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Today I will begin by talking
about a subject that we were

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just getting to at the end of
class last time,

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and that will be electrolysis.
And this is an interesting

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topic.
One of the examples of

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electrolysis that is commonly
used is that which forms the

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basis for the so-called Dow
process.

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Here, what I am representing is
a container.

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And a container that would be
used to house molten magnesium

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chloride.
If you take magnesium chloride,

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which is a typical salt,
and get it hot enough,

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then it will become liquid.
And when it becomes liquid in

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the melt, what you can do is
carry out a non-spontaneous

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reaction by applying an external
voltage.

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So, the idea is that this is
some source of a potential

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difference.
And you can take a positive

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electrode and a negative
electrode and immerse it into

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the melt.
And if the potential is

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sufficiently great,
you can force a non-spontaneous

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reaction to run,
because overall,

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the reaction will be
spontaneous when you factor in

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the potential difference of the
external source.

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And, in this particular case,
the reaction that would be

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taking place is as follows.
At one side we would have

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magnesium two plus
ions being reduced to,

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at this temperature,
liquid elemental magnesium.

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So, metallic magnesium at one
electrode.

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And, at the same time,
at the other electrode,

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what would be taking place is
the oxidation of chloride ions,

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two of which counterbalance the
magnesium two plus

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in the salt magnesium
dichloride.

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And we put in a couple of
electrons.

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Sorry, we are oxidizing
chloride to Cl two,

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which would be bubbling out as
a gas at the temperatures of

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this melt.
You initially would have this

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molten salt.
You would immerse your two

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electrodes.
You would apply an external

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potential.
And if that potential is great

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enough, then these two reactions
start taking place at the two

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electrodes.
And you can see that in order

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for this to happen,
chloride is being oxidized.

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And so we can see then that
this must be the source of the

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positive electrode and electrons
will be going this way and going

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down here.
And when they collect in this

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electrode on the left,
they are used to reduce

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magnesium two plus
to liquid magnesium.

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So, that is an electrolysis
reaction in which something that

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does not want to happen,
namely the reduction of

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magnesium two plus
to magnesium zero by chloride

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ions, that is non-spontaneous.
And we are forcing it to happen

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by applying this external
potential.

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And just how great must that
external potential be?

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Well, Christine,
could you locate for me on the

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table, there,
the potential for magnesium two

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plus plus two electrons going to
magnesium

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and the potential for
Cl two plus two electrons going

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to two Cl minus?

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Note that our standard
potentials that we defined last

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time, relative to the standard
hydrogen electrode,

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are always written as reduction
potentials by convention.

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What is it here?
Can we zoom in on it?

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Zoom in our finger.

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Okay.
By the way, I had her do this

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in part because I wanted to draw
your attention to this table

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that is in one of the appendices
of the textbook.

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So, it is in the back.
There is an abbreviated form of

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this table right in Chapter 12,
which you should be reading.

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But when you want to find a
standard potential that is not

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in that abbreviated table,
you will now know that you can

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go to your appendix to find that
reduction potential.

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Christine, please read out to
me the value for the chloride.

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Is that the one that you have
there?

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+1.36.
Okay.

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And this is positive of the
standard hydrogen electrode,

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showing you that Cl two
is an oxidizing substance.

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And then what about the
magnesium plus two

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electrons?
-2.36.

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Remember I mentioned last time
that elemental sodium was around

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-2.7 volts, so it was a very
strong reducing agent,

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magnesium.
It is also a very strong

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reducing agent.
Chlorine is an oxidant,

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so clearly the direction of
spontaneity in this system is

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for a magnesium metal to be
reducing Cl two to make

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chloride ions.
And that is why the form of the

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salt that Henry Dow as a young
man was extracting from the

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briny marshes near his home in
Michigan, and later founded the

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Dow company on the basis of
chemistries like this.

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That is how you found it,
as magnesium two plus,

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2 Cl minus.
And so, overall,

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if you reverse the sign of the
anode reaction and add it to the

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cathode reaction,
you are going to see that this

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one is downhill,
sorry, is non-spontaneous by an

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absolute magnitude of 3.72
volts, which is the difference

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between these two values,
this absolute difference in

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potential.
That means that we are going to

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have to apply an external
voltage of greater than 3.72

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volts, positive,
in order to get this thing to

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run and to be able to get
gaseous Cl two bubbling

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out and in order to get liquid
molten magnesium to be formed

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and separating from the molten
magnesium chloride.

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That is a large external
potential.

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That is the price that you have
to pay to make this

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non-spontaneous reaction go.
And that is the minimum price.

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If we could just briefly switch
to the Athena from the document

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camera.
I want to mention this one

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word, overpotential.

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You will see an explicit
definition of the word

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overpotential given in my notes
and in your book,

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but what overpotential is,
in fact, a number.

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And it is the amount of voltage
that is greater than this

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non-spontaneous cell potential
that you have to apply in order

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for the reaction really to start
kicking in and working.

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In practice,
you choose maybe a couple

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platinum electrodes,
and you dip them into the thing

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that you want to electrolyze,
and you apply an external

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potential.
And, in practice,

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that external potential is
something much greater than the

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cell voltage of interest.
And sometimes it is even a

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large amount in excess.
And the reason has to do with

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the barrier to the reaction,
that is introduced by the way

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that it takes place at the
interface on the surface of

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these electrodes,
so you have to apply some

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overpotential.
That is energy that you have to

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put into the system over and
above the inherent energy of the

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electrolysis to make it actually
work in practice.

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And so, the substance that we
would most like to be able to

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electrolyze is water.
Because, if you think about it,

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electrolysis of water would
produce hydrogen and oxygen.

