36.420
we're only gaining one electron.

00:02:36.420 --> 00:02:39.060
Therefore we need to
multiply everything through

00:02:39.060 --> 00:02:42.550
in our reduction half reaction by three,

00:02:42.550 --> 00:02:47.550
that gives us 3Ag+ plus
three electrons goes to 3Ag.

00:02:48.020 --> 00:02:50.600
Notice that even though
we multiplied everything

00:02:50.600 --> 00:02:54.010
through in our reduction half
reaction by a factor of three,

00:02:54.010 --> 00:02:57.100
we did not multiply the
standard reduction potential

00:02:57.100 --> 00:02:58.660
by a factor of three.

00:02:58.660 --> 00:03:02.560
And that's because voltage
is an intensive property

00:03:02.560 --> 00:03:06.220
and doesn't depend on
the amount of substance.

00:03:06.220 --> 00:03:08.820
So it doesn't matter if we're
talking about the reduction

00:03:08.820 --> 00:03:10.320
of one mole of silver cations,

00:03:10.320 --> 00:03:13.660
or three moles of silver cations.

00:03:13.660 --> 00:03:16.380
The standard reduction
potential is the same

00:03:16.380 --> 00:03:19.240
for both half reactions.

00:03:19.240 --> 00:03:22.100
And when we add our two
half reactions together,

00:03:22.100 --> 00:03:24.040
so here are all of the reactants

00:03:24.040 --> 00:03:27.410
and then over here would
be all of the products.

00:03:27.410 --> 00:03:30.540
The three electrons would
cancel out on both sides

00:03:30.540 --> 00:03:33.680
and give us 3Ag+ plus solid chromium

00:03:33.680 --> 00:03:37.440
goes to 3Ag plus Cr3+,

00:03:37.440 --> 00:03:41.390
which gives us back our
original redox reaction.

00:03:41.390 --> 00:03:43.960
And since we were able to
add our two half reactions

00:03:43.960 --> 00:03:47.380
together and get our
overall redox reaction,

00:03:47.380 --> 00:03:51.420
to find the standard
voltage for this reaction,

00:03:51.420 --> 00:03:54.250
we should be able to add
together the voltages

00:03:54.250 --> 00:03:56.760
for the two half reactions.

00:03:56.760 --> 00:04:01.760
So positive 0.8 plus 0.74 is
equal to positive 1.54 volts.

00:04:05.450 --> 00:04:08.700
So the standard potential
for this redox reaction

00:04:08.700 --> 00:04:13.623
at 25 degrees Celsius is
equal to positive 1.54 volts.

00:04:14.580 --> 00:04:17.470
There's another way to
calculate the standard potential

00:04:17.470 --> 00:04:19.250
for this redox reaction,

00:04:19.250 --> 00:04:23.200
and this way only uses
standard reduction potentials.

00:04:23.200 --> 00:04:26.050
So to calculate the standard potential

00:04:26.050 --> 00:04:27.890
for the redox reaction,

00:04:27.890 --> 00:04:30.290
we take the standard reduction potential

00:04:30.290 --> 00:04:32.170
for the reduction process

00:04:32.170 --> 00:04:35.920
and from that we subtract the
standard reduction potential

00:04:35.920 --> 00:04:38.550
for the oxidation process.

00:04:38.550 --> 00:04:40.850
So for our reduction half reaction

00:04:40.850 --> 00:04:42.970
the standard reduction potential

00:04:42.970 --> 00:04:45.860
is equal to positive 0.80 volts.

00:04:45.860 --> 00:04:48.260
So we'd plug that into our equation

00:04:48.260 --> 00:04:50.210
for the reduction process.

00:04:50.210 --> 00:04:53.330
And for our oxidation half reaction,

00:04:53.330 --> 00:04:55.530
the standard reduction potential,

00:04:55.530 --> 00:04:58.830
if you remember from the standard
reduction potential table

00:04:58.830 --> 00:05:01.900
is equal to negative 0.74 volts.