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And then we could have those in
our fuel tank,

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and we can go ahead and let
combustion take place,

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which produces what?
It would produce water.

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You would get a beautiful
complete cycle,

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if you could do a good job of
electrolyzing water into its

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components, H two and O
two.

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And the reason that that is
difficult is because you have to

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put in energy,
somehow, in order to make that

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non-spontaneous reaction go
uphill.

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We would like to make that the
basis for the hydrogen economy.

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I really love Iceland.
This is my favorite country,

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I think.
Although, I have never been

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there.
And it is, because they do this

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reaction.
They do this electrolysis of

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water every night,
so that they will have hydrogen

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and oxygen for combustion
purposes the next day in their

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buses in the city.
And they are actually doing

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this.
And why are they able to do

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that?
Well, their special location on

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the planet gives them a lot of
free electricity that they can

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use to run the electrolysis of
water to make hydrogen and

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oxygen every night.
And then they can combust that

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in their buses.
And you essentially have free

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hydrogen, and it is not a
problem.

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And in places like Iceland or
other places in the world,

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where you have freely available
energy from geothermal sources

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or from hydroelectric power,
then you can do that.

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You can use that energy,
that electricity that is free,

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that the earth is just giving
you, to go ahead and carry out

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the electrolysis of water and
make hydrogen that way.

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And we would like to be able to
do that using photochemistry.

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One possible way to get the
positive holes in the electrons

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apart is to use h nu.
A photon can impinge on a

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material and lead to the
separation of charge in what is

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called a photovoltaic device.
And, if you can interface a

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photovoltaic device to a couple
of electrodes that have surfaces

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that are appropriate,
chemically, for minimizing the

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overpotential,
then you can use light coming

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in from the sun to make
electrolysis of water happen to

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get hydrogen and oxygen from
light that is coming in incident

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on the earth.
And so, one thing that is

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currently ongoing research in a
partnership at Caltech and MIT,

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and I am involved with this,
this is Professor Nocera,

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here.
And you will see this furry

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face walking around the hallways
at MIT.

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And he is an amazing guy.
And if he does what he is

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trying to do right,
he is probably going to be the

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next one in our department to
win the Nobel Prize.

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So, Professor Nocera is
currently the professor of

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chemistry and professor of
energy here at MIT.

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He is the energy professor.
And he is a lot more voluble

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and vigorous,
I think, than I am.

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He is really bouncing off the
walls with energy.

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And that is good because the
technical challenges that we

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have to surmount in order to be
able to use sunlight to carry

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out water oxidation,
water electrolysis are very

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great, but they come down to
chemistry and to molecules.

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And starting on Wednesday,
we are going to enter a part of

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the semester in which we treat
the subject of bonding in

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chemistry in quite a bit of
detail.

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We are going to be talking
about molecular orbitals and

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things like that.
And if you get a good

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understanding of molecular
orbital theory and electronic

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structure and the shapes of
molecules, you may be able to

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contribute to this problem by
helping to design molecules that

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can go on the surface of the
electrodes.

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And the purpose of those
molecules is to reduce the

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overpotential for water
electrolysis with sunlight as

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the source of the electrons and
the holes.

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And this man's entire life is
defined by one number,

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and that is the overpotential.
We are making electrode

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molecules that will bring down
the overpotential and give us a

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free source of clean energy from
water.

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00:14:13,000 --> 00:14:17,000
One week from today,
you are going to have a little

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00:14:17,000 --> 00:14:21,000
introduction to undergraduate
research here at MIT.

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00:14:21,000 --> 00:14:26,000
We are going to have some
visitors who will come in and

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talk to you about the 5
subject that is given during

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IAP.
And the people who successfully

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complete the 5.301 subject are
then guaranteed a position in a

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research laboratory in chemistry
as an undergraduate research

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00:14:43,000 --> 00:14:47,000
opportunities program student.
If you want to get involved in

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00:14:47,000 --> 00:14:51,000
the molecular chemistry of
renewable energy,

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00:14:51,000 --> 00:14:55,000
then you might want to make
sure you are here on time for

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class on Monday,
and you will get that

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introduction.
And you can think about what to

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00:15:02,000 --> 00:15:07,000
do in terms of 5.301.
Now, I want to bring you up to

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speed on some more important
concepts that pertain to

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electrochemistry.
Let's talk about electrical

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work.

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Here, I want to make an analogy
to the way that electrons can do

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work when they fall through a
potential difference and

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experience a potential
difference.

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00:15:45,000 --> 00:15:49,000
I want to make an analogy
between that and the way that

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things go downhill by virtue of
the force of gravity.

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And so, you can think of a
waterfall of electrons flowing

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from a place of high potential
to a place of low potential if

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00:16:05,000 --> 00:16:11,000
you want to get an idea for our
expression for the way that the

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00:16:11,000 --> 00:16:15,000
free energy of reaction is
related to electrical work.

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00:16:15,000 --> 00:16:20,000
And it is the total charge that
is passed multiplied by the

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00:16:20,000 --> 00:16:27,000
potential difference through
which the charge is passing.

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00:16:32,000 --> 00:16:37,000
We have been talking about this
potential difference and calling

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it E.
And if it is standard

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00:16:39,000 --> 00:16:43,000
potential, then we will use the
superscript zero,

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00:16:43,000 --> 00:16:47,000
or the little degree symbol for
a standard.

236
00:16:47,000 --> 00:16:51,000
And now, we want to know what
is the total charge?

237
00:16:51,000 --> 00:16:56,000
It is going to be negative n
times the charge on the electron

238
00:16:56,000 --> 00:17:02,000
times N sub A,
which is Avogadro's number.