00:05:01.900 --> 00:05:04.440
So that would get
plugged into our equation

00:05:04.440 --> 00:05:06.900
for the oxidation process.

00:05:06.900 --> 00:05:08.920
And when we plug in our voltages,

00:05:08.920 --> 00:05:11.620
we get the same answer that we got before.

00:05:11.620 --> 00:05:14.480
So the standard potential for our reaction

00:05:14.480 --> 00:05:19.480
is equal to positive 1.54
volts at 25 degrees Celsius.


Standard cell potential

WEBVTT
Kind: captions
Language: en

00:00:00.330 --> 00:00:02.280
- [Instructor] Standard Cell Potential,

00:00:03.557 --> 00:00:04.883
which is also called
Standard Cell Voltage,

00:00:05.793 --> 00:00:07.890
refers to the voltage of
an electrochemical cell

00:00:08.992 --> 00:00:11.013
when reactants and products
are in their standard states,

00:00:12.051 --> 00:00:12.884
at a particular temperature.

00:00:14.043 --> 00:00:15.033
For a zinc copper galvanic cell,

00:00:16.236 --> 00:00:18.663
solid zinc reacts with
copper two plus ions,

00:00:20.104 --> 00:00:22.793
to form solid copper,
and zinc two plus ions.

00:00:24.777 --> 00:00:27.227
Standard Cell Potential
is symbolized by E naught

00:00:28.660 --> 00:00:30.710
of the cell, where the superscript naught

00:00:32.276 --> 00:00:34.611
refers to the fact that
reactants and products

00:00:34.611 --> 00:00:35.663
are in their standard states.

00:00:36.819 --> 00:00:39.060
For a solid, standard state
refers to the pure solid

00:00:40.096 --> 00:00:41.333
under a pressure of one atmosphere.

00:00:42.435 --> 00:00:43.383
So we're talking about pure solid zinc,

00:00:44.397 --> 00:00:45.230
and pure solid copper.

00:00:46.194 --> 00:00:47.547
And for a solution,

00:00:47.547 --> 00:00:50.063
standard state refers to
a one molar concentration.

00:00:51.213 --> 00:00:52.140
So our concentration of copper two

00:00:53.121 --> 00:00:54.323
plus ions in solution is one molar,

00:00:55.531 --> 00:00:58.073
and so is our concentration
of zinc two plus ions.

00:01:00.163 --> 00:01:02.593
And for this zinc copper
cell, at 25 degrees Celsius,

00:01:03.579 --> 00:01:06.843
the Standard Cell Potential is
equal to positive 1.10 volts.

00:01:08.819 --> 00:01:10.320
Now that we've talked about
the Standard Cell Potential

00:01:11.227 --> 00:01:12.213
for a zinc copper cell,

00:01:13.241 --> 00:01:15.353
let's look at a diagram
of this galvanic cell.

00:01:16.858 --> 00:01:18.510
For this cell, the anode
compartment contains

00:01:19.639 --> 00:01:22.150
solid zinc metal, and
one molar concentration

00:01:23.643 --> 00:01:25.633
of zinc two plus ions,
in aqueous solution.

00:01:27.158 --> 00:01:29.063
The anode is where oxidation takes place.

00:01:30.122 --> 00:01:31.123
So at the zinc electrode,

00:01:32.074 --> 00:01:34.674
the solid zinc is converted
into zinc two plus ions,

00:01:35.732 --> 00:01:36.893
and two electrons are lost.

00:01:38.475 --> 00:01:40.883
So this is the oxidation
half reaction for this cell.

00:01:42.092 --> 00:01:45.029
The electrons that are lost
in the oxidation half reaction

00:01:45.029 --> 00:01:46.183
travel through the wire.

00:01:47.598 --> 00:01:50.398
So there's a flow of electrons,
from the zinc electrode,

00:01:51.763 --> 00:01:52.663
toward the copper electrode.

00:01:54.221 --> 00:01:55.054
The cathode compartment

00:01:55.980 --> 00:01:58.580
contains the solid copper
electrode and in solution,

00:01:59.861 --> 00:02:02.153
one molar concentration
of copper two plus ions.

00:02:03.530 --> 00:02:05.273
And the cathode is where
reduction takes place.

00:02:06.424 --> 00:02:07.933
So at the surface of the copper electrode,

00:02:09.077 --> 00:02:10.540
the copper two plus ions in solution

00:02:11.483 --> 00:02:13.883
gain two electrons and
turn into solid copper.

00:02:15.709 --> 00:02:18.253
So this is the reduction
half reaction for this cell.

00:02:19.845 --> 00:02:22.463
If we were to attach a volt
meter to our two electrodes,

00:02:23.511 --> 00:02:25.773
the volt meter would read 1.10 volts.

00:02:27.708 --> 00:02:30.108
And this voltage is the
Standard Cell Potential.

00:02:31.039 --> 00:02:32.189
So E naught of the cell

00:02:33.690 --> 00:02:36.473
is equal to positive 1.10 volts.

00:02:38.709 --> 00:02:40.940
And the Standard Cell Potential
depends on the potentials

00:02:42.113 --> 00:02:44.563
for the two half reactions
that make up the cell.

00:02:46.165 --> 00:02:48.173
So if we know the potentials
for the half reactions

00:02:48.173 --> 00:02:49.469
that make up the cell,

00:02:49.469 --> 00:02:50.920
we can calculate the standard potential

00:02:51.848 --> 00:02:52.681
of any galvanic cell.

00:02:54.020 --> 00:02:55.813
And let's see how we could do
that for this particular cell.

00:02:57.840 --> 00:02:58.673
To calculate these standard cell potential

00:02:59.722 --> 00:03:00.555
for our zinc copper cell,

00:03:01.591 --> 00:03:03.176
we're going to use what are called

00:03:03.176 --> 00:03:04.676
Standard Reduction Potentials.

00:03:06.468 --> 00:03:08.360
Standard Reduction
Potentials are symbolized by

00:03:09.251 --> 00:03:10.351
E naught of reduction.

00:03:11.982 --> 00:03:13.689
And these refer to the voltages

00:03:13.689 --> 00:03:14.522
for half reactions that are

00:03:15.457 --> 00:03:17.003
written as reduction half reactions.

00:03:18.136 --> 00:03:21.086
For example, if copper two plus
is reduced by two electrons

00:03:22.405 --> 00:03:23.238
to form solid copper,

00:03:24.306 --> 00:03:25.703
the voltage for this half reaction,

00:03:26.548 --> 00:03:28.228
or the Standard Reduction Potential,

00:03:28.228 --> 00:03:30.063
is equal to positive 0.34 volts.

00:03:32.047 --> 00:03:33.650
And for the reduction
of zinc two plus ions

00:03:34.831 --> 00:03:36.383
by two electrons to form solid zinc,

00:03:37.317 --> 00:03:38.150
the Standard Reduction Potential

00:03:38.150 --> 00:03:41.263
is equal to negative 0.76 volts.

00:03:43.271 --> 00:03:45.003
We just saw from our
zinc copper cell diagram,

00:03:45.861 --> 00:03:47.711
that copper two plus ions are reduced

00:03:48.818 --> 00:03:49.651
to form solid copper.

00:03:50.597 --> 00:03:52.373
So, we're going to leave this
half reaction as it's written.

00:03:53.767 --> 00:03:56.051
However, in our diagram
for our zinc copper cell,

00:03:56.051 --> 00:03:59.688
we saw that solid zinc was
actually being oxidized to form

00:03:59.688 --> 00:04:00.638
zinc two plus ions.

00:04:02.167 --> 00:04:04.050
So we need to rewrite
the second half reaction

00:04:05.011 --> 00:04:06.463
as an oxidation half reaction.

00:04:07.583 --> 00:04:08.873
And that just means reversing everything.

00:04:10.048 --> 00:04:11.353
And if we're reversing our half reaction,

00:04:12.438 --> 00:04:14.393
we have to change the sign on the voltage.