239
00:17:02,000 --> 00:17:06,000
And we are going to be
interested in getting the total

240
00:17:06,000 --> 00:17:10,000
charge expressed in terms of a
quantity per mole.

241
00:17:10,000 --> 00:17:15,000
Actually, we are going to
express it in Coulombs per mole.

242
00:17:15,000 --> 00:17:18,000
And so, we combine e and (N)A
together.

243
00:17:18,000 --> 00:17:22,000
And this is one of our
fundamental constants.

244
00:17:22,000 --> 00:17:30,000
This is Faraday's constant.
And you will remember that that

245
00:17:30,000 --> 00:17:37,000
is approximately 96,485.3
Coulombs per mole.

246
00:17:37,000 --> 00:17:45,000
That is Faraday's constant.
And we will use that as our way

247
00:17:45,000 --> 00:17:53,000
of expressing charge here.
And so, we get the important

248
00:17:53,000 --> 00:18:00,000
expression, delta Gr is equal to
negative nFE.

249
00:18:00,000 --> 00:18:07,000
And we will use that a couple

250
00:18:07,000 --> 00:18:12,000
of times today in getting access
to other quantities.

251
00:18:12,000 --> 00:18:15,000
And what is n?
That is going to be the

252
00:18:15,000 --> 00:18:19,000
stoichiometric number of
electrons passed in the

253
00:18:19,000 --> 00:18:23,000
half-reaction of interest.
I will show you a couple of

254
00:18:23,000 --> 00:18:30,000
examples of how you figure out
what n is in this expression.

255
00:18:30,000 --> 00:18:36,000
So, it is the total charge
times the potential difference.

256
00:18:36,000 --> 00:18:42,000
One of the things that this is
useful for is for finding a

257
00:18:42,000 --> 00:18:47,000
relationship between an
equilibrium constant and a

258
00:18:47,000 --> 00:18:52,000
standard cell potential.
And we can do that by recalling

259
00:18:52,000 --> 00:18:59,000
from chemical equilibrium the
expression that delta G is equal

260
00:18:59,000 --> 00:19:05,000
to minus RT ln K.

261
00:19:05,000 --> 00:19:12,000
And now we have a new
expression, down here,

262
00:19:12,000 --> 00:19:18,000
for delta G.
And let's go ahead and insert

263
00:19:18,000 --> 00:19:30,000
that, and we will see that nFE
is going to be equal to RT ln K.

264
00:19:30,000 --> 00:19:35,000
And so, we can just rearrange
that and solve for quantity of

265
00:19:35,000 --> 00:19:39,000
interest here.
We are going to see that the

266
00:19:39,000 --> 00:19:44,000
equilibrium constant,
K, is equal to an exponential

267
00:19:44,000 --> 00:19:49,000
of nFE divided by RT. And

268
00:19:49,000 --> 00:19:53,000
this should be for a standard
potential.

269
00:19:53,000 --> 00:19:57,000
And then this is the
equilibrium constant,

270
00:19:57,000 --> 00:20:02,000
that special value of Q that
applies when the system is at

271
00:20:02,000 --> 00:20:06,000
equilibrium.
So, let me put the little

272
00:20:06,000 --> 00:20:11,000
superscript up here to make it
clear that we are talking here

273
00:20:11,000 --> 00:20:16,000
about a standard potential.
The idea is then if you have a

274
00:20:16,000 --> 00:20:20,000
reaction and you can express it
in terms of half-reactions,

275
00:20:20,000 --> 00:20:24,000
then you can go to our table in
the appendix,

276
00:20:24,000 --> 00:20:29,000
if those half-reactions happen
to be in the table.

277
00:20:29,000 --> 00:20:33,000
And then you can calculate the
value of the standard potential

278
00:20:33,000 --> 00:20:37,000
for the cell that would be
comprised of those two

279
00:20:37,000 --> 00:20:40,000
half-reactions,
as we described last time.

280
00:20:40,000 --> 00:20:45,000
And then, given these other
quantities, we can calculate the

281
00:20:45,000 --> 00:20:48,000
equilibrium constant for the
reaction.

282
00:20:48,000 --> 00:20:51,000
So, it is a useful thing to be
able to do.

283
00:20:51,000 --> 00:20:54,000
You take a standard cell
potential and then take a

284
00:20:54,000 --> 00:21:00,000
temperature like the standard
temperature, 298.15K.

285
00:21:00,000 --> 00:21:03,000
And you have your value of F
that I gave you.

286
00:21:03,000 --> 00:21:06,000
And you can look up,
of course, the value for the

287
00:21:06,000 --> 00:21:08,000
gas constant.
And you have n,

288
00:21:08,000 --> 00:21:12,000
from the stoichiometry of the
number of electrons passing

289
00:21:12,000 --> 00:21:14,000
through this potential
difference.

290
00:21:14,000 --> 00:21:17,000
And then you can get K.
And you see that this

291
00:21:17,000 --> 00:21:21,000
equilibrium constant K goes up
exponentially as the standard

292
00:21:21,000 --> 00:21:25,000
cell potential increases.
That is quite an interesting

293
00:21:25,000 --> 00:21:26,000
relationship,
too.

294
00:21:26,000 --> 00:21:32,000
And I would like to give you an
example of how we can use this.

295
00:21:40,000 --> 00:21:45,000
Let's consider a possible
reaction of this sort.

296
00:21:50,000 --> 00:21:52,000
This reaction,
of course, as you would see in

297
00:21:52,000 --> 00:21:57,000
the table of standard reduction
potentials, is assumed to be

298
00:21:57,000 --> 00:22:00,000
taking place in dilute aqueous
solution.