00:04:15.738 --> 00:04:17.838
So if the potential is negative 0.76 volts

00:04:18.958 --> 00:04:20.013
for the reduction half reaction,

00:04:20.972 --> 00:04:22.522
it would be positive 0.76 volts

00:04:23.540 --> 00:04:24.923
for the oxidation half reaction.

00:04:26.518 --> 00:04:28.553
So now I've changed it to show
the oxidation half reaction,

00:04:29.595 --> 00:04:31.923
solid zinc turning into zinc
two plus ions in solution,

00:04:33.122 --> 00:04:33.955
and the loss of two electrons.

00:04:34.854 --> 00:04:37.604
The Standard Oxidation
Potential for this half reaction

00:04:38.906 --> 00:04:40.163
is positive 0.76 volts.

00:04:42.349 --> 00:04:44.683
The next step is to add the
two half reactions together.

00:04:45.611 --> 00:04:47.163
So here are all of our reactants,

00:04:48.397 --> 00:04:50.047
and here are all of the products.

00:04:51.749 --> 00:04:53.273
Notice how two electrons would cancel out,

00:04:54.295 --> 00:04:57.670
and that gives us copper two
plus ions in aqueous solution,

00:04:57.670 --> 00:05:00.403
plus solid zinc forms solid copper,

00:05:01.806 --> 00:05:03.556
and zinc two plus ions in solution.

00:05:05.484 --> 00:05:06.960
Since we were able to
get the overall equation

00:05:07.899 --> 00:05:09.623
by adding the two half reactions together,

00:05:10.701 --> 00:05:13.480
we should be able to get the
overall Standard Cell Potential

00:05:14.705 --> 00:05:16.330
E naught of the cell,
by adding together the

00:05:17.184 --> 00:05:19.534
standard potentials for
our two half reactions.

00:05:21.154 --> 00:05:22.050
So E naught of the cell is equal to

00:05:23.055 --> 00:05:24.493
E naught of the reduction half reaction,

00:05:25.514 --> 00:05:27.463
plus E naught of the
oxidation half reaction,

00:05:28.561 --> 00:05:31.893
which is equal to 0.34 plus 0.76,

00:05:33.400 --> 00:05:35.023
or positive 1.10 volts.

00:05:36.550 --> 00:05:38.780
So, the Standard Cell Potential
of our zinc copper cell

00:05:39.930 --> 00:05:43.313
is equal to positive 1.10
volts at 25 degrees Celsius.

00:05:45.022 --> 00:05:46.380
We can also calculate
Standard Cell Potential

00:05:47.215 --> 00:05:48.060
using a slightly different form

00:05:48.984 --> 00:05:49.963
of the equation that we just learned.

00:05:51.431 --> 00:05:52.264
Remember that we flipped the sign

00:05:53.198 --> 00:05:54.390
of the Standard Reduction Potential

00:05:55.465 --> 00:05:57.033
to get the Standard Oxidation Potential,

00:05:58.113 --> 00:05:59.863
which we plugged into this equation,

00:06:00.783 --> 00:06:02.348
and were able to calculate

00:06:02.348 --> 00:06:04.098
the standard potential of the cell.

00:06:05.755 --> 00:06:06.593
Instead of flipping the sign ourselves,

00:06:07.633 --> 00:06:08.793
and getting a Standard
Oxidation Potential,

00:06:09.719 --> 00:06:11.850
we could use this new equation to just use

00:06:13.182 --> 00:06:15.347
reduction potentials to calculate

00:06:15.347 --> 00:06:16.953
the standard potential of the cell.

00:06:18.282 --> 00:06:20.390
Notice that this new
equation has a negative sign

00:06:21.299 --> 00:06:22.990
that essentially flips the sign of the

00:06:23.981 --> 00:06:26.353
Standard Reduction Potential
for the oxidation process.

00:06:27.402 --> 00:06:29.470
So, it accomplishes the same
thing that we did ourselves

00:06:30.341 --> 00:06:31.174
in the first equation.