299
00:22:00,000 --> 00:22:02,000
Actually, if it is in a
standard state,

300
00:22:02,000 --> 00:22:07,000
the concentrations of the ions
will be taken to be 1.0 molar,

301
00:22:07,000 --> 00:22:09,000
which is the standard
concentration.

302
00:22:09,000 --> 00:22:13,000
One thing that I regret about
my lecture on Friday and about

303
00:22:13,000 --> 00:22:17,000
my lecture today is that I am
not really talking too much

304
00:22:17,000 --> 00:22:21,000
about molecules and molecular
shapes and what these things

305
00:22:21,000 --> 00:22:25,000
actually look like in solution.
But normally,

306
00:22:25,000 --> 00:22:28,000
when you have metal or
metalloid ions dissolved in

307
00:22:28,000 --> 00:22:31,000
aqueous solution,
there is some number of water

308
00:22:31,000 --> 00:22:35,000
molecules that are interacting
directly with the metal ions.

309
00:22:35,000 --> 00:22:37,000
It is just not a so-called
naked metal ion,

310
00:22:37,000 --> 00:22:40,000
but it is bonded to water
molecules in solution.

311
00:22:40,000 --> 00:22:44,000
And this is through the type of
Lewis acid, Lewis base chemistry

312
00:22:44,000 --> 00:22:47,000
that we talked about earlier.
There would really be some

313
00:22:47,000 --> 00:22:51,000
number of water molecules bonded
here.

314
00:22:51,000 --> 00:22:55,000
And we will look at so-called
coordination complexes in quite

315
00:22:55,000 --> 00:23:00,000
a bit more detail in the
upcoming section on chemical

316
00:23:00,000 --> 00:23:04,000
bonding, but this is how you
will find it typically written

317
00:23:04,000 --> 00:23:07,000
in a book.
And that is not representative

318
00:23:07,000 --> 00:23:12,000
of the structure in solution,
which you do need to know in

319
00:23:12,000 --> 00:23:17,000
order to have a little bit of an
understanding of the system.

320
00:23:17,000 --> 00:23:22,000
And we are going to consider
the possibility that indium may

321
00:23:22,000 --> 00:23:28,000
react with uranium to provide
indium two plus.

322
00:23:28,000 --> 00:23:31,000
We are going from trivalent to
divalent indium.

323
00:23:31,000 --> 00:23:35,000
And the source of that electron
is the uranium three plus

324
00:23:35,000 --> 00:23:39,000
ion going to the
uranium four plus ion.

325
00:23:39,000 --> 00:23:41,000
There is a hypothetical

326
00:23:41,000 --> 00:23:43,000
reaction that we want to
consider.

327
00:23:43,000 --> 00:23:48,000
And we are going to need to be
able to express that in terms of

328
00:23:48,000 --> 00:23:50,000
half-reactions.

329
00:23:55,000 --> 00:24:01,000
And one of these half-reactions
is simply indium two plus

330
00:24:01,000 --> 00:24:08,000
plus the electron.
Sorry, indium three plus plus

331
00:24:08,000 --> 00:24:14,000
an electron going to indium two
plus.

332
00:24:14,000 --> 00:24:16,000
And, similarly,

333
00:24:16,000 --> 00:24:20,000
over here, we have uranium
three plus.

334
00:24:20,000 --> 00:24:27,000
I am writing both of these as
reductions, as you will find

335
00:24:27,000 --> 00:24:33,000
them in the table.
Four plus, plus an electron,

336
00:24:33,000 --> 00:24:39,000
going to uranium three plus.

337
00:24:39,000 --> 00:24:42,000
Those are our two
half-reactions.

338
00:24:42,000 --> 00:24:46,000
Christine, can you zero in on
them?

339
00:24:46,000 --> 00:24:49,000
Yes, you are doing that
already.

340
00:24:49,000 --> 00:24:52,000
Thank you.
-0.49 for indium.

341
00:24:52,000 --> 00:24:57,000
And then, when we look at
uranium, these are arranged

342
00:24:57,000 --> 00:25:01,000
alphabetically.

343
00:25:07,000 --> 00:25:10,000
Here they are in alphabetical
order.

344
00:25:10,000 --> 00:25:15,000
In the tables they can be
arranged by potential or they

345
00:25:15,000 --> 00:25:21,000
can be arranged alphabetically.
Sometimes it is easier to find

346
00:25:21,000 --> 00:25:25,000
a potential by taking advantage
of alphabetization.

347
00:25:25,000 --> 00:25:30,000
And so, if we can focus in on
that --

348
00:25:30,000 --> 00:25:34,000
We focus in on that,
and you will see that the 4+

349
00:25:34,000 --> 00:25:37,000
going to the 3+ is -0.61.

350
00:25:42,000 --> 00:25:48,000
And then you will remember that
to find the standard cell

351
00:25:48,000 --> 00:25:55,000
potential, E zero of the
cell is going to be equal to E

352
00:25:55,000 --> 00:26:00,000
zero of the cathode minus E zero
of the anode.

353
00:26:06,000 --> 00:26:10,000
And what we can see here is
that this reaction is going to

354
00:26:10,000 --> 00:26:14,000
proceed spontaneously as written
because in going from left to

355
00:26:14,000 --> 00:26:18,000
right, as written,
uranium plus three is going to

356
00:26:18,000 --> 00:26:22,000
plus four.
So, uranium is the reducing

357
00:26:22,000 --> 00:26:24,000
agent.
And it is a stronger reducing

358
00:26:24,000 --> 00:26:28,000
agent than the other possible
reducing agent,

359
00:26:28,000 --> 00:26:32,000
indium two plus,
because it is more negative

360
00:26:32,000 --> 00:26:38,000
than the indium two plus.
And how much more negative is

361
00:26:38,000 --> 00:26:41,000
it?
It is more negative by +0.12

362
00:26:41,000 --> 00:26:43,000
volts.