00:06:32.402 --> 00:06:33.623
So to find the standard
potential of the cell,

00:06:34.506 --> 00:06:36.599
we take the Standard Reduction Potential

00:06:36.599 --> 00:06:37.899
for the reduction process,

00:06:38.915 --> 00:06:41.160
and from that, we subtract the
Standard Reduction Potential

00:06:42.164 --> 00:06:43.123
for the oxidation process.

00:06:44.239 --> 00:06:45.333
So, this equation is a bit of a shortcut.

00:06:46.819 --> 00:06:48.090
So when we plug in our
Standard Reduction Potentials

00:06:49.054 --> 00:06:52.123
of positive 0.34 volts
and negative 0.76 volts,

00:06:53.333 --> 00:06:55.330
the negative sign means we
end up with the same answer

00:06:56.264 --> 00:06:58.553
we got before, a positive 1.10 volts.


Electrolytic cells

WEBVTT
Kind: captions
Language: en

00:00:00.420 --> 00:00:01.370
- [Instructor] Electrolytic cells

00:00:01.370 --> 00:00:03.200
use an electric current to drive

00:00:03.200 --> 00:00:06.320
a thermodynamically unfavorable reaction.

00:00:06.320 --> 00:00:09.190
Before we look at a diagram
of electrolytic cell,

00:00:09.190 --> 00:00:10.930
let's look at the half reactions

00:00:10.930 --> 00:00:13.060
that will occur in the cell.

00:00:13.060 --> 00:00:16.550
In one half reaction, liquid sodium ions

00:00:16.550 --> 00:00:20.980
react with an electron to
form liquid sodium metal.

00:00:20.980 --> 00:00:24.530
Because the liquid sodium
ion is gaining an electron,

00:00:24.530 --> 00:00:28.180
this represents the
reduction half reaction.

00:00:28.180 --> 00:00:30.210
The other way to tell
that this is a reduction

00:00:30.210 --> 00:00:33.460
half reaction is to
assign oxidation numbers.

00:00:33.460 --> 00:00:36.980
Liquid sodium ions have
an oxidation number

00:00:36.980 --> 00:00:41.060
of plus one and liquid sodium metal

00:00:41.060 --> 00:00:43.510
has an oxidation number of zero.

00:00:43.510 --> 00:00:45.520
So going from plus one to zero

00:00:45.520 --> 00:00:49.610
is a decrease or a reduction
in the oxidation number.

00:00:49.610 --> 00:00:53.020
Therefore, liquid sodium ions are reduced

00:00:53.020 --> 00:00:56.720
to form liquid sodium metal
in this half reaction.

00:00:56.720 --> 00:00:59.410
In our next unbalanced half reaction,

00:00:59.410 --> 00:01:03.940
liquid chloride anions
turn into chlorine gas.

00:01:03.940 --> 00:01:06.260
If we assign oxidation numbers,

00:01:06.260 --> 00:01:10.540
liquid chloride anions have a
oxidation number of minus one,

00:01:10.540 --> 00:01:13.710
and chlorine gas has an
oxidation number of zero.

00:01:13.710 --> 00:01:15.490
So going from minus one to zero

00:01:15.490 --> 00:01:18.270
is an increase in the oxidation number.

00:01:18.270 --> 00:01:20.100
Therefore, the liquid chloride anions

00:01:20.100 --> 00:01:22.670
are oxidized to chlorine gas.

00:01:22.670 --> 00:01:25.830
So this is the oxidation half reaction,

00:01:25.830 --> 00:01:27.190
and we need to balance it.

00:01:27.190 --> 00:01:29.804
Since we have two chlorines on the right,

00:01:29.804 --> 00:01:32.038
we need to put two as a coefficient

00:01:32.038 --> 00:01:34.810
in front of the chloride
anions on the left.

00:01:34.810 --> 00:01:37.120
Loss of electrons is oxidation.