363
00:26:48,000 --> 00:26:51,000
And we, therefore,
because this cell potential is

364
00:26:51,000 --> 00:26:56,000
positive, we know that this
reaction is spontaneous as

365
00:26:56,000 --> 00:26:59,000
written.
And it is positive by 0.12

366
00:26:59,000 --> 00:27:02,000
volts.
That is a spontaneous reaction.

367
00:27:02,000 --> 00:27:07,000
And now we have all the
information that we would need

368
00:27:07,000 --> 00:27:11,000
in order to get the equilibrium
constant for this reaction.

369
00:27:11,000 --> 00:27:15,000
We know that,
in terms of standard potential,

370
00:27:15,000 --> 00:27:17,000
it is 0.12.
Let's see, here,

371
00:27:17,000 --> 00:27:21,000
if we can just calculate this.

372
00:28:05,000 --> 00:28:08,000
All right.
I am going to take this and

373
00:28:08,000 --> 00:28:10,000
zoom it a little bit.

374
00:28:27,000 --> 00:28:37,000
We are going to put in
Faraday's constant of 96,485.3,

375
00:28:37,000 --> 00:28:46,000
and also we are going to need
the gas constant.

376
00:28:46,000 --> 00:28:55,000
Everyone remember the value of
the gas constant,

377
00:28:55,000 --> 00:29:00,000
8.31447?
And now, if we look over at our

378
00:29:00,000 --> 00:29:05,000
relationship between the
equilibrium constant and the

379
00:29:05,000 --> 00:29:10,000
free energy over there.
We took a free energy and we

380
00:29:10,000 --> 00:29:13,000
converted it into a standard
cell potential.

381
00:29:13,000 --> 00:29:17,000
Now we have it,
so we want to solve for an

382
00:29:17,000 --> 00:29:21,000
exponential of nFE.
n is one here because,

383
00:29:21,000 --> 00:29:24,000
in both cases,
we have one electron being

384
00:29:24,000 --> 00:29:29,000
transferred in the
half-reaction.

385
00:29:29,000 --> 00:29:36,000
We have the exponential of F
times 0.12.

386
00:29:36,000 --> 00:29:40,000
And then, that is divided by
RT.

387
00:29:40,000 --> 00:29:45,000
And so we have R times 298.15
K.

388
00:29:45,000 --> 00:29:53,000
And let me just make sure I get
back over here and get another

389
00:29:53,000 --> 00:30:00,000
parenthesis in there and see
what it is.

390
00:30:00,000 --> 00:30:03,000
Our equilibrium constant,
for that reaction,

391
00:30:03,000 --> 00:30:07,000
with a separation for the
standard cell potential of only

392
00:30:07,000 --> 00:30:12,000
0.12 volts, is about 106.
The equilibrium constant says

393
00:30:12,000 --> 00:30:17,000
that the concentration of these
ions multiplied together divided

394
00:30:17,000 --> 00:30:22,000
by the concentration of these
ions multiplied together is

395
00:30:22,000 --> 00:30:26,000
based on the information
contained in the half-cell

396
00:30:26,000 --> 00:30:30,000
potentials.
So, that was pretty cool.

397
00:30:30,000 --> 00:30:34,000
And we will now go ahead and
talk about another relationship

398
00:30:34,000 --> 00:30:36,000
that we can get.

399
00:30:42,000 --> 00:30:46,000
Over there, I was able to give
you that relationship between

400
00:30:46,000 --> 00:30:50,000
the equilibrium constant and
standard cell potential by

401
00:30:50,000 --> 00:30:53,000
remembering something from
chemical equilibrium.

402
00:30:53,000 --> 00:30:57,000
And now, we are going to do
that again.

403
00:30:57,000 --> 00:31:02,000
And we are going to remember
that delta G for a reaction is

404
00:31:02,000 --> 00:31:05,000
equal to delta G nought plus RT
ln Q.

405
00:31:05,000 --> 00:31:11,000
That is something that

406
00:31:11,000 --> 00:31:16,000
you should remember from your
studies of chemical equilibrium.

407
00:31:16,000 --> 00:31:21,000
And the neat thing now is we
know we have another

408
00:31:21,000 --> 00:31:27,000
relationship for the free energy
of a reaction that is related to

409
00:31:27,000 --> 00:31:31,000
electrical work.
We got that up there.

410
00:31:31,000 --> 00:31:34,000
Now we can go,
minus nFE.

411
00:31:34,000 --> 00:31:39,000
Notice that what we are going
here is we are considering what

412
00:31:39,000 --> 00:31:45,000
happens when we perturb a system
from equilibrium conditions.

413
00:31:45,000 --> 00:31:50,000
And when might that happen?
Well, I showed you how to set

414
00:31:50,000 --> 00:31:55,000
up a Galvanic cell last time,
and I showed you how to

415
00:31:55,000 --> 00:32:00,000
calculate its standard cell
potential.

416
00:32:00,000 --> 00:32:03,000
You could measure the standard
cell potential by using,

417
00:32:03,000 --> 00:32:07,000
externally, a volt meter.
And, as long as you did not let

418
00:32:07,000 --> 00:32:11,000
the reaction proceed to any
significant extent,

419
00:32:11,000 --> 00:32:14,000
then you would know the
standard potential for that

420
00:32:14,000 --> 00:32:17,000
cell.
And what we are going to do now

421
00:32:17,000 --> 00:32:21,000
is we are going to say,
let's let the reaction proceed.

422
00:32:21,000 --> 00:32:26,000
Let's let a battery become
discharged, for example.

423
00:32:26,000 --> 00:32:30,000
And, as the reaction is
proceeding, we might want,

424
00:32:30,000 --> 00:32:34,000
for example,
to be able to make a plot of

425
00:32:34,000 --> 00:32:40,000
the potential at any particular
extent of a reaction.

426
00:32:40,000 --> 00:32:45,000
And, in order to do that,
we will develop this equation,

427
00:32:45,000 --> 00:32:50,000
here, that shows how the
potential at any particular

428
00:32:50,000 --> 00:32:56,000
point along the reaction going
to completion is related to the

429
00:32:56,000 --> 00:33:00,000
standard potential.

430
00:33:07,000 --> 00:33:12,000
And all I have done now is
substitute minus nFE

431
00:33:12,000 --> 00:33:17,000
and minus nFE zero
in, respectively,

432
00:33:17,000 --> 00:33:23,000
for delta G and delta G zero
for the

433
00:33:23,000 --> 00:33:27,000
reaction.
And now I want to go through

434
00:33:27,000 --> 00:33:31,000
and divide by minus nF.
And when I do that,

435
00:33:31,000 --> 00:33:38,000
I am going to get E is equal to
E zero minus (RT over nF) ln Q.

436
00:33:38,000 --> 00:33:45,000
And this equation is very

437
00:33:45,000 --> 00:33:51,000
central to electrochemistry.
It is the Nernst equation.

438
00:33:51,000 --> 00:33:57,000
And you will be able to use the
Nernst equation to predict how

439
00:33:57,000 --> 00:34:05,000
systems will behave when they
are perturbed from equilibrium.

440
00:34:05,000 --> 00:34:09,000
And one example would be that
of a battery discharge,

441
00:34:09,000 --> 00:34:12,000
like I was mentioning a moment
ago.

442
00:34:20,000 --> 00:34:26,000
And so, let's kind of pick some
arbitrary material,

443
00:34:26,000 --> 00:34:31,000
out of which to make a battery.

444
00:34:37,000 --> 00:34:42,000
And we will use the Nernst
equation to get information

445
00:34:42,000 --> 00:34:48,000
about how it will discharge.
And let's consider the

446
00:34:48,000 --> 00:34:53,000
following cell.
It will be one in which the two

447
00:34:53,000 --> 00:34:59,000
half-reactions are as follows.
Iron two plus plus two

448
00:34:59,000 --> 00:35:07,000
electrons going to Fe metal.

449
00:35:07,000 --> 00:35:12,000
And the other material will be
cadmium two plus plus two

450
00:35:12,000 --> 00:35:15,000
electrons going to cadmium
metal.

451
00:35:15,000 --> 00:35:21,000
Now we need the two standard

452
00:35:21,000 --> 00:35:26,000
reduction potentials for iron
two plus,

453
00:35:26,000 --> 00:35:27,000
--

454
00:35:33,000 --> 00:35:37,000
-- which is coordinated in
aqueous solution to six water

455
00:35:37,000 --> 00:35:42,000
molecules in an octahedral
array, but we will get to that

456
00:35:42,000 --> 00:35:45,000
later.
And I can see the equation but

457
00:35:45,000 --> 00:35:50,000
not the value.
And there it is at -0.44.

458
00:35:58,000 --> 00:36:01,000
And now we need cadmium two
plus going to

459
00:36:01,000 --> 00:36:07,000
cadmium, elemental form.
And we see that it is up there

460
00:36:07,000 --> 00:36:09,000
at -0.40.

461
00:36:22,000 --> 00:36:27,000
Both cadmium metal and iron
metal are negative of the

462
00:36:27,000 --> 00:36:32,000
standard hydrogen electrode.
They are both reducing systems,

463
00:36:32,000 --> 00:36:35,000
as it were.
However, we see that iron is

464
00:36:35,000 --> 00:36:38,000
more negative.
And, since iron is more

465
00:36:38,000 --> 00:36:43,000
negative, if we set the cell up,
if we have a piece of cadmium

466
00:36:43,000 --> 00:36:48,000
on one side and a piece of iron
on the other side in this cell,

467
00:36:48,000 --> 00:36:53,000
then this piece of iron metal
is going to be providing the

468
00:36:53,000 --> 00:36:58,000
electrons in this system,
when we go in the spontaneous

469
00:36:58,000 --> 00:37:02,000
direction.
The electron flow in the

470
00:37:02,000 --> 00:37:08,000
external circuit will be from
the iron metal electrode to the

471
00:37:08,000 --> 00:37:13,000
cadmium metal electrode.
And things happen in solution,

472
00:37:13,000 --> 00:37:18,000
as we've discussed,
but that makes iron the anode.

473
00:37:18,000 --> 00:37:24,000
And for the reaction going that
way, as we would have set it up,

474
00:37:24,000 --> 00:37:30,000
then we have a standard cell
potential E zero of +0.04 volts.

475
00:37:30,000 --> 00:37:35,000
And now, in order to apply the

476
00:37:35,000 --> 00:37:40,000
Nernst equation to answer a
question about this battery --

477
00:37:40,000 --> 00:37:43,000
and this battery is not a
high-voltage battery,

478
00:37:43,000 --> 00:37:47,000
it is only 0.04 volts,
not high-voltage at all,

479
00:37:47,000 --> 00:37:51,000
so you probably would not use
it for a battery,

480
00:37:51,000 --> 00:37:56,000
unless you really only needed a
small potential difference in

481
00:37:56,000 --> 00:37:58,000
the first place -- but,
nonetheless,

482
00:37:58,000 --> 00:38:04,000
we want to know how we can use
the Nernst equation.