00:01:37.120 --> 00:01:40.530
And since we're oxidizing
two chloride anions,

00:01:40.530 --> 00:01:43.993
we're going to lose two electrons.

00:01:44.850 --> 00:01:45.850
As a quick reminder,

00:01:45.850 --> 00:01:46.970
one way to remember that

00:01:46.970 --> 00:01:49.850
loss of electrons is oxidation

00:01:49.850 --> 00:01:53.220
and gain of electrons is reduction

00:01:53.220 --> 00:01:56.520
is to think about LEO the lion goes GER.

00:01:56.520 --> 00:01:59.810
So loss of electrons is oxidation

00:01:59.810 --> 00:02:03.320
and gain of electrons is reduction.

00:02:03.320 --> 00:02:05.870
Next let's add our two
half reactions together

00:02:05.870 --> 00:02:08.520
to get the overall redox reaction.

00:02:08.520 --> 00:02:09.620
And before we do that,

00:02:09.620 --> 00:02:11.640
we have to make sure
the number of electrons

00:02:11.640 --> 00:02:14.090
are equal in both half reactions.

00:02:14.090 --> 00:02:15.670
Since we have two electrons

00:02:15.670 --> 00:02:17.380
in the oxidation half reaction

00:02:17.380 --> 00:02:20.710
and only one electron in
the reduction half reaction,

00:02:20.710 --> 00:02:22.320
we need to multiply everything

00:02:22.320 --> 00:02:25.430
in our reduction half
reaction through by two.

00:02:25.430 --> 00:02:28.470
So that'd be two liquid sodium ions,

00:02:28.470 --> 00:02:32.970
two electrons, and two liquid sodiums.

00:02:32.970 --> 00:02:35.100
When we add the two
half reactions together,

00:02:35.100 --> 00:02:37.420
the two electrons cancel out

00:02:37.420 --> 00:02:41.350
and we get two liquid sodium cations

00:02:41.350 --> 00:02:44.630
plus two liquid chloride anions

00:02:44.630 --> 00:02:48.780
goes to two liquid
sodiums and chlorine gas.

00:02:48.780 --> 00:02:51.560
So if we had some molten sodium chloride,

00:02:51.560 --> 00:02:55.460
we could form sodium and chlorine gas.

00:02:55.460 --> 00:02:57.970
However, delta G naught for this reaction

00:02:57.970 --> 00:02:59.750
is greater than zero,

00:02:59.750 --> 00:03:01.200
which means this reaction

00:03:01.200 --> 00:03:04.220
is thermodynamically unfavorable.

00:03:04.220 --> 00:03:06.460
So in order for this reaction to occur,

00:03:06.460 --> 00:03:08.200
we need some sort of a power source

00:03:08.200 --> 00:03:10.770
to provide an electric current,

00:03:10.770 --> 00:03:15.300
to drive this thermodynamically
unfavorable reaction.

00:03:15.300 --> 00:03:18.310
Here's a diagram showing
our electrolytic cell

00:03:18.310 --> 00:03:22.320
for the electrolysis of
molten sodium chloride,

00:03:22.320 --> 00:03:24.390
and let's start with the power source.

00:03:24.390 --> 00:03:26.910
So the negative terminal
of the power source

00:03:26.910 --> 00:03:28.660
is where the electrons come from.

00:03:28.660 --> 00:03:32.460
So imagine we have electrons
moving in this wire

00:03:32.460 --> 00:03:35.460
toward the inert electrode on the right.

00:03:35.460 --> 00:03:36.860
Inert means the electrodes

00:03:36.860 --> 00:03:39.970
aren't going to participate
in this reaction.

00:03:39.970 --> 00:03:41.360
For example, the electrode

00:03:41.360 --> 00:03:43.150
could be a piece of platinum metal,

00:03:43.150 --> 00:03:44.990
which is very unreactive.

00:03:44.990 --> 00:03:47.470
At the surface of the electrode,

00:03:47.470 --> 00:03:51.400
the electrons reduce
the liquid sodium ions

00:03:51.400 --> 00:03:53.680
into liquid sodium.