483
00:38:04,000 --> 00:38:18,000
We can say, what is E at 80%
completion?

484
00:38:23,000 --> 00:38:27,000
That is the type of question we
can answer using the Nernst

485
00:38:27,000 --> 00:38:30,000
equation.
What is the potential drop to

486
00:38:30,000 --> 00:38:35,000
from an initial positive 0.04
volts when the reaction is 80%

487
00:38:35,000 --> 00:38:39,000
complete?
Under standard conditions,

488
00:38:39,000 --> 00:38:44,000
we have one molar solution on
both sides of the ions.

489
00:38:44,000 --> 00:38:49,000
So, in this solution,
we would have 1.0 molar candium

490
00:38:49,000 --> 00:38:53,000
two plus,
and over here,

491
00:38:53,000 --> 00:38:57,000
in this solution, we have
1.0 molar Fe two plus.

492
00:38:57,000 --> 00:39:02,000
Because those are the standard

493
00:39:02,000 --> 00:39:06,000
conditions, as you will see
written in the header to that

494
00:39:06,000 --> 00:39:10,000
long table in your appendix,
because you have to define

495
00:39:10,000 --> 00:39:14,000
standard conditions in order to
have a standard potential.

496
00:39:14,000 --> 00:39:17,000
And the standard conditions are
1.0 molar of the ions,

497
00:39:17,000 --> 00:39:19,000
respectively,
in solution.

498
00:39:19,000 --> 00:39:23,000
At 80% completion,
what will happen is we will

499
00:39:23,000 --> 00:39:27,000
have gone in terms of producing
iron two ions in

500
00:39:27,000 --> 00:39:31,000
solution.
We are producing them in

501
00:39:31,000 --> 00:39:36,000
solution as the electrons flow
through the external circuit

502
00:39:36,000 --> 00:39:41,000
because the piece of iron metal
is the anode.

503
00:39:41,000 --> 00:39:45,000
And so, what we are interested
in here is Q.

504
00:39:45,000 --> 00:39:50,000
I am writing backwards.
Pretty soon I will be writing

505
00:39:50,000 --> 00:39:54,000
straight down or sideways.
In the reaction,

506
00:39:54,000 --> 00:39:59,000
as written in the forward
direction, we are producing Fe

507
00:39:59,000 --> 00:40:05,000
two plus ions such
that we can define our Q at this

508
00:40:05,000 --> 00:40:10,000
80% completion value as 1.8
molar divided by 0.2 molar.

509
00:40:10,000 --> 00:40:16,000
And that is equal to,

510
00:40:16,000 --> 00:40:20,000
of course, just nine.
Q is equal to nine.

511
00:40:20,000 --> 00:40:24,000
We have RT, that will be 298.15
Kelvin.

512
00:40:24,000 --> 00:40:29,000
RF is our number of Coulombs
per mole.

513
00:40:29,000 --> 00:40:34,000
That is the Faraday constant.
And, in this case,

514
00:40:34,000 --> 00:40:40,000
because in the half-reactions,
you have two electrons rather

515
00:40:40,000 --> 00:40:44,000
than one being transferred,
n is equal to two.

516
00:40:44,000 --> 00:40:49,000
If you plug in now,
you have E zero is 0.04 and you

517
00:40:49,000 --> 00:40:54,000
have RT over nF ln of 9,
that will give you the

518
00:40:54,000 --> 00:40:59,000
potential at 80% completion,
which is somewhere in between

519
00:40:59,000 --> 00:41:05,000
0.04 and zero,
as this battery is discharged.

520
00:41:05,000 --> 00:41:09,000
If you want to model the way in
which a battery is discharged,

521
00:41:09,000 --> 00:41:13,000
you can use the Nernst equation
for that purpose.

522
00:41:13,000 --> 00:41:17,000
But, there is another type of
cell that is interesting,

523
00:41:17,000 --> 00:41:22,000
that you can also describe very
well using the Nernst equation.

524
00:41:22,000 --> 00:41:27,000
This will be what is called a
concentration cell.

525
00:41:43,000 --> 00:41:45,000
Concentration cells are pretty
neat.

526
00:41:45,000 --> 00:41:50,000
In a concentration cell you set
up a potential difference with

527
00:41:50,000 --> 00:41:55,000
the same electrode materials on
both sides, just different

528
00:41:55,000 --> 00:42:00,000
concentrations.
If I were writing this in terms

529
00:42:00,000 --> 00:42:06,000
of the type of diagram I used
last time, I could show you that

530
00:42:06,000 --> 00:42:13,000
we could connect two beakers
with a salt bridge like that and

531
00:42:13,000 --> 00:42:18,000
we could have electrode
materials here dipping into

532
00:42:18,000 --> 00:42:24,000
solution on both sides.
And, in solution on one side,

533
00:42:24,000 --> 00:42:29,000
we are going to have Sn two
plus,

534
00:42:29,000 --> 00:42:33,000
divalent tin ions,
at a concentration of 0.01

535
00:42:33,000 --> 00:42:39,000
molar.
And, on the other side,

536
00:42:39,000 --> 00:42:46,000
we will have Sn two plus
at a concentration

537
00:42:46,000 --> 00:42:51,000
of 0.10 molar.
This one on the left is dilute

538
00:42:51,000 --> 00:42:58,000
by a factor of ten.
That one is the more dilute one

539
00:42:58,000 --> 00:43:04,000
by a factor of ten.
And we can ask what kind of a

540
00:43:04,000 --> 00:43:09,000
voltage is set up as a
consequence of just simply

541
00:43:09,000 --> 00:43:15,000
having different concentrations
of tin ions in the two beakers.

542
00:43:15,000 --> 00:43:21,000
And we can ask in which
direction the electrons would

543
00:43:21,000 --> 00:43:26,000
flow in a system like this.
Well, in this reaction,

544
00:43:26,000 --> 00:43:32,000
the system wants to go from a
state where tin two plus ions at

545
00:43:32,000 --> 00:43:38,000
0.10 molar
go to the beaker where we have

546
00:43:38,000 --> 00:43:45,000
Sn two plus ions at 0.01 molar.

547
00:43:45,000 --> 00:43:49,000
We have a potential setup here
that is similar to the way

548
00:43:49,000 --> 00:43:52,000
osmosis takes place.
The ions want to move from a

549
00:43:52,000 --> 00:43:57,000
region of high concentration to
a region of low concentration.

550
00:43:57,000 --> 00:44:01,000
And the system would come to
equilibrium at that time when

551
00:44:01,000 --> 00:44:05,000
the concentration is the same on
both sides.

552
00:44:05,000 --> 00:44:09,000
And then there would no longer
be any potential.

553
00:44:09,000 --> 00:44:14,000
The value of E nought for this
cell is equal to zero,

554
00:44:14,000 --> 00:44:19,000
because we have the same
half-reactions that would take

555
00:44:19,000 --> 00:44:23,000
place on both sides.
So, E nought is equal to zero

556
00:44:23,000 --> 00:44:30,000
volts in the concentration cell.
And which way do electrons go?

557
00:44:30,000 --> 00:44:32,000
Well, here is how you can think
of it.

558
00:44:32,000 --> 00:44:37,000
The electrode material can be
made out of tin on both sides.

559
00:44:37,000 --> 00:44:42,000
And, since tin ions are going
to appear in solution on the

560
00:44:42,000 --> 00:44:46,000
dilute side, they are going to
jump into solution from the

561
00:44:46,000 --> 00:44:50,000
electrode itself.
A tin atom that is a piece of

562
00:44:50,000 --> 00:44:55,000
the metal on the surface of the
electrode will jump into

563
00:44:55,000 --> 00:44:59,000
solution, increasing the
concentration of tin two plus

564
00:44:59,000 --> 00:45:05,000
on the dilute side.
And, every time it does that,

565
00:45:05,000 --> 00:45:08,000
we get two electrons racing
through the external circuit,

566
00:45:08,000 --> 00:45:12,000
going over here,
where those two electrons can

567
00:45:12,000 --> 00:45:15,000
be used to reduce tin two
ions in solution

568
00:45:15,000 --> 00:45:19,000
that become a new tin atom on
the surface of the electrode on

569
00:45:19,000 --> 00:45:23,000
the right-hand side.
That is physically what happens

570
00:45:23,000 --> 00:45:27,000
to equilibrate the concentration
of tin ions on both sides as a

571
00:45:27,000 --> 00:45:33,000
function of time.
And so, what we can find is

572
00:45:33,000 --> 00:45:41,000
that these two values of the
concentrations go into

573
00:45:41,000 --> 00:45:47,000
determining our Q in the
following way.

574
00:46:10,000 --> 00:46:13,000
We have Q of 0.1.
Because the tin ions are going

575
00:46:13,000 --> 00:46:16,000
from a place of high
concentration to a place of low

576
00:46:16,000 --> 00:46:19,000
concentration,
and electrons are traveling

577
00:46:19,000 --> 00:46:23,000
through the external circuit
defining the cathode and the

578
00:46:23,000 --> 00:46:27,000
anode in the direction that I
mentioned, the electrons are

579
00:46:27,000 --> 00:46:32,000
coming out over here.
So, the dilute side is going to

580
00:46:32,000 --> 00:46:35,000
be your anode.
And we have n equal to two,

581
00:46:35,000 --> 00:46:39,000
in this particular case,
because it is a two electron

582
00:46:39,000 --> 00:46:43,000
process that converts tin two
plus into metallic

583
00:46:43,000 --> 00:46:45,000
tin.
You can take these numbers and

584
00:46:45,000 --> 00:46:48,000
plug them into the Nernst
equation.

585
00:46:48,000 --> 00:46:51,000
And where you have E nought is
equal to zero,

586
00:46:51,000 --> 00:46:54,000
n is equal to two,
T would be 298.15.

587
00:46:54,000 --> 00:46:59,000
We know the Faraday constant
and the gas constant.

588
00:46:59,000 --> 00:47:03,000
We have Q is equal to 0.1,
which is our concentration

589
00:47:03,000 --> 00:47:07,000
ratio between the two sides.
And, when we solve that,

590
00:47:07,000 --> 00:47:12,000
we are going to find that some
small voltage is produced

591
00:47:12,000 --> 00:47:16,000
initially at this set of
concentrations.

592
00:47:16,000 --> 00:47:21,000
I hope that in the last couple
of lectures, today and last

593
00:47:21,000 --> 00:47:25,000
Friday, I have hammered home the
relationship between

594
00:47:25,000 --> 00:47:31,000
electrochemistry and energy.
And next time we will go

595
00:47:31,000 --> 00:47:35,000
forward in talking about
molecular structure and see how

596
00:47:35,831 --> 00:47:38,000
that is going to play a role.