00:03:53.680 --> 00:03:58.060
Therefore, liquid sodium
will form at this electrode.

00:03:58.060 --> 00:04:00.350
The melting point of sodium chloride

00:04:00.350 --> 00:04:04.110
is higher than the
melting point of sodium.

00:04:04.110 --> 00:04:06.210
Therefore, at this high temperature

00:04:06.210 --> 00:04:08.900
sodium will remain a liquid.

00:04:08.900 --> 00:04:11.760
Eventually when the
liquid sodium cools down,

00:04:11.760 --> 00:04:14.350
you would have some solid sodium metal

00:04:14.350 --> 00:04:16.130
because reduction is occurring

00:04:16.130 --> 00:04:18.090
at the electrode on the right,

00:04:18.090 --> 00:04:21.200
the electrode on the
right must be the cathode.

00:04:21.200 --> 00:04:22.720
Next, let's think about the other

00:04:22.720 --> 00:04:24.940
inert electrode on the left.

00:04:24.940 --> 00:04:27.780
At this electrode,
oxidation is taking place

00:04:27.780 --> 00:04:29.780
because liquid chloride anions

00:04:29.780 --> 00:04:32.240
are turning into chlorine gas.

00:04:32.240 --> 00:04:33.860
So we would see bubbles

00:04:33.860 --> 00:04:37.560
of chlorine gas at this electrode.

00:04:37.560 --> 00:04:41.200
And as the liquid chloride
anions are oxidized,

00:04:41.200 --> 00:04:42.920
that's loss of electrons.

00:04:42.920 --> 00:04:46.770
So the electrons would
flow through this wire

00:04:46.770 --> 00:04:51.110
back toward the positive
terminal of the power source.

00:04:51.110 --> 00:04:52.910
Since oxidation is occurring

00:04:52.910 --> 00:04:55.730
at the inert electrode on the left,

00:04:55.730 --> 00:04:58.660
this electrode must be the anode.

00:04:58.660 --> 00:05:00.950
As a quick review,
remember that a good way

00:05:00.950 --> 00:05:04.500
to remember this is to
think about an ox and a cat.

00:05:04.500 --> 00:05:06.470
So oxidation occurs at the anode.

00:05:06.470 --> 00:05:08.110
That's an ox.

00:05:08.110 --> 00:05:10.770
And reduction occurs at the cathode.

00:05:10.770 --> 00:05:12.960
That's a red cat.

00:05:12.960 --> 00:05:14.280
And as a quick summary,

00:05:14.280 --> 00:05:17.540
this diagram shows the electrolytic cell

00:05:17.540 --> 00:05:20.790
for the electrolysis of
molten sodium chloride

00:05:20.790 --> 00:05:24.550
to form liquid sodium and chlorine gas.

00:05:24.550 --> 00:05:28.170
Let's look at another example
of an electrolytic cell.

00:05:28.170 --> 00:05:29.850
This electrolytic cell shows

00:05:29.850 --> 00:05:32.540
the process of electroplating.

00:05:32.540 --> 00:05:34.670
So let's say you have
a pendant made of steel

00:05:34.670 --> 00:05:36.960
and you do some engraving on it.

00:05:36.960 --> 00:05:38.280
I do a lot of engraving.

00:05:38.280 --> 00:05:39.890
And so I drew a little picture

00:05:39.890 --> 00:05:43.040
of some flowers on this steel pendant,

00:05:43.040 --> 00:05:45.180
however, steel rusts pretty easily.

00:05:45.180 --> 00:05:47.340
So if we want to protect the engraving,

00:05:47.340 --> 00:05:48.680
we could electroplate it

00:05:48.680 --> 00:05:52.120
and put a thin layer of
nickel on top of the steel.

00:05:52.120 --> 00:05:54.790
So electroplating uses electrolysis

00:05:54.790 --> 00:05:58.180
to plate one metal onto another metal,

00:05:58.180 --> 00:06:01.130
either to protect against
rust or corrosion,

00:06:01.130 --> 00:06:03.290
or if you want to plate silver or gold

00:06:03.290 --> 00:06:05.870
onto another metal to
make it more beautiful.

00:06:05.870 --> 00:06:07.560
So for this electrolytic cell,

00:06:07.560 --> 00:06:09.640
we have a steel electrode on the right

00:06:09.640 --> 00:06:12.050
and a nickel electrode on the left.

00:06:12.050 --> 00:06:16.210
And we have an aqueous
solution of nickel sulfate.

00:06:16.210 --> 00:06:17.760
We know that electrons come

00:06:17.760 --> 00:06:20.250
from the negative terminal
of the power source.

00:06:20.250 --> 00:06:23.540
So imagine that electrons move in the wire

00:06:23.540 --> 00:06:26.720
toward the steel electrode on the right.

00:06:26.720 --> 00:06:29.070
When the nickel two
plus ion is in solution

00:06:29.070 --> 00:06:31.180
come in contact with the electrons

00:06:31.180 --> 00:06:33.320
at the surface of the steel electrode,

00:06:33.320 --> 00:06:35.630
reduction takes place.

00:06:35.630 --> 00:06:38.570
So reduction takes place at the cathode

00:06:38.570 --> 00:06:40.000
and for the half reaction,

00:06:40.000 --> 00:06:42.640
it'd be nickel two
plus, plus two electrons

00:06:42.640 --> 00:06:44.620
turns into solid nickel.

00:06:44.620 --> 00:06:47.040
So a thin layer of nickel is now

00:06:47.040 --> 00:06:49.610
plated on the steel object.

00:06:49.610 --> 00:06:53.020
The electrode on the left
is a piece of solid nickel.

00:06:53.020 --> 00:06:55.370
And when solid nickel is oxidized,

00:06:55.370 --> 00:06:58.540
it turns into nickel two plus ions.

00:06:58.540 --> 00:06:59.880
When solid nickel turns

00:06:59.880 --> 00:07:03.000
into nickel two plus ions in solution,

00:07:03.000 --> 00:07:04.990
two electrons are lost.

00:07:04.990 --> 00:07:07.020
And so we can think about electrons

00:07:07.020 --> 00:07:08.760
moving in this wire back toward

00:07:08.760 --> 00:07:11.930
the positive terminal of the power source.

00:07:11.930 --> 00:07:14.010
So for the oxidation half reaction,

00:07:14.010 --> 00:07:17.700
solid nickel turns into
nickel two plus ions,

00:07:17.700 --> 00:07:19.920
and two electrons are lost.

00:07:19.920 --> 00:07:22.510
And because oxidation occurs at the anode,

00:07:22.510 --> 00:07:25.630
the nickel electrode is the anode

00:07:25.630 --> 00:07:27.520
for this electrolytic cell.

00:07:27.520 --> 00:07:30.810
Let's look at a quick summary
for an electrolytic cell.

00:07:30.810 --> 00:07:32.330
In an electrolytic cell,

00:07:32.330 --> 00:07:36.720
the redox reaction is
thermodynamically unfavorable.

00:07:36.720 --> 00:07:38.170
Because the redox reaction

00:07:38.170 --> 00:07:40.430
is thermodynamically unfavorable,

00:07:40.430 --> 00:07:42.330
an electrolytic cell requires

00:07:42.330 --> 00:07:45.550
a power source to supply a current

00:07:45.550 --> 00:07:48.560
to drive the unfavorable redox reaction.

00:07:48.560 --> 00:07:50.440
Oxidation occurs at the anode

00:07:50.440 --> 00:07:53.210
and reduction occurs at the cathode.

00:07:53.210 --> 00:07:54.790
And finally, an electrolytic cell

00:07:54.790 --> 00:07:56.410
is used for electroplating

00:07:56.410 --> 00:07:58.510
where a thin layer of one metal

00:07:58.510 --> 00:08:00.713
is plated onto another metal.