So in this lecture what we'll look at a relationship between the Equilibrium constant of a redox
reaction found in a electrochemical cell,
and that electrochemical cell's Cell voltage or it's electromotive force.
Now before we look at their relationship let's remember
what equilibrium constant is.
Suppose we have the following expression or reaction
in which we have two reactants and two products, now our A reactant is in the gas state, our
B reactant is in the liquid state our C product is in
the aqueous state and the D product is in the
aqueous state, so let's write the equilibrium constant expression for this reaction
and let's assume that the equilibrium has been reached.
So our equilibrium constant under standard conditions is equal to the concentration
of product C times the concentration of product
D divided by the concentration of product A. Now remember whenever we're writing
our equilibrium constant expression only
aqueous molecules and gas molecules kept in our equilibrium expression, we don't include
liquid molecules or solid molecules and that's exactly why
we don't include the B molecule, because it's in the liquid
state.
So we only get this expression, no B is included.
Now noticed that this K is simply a ratio between the product
concentration and the reactant concentration, so we say
that if our K is greater than 1 that means our reactions favorable, if our
k is less than one, our reaction is unfavorable,
in other words if it is greater than one that means our equilibrium lies
to the right because these guys completely react to form our products, in other words
if K is much larger than one we have very little
value of concentration of A and a lot of concentration of
products namely C and D, likewise if K is much less than 1 that means
we have lots of this guy leftover an equilibrium, and very little formation of our product that
means equilibrium lies to the left which also means we have very
little of this guy and a lot of this guy.
Now let's look at B so we just learn that the change
in Gibb's fee energy is equal to negative n
the number of moles of electrons times <A HREF="event:GLSPDF:gls_CH101_Faradays constant.pdf">Faraday's
constant</A> times the cell voltage.
Now before we know that the change in Gibb's free energy can also
be expressed as -R times T, where R is the gas constant
and T is the temperature in kelvin times our natural log of our k, our equilibrium
constants expression under standard conditions.
Now what we can do is set these guys equal, why?
Will because this and this are the same expression because
they both equal the same thing namely change in Gibb's
free energy.
so let's set these guys equal, so we get negative nF times E equals -RT
natural log of K.
So we can basically rearrange this guy and solve our cell voltage,
and what we get is the negative cancel and we bring
the n and F to this side and we get our cell voltage equal to
gas constant times temperature in kelvin divided by moles of electrons times Faraday's
constant times natural log of our equilibrium constant, so
this expression basically relates our cell's voltage
and the concentration of our reactants and products with the ratio
of the concentration.
Now notice in this equation we have R is our constant and that is also
constant and if we assume constant temperature
say 25°Celsius you can simplify this expression into the following expression,
so let's rewrite this expression cell voltage E is
equal to a constant times a constant in our case
because T is 298 in Kelvin since we assume that temperature is
constant and temperature is 25°Celsius.
That is our assumption that means in Kelvin 25 plus 273
is 298, so a constant times our temperatures thats constant
divided by another constant call Faraday's constant
we get something called 0.0257 a number right, so this expression
under these conditions gives us the following simplified expression and this
expression is nice because here we have three unknowns
this is our unknown, this is our unknown and this is our unknown,
so for example we know our cell voltage and we know our number of moles we can
rearrange this equation and solve for K in the following
way.
Natural log of k is equal to n times cell voltage divided by
0.0257 and suppose that we know our
K and we know our n, that means we can use our equilibrium constant,
and our n define our cell voltage, so these equations
become very useful and these two equations build a relationship between the concentration
of reactants and products and the cell's voltage.
Now notice one interesting thing whenever we take natural log
of a number that's bigger than one this expression becomes
positive and that means if this expression is positive
then this is positive and recall that if our cell voltage
is positive that means we have a product favored reaction, it's spontaneous,
and that makes sense because earlier we said if we have
a K greater than one that means we have a favorable reaction, a reaction favors this
going this way likewise if this K is less than one if it say 0.9 then
this guy or natural log of the number that's less than one becomes negative, so
our E becomes negative and that means if E is
negative our reaction is not product favored, it's
reactant favored that means it's not favorable and just like
our k is less than one means that our reaction is
unfavorable.
The following thing what I talked about is the following
I wanna show you how we can convert the formula we just found to
something simpler, now notice we are dealing with natural logs
that means if we want to convert this log to exponents, you'll have to deal with
bases of e.
Now bases of e are very easy to calculate you need a calculator to calculate bases of
e, for example you don't know what e1, e2, e3 or may be
e1 you do but for e2, e3, e4 we don't really know
what that is without using calculator but bases like base 10, that's easy to calculate
like 101, 102, 103 that's easy to calculate
you don't need a calculator.
Therefore our goal is to convert this natural log into base 10
log and the way we do it we have to remember from
algebra what the base conversion formula is for logs.
In other words this is a formula that basically tells us we can convert
any log of any base of any inside to the base of 10 so the way we do it
is log of base x of y is equal to log of base 10 of y divided by log of base
10 of x.
Now in our case we have a natural log of some y that's equivalent to saying log
of base e of y equals that means we have to divide
log base 10 of y divided by log of base 10 of e, so this exactly what
we follow in this process here.
In this step natural log of K is equal to log of base e of k
is equal to seems that process this guy over this guy
same thing that we did here is equal to now this is something we know we know what e is,
e1 is some number, so basically plug this guy
into the calculator and we find that it's 0.434.
So now we have log of base 10 of k is equal to
this whole guy and this guy is over this guy, so what we do next is we bring this coming
over to this side that's what we do in step e, we get log of
base 10 of k equals this guy's priority here what we get
is 0.434 divided by 0.0257
times n times e, now we plug this guy into the calculator and we get
16.89 of moles of electrons times cell voltage.
Now in textbooks you'll find this formula but this guy and this guy are the same now
the way you go from this formula to this formula is simply take this number and you reach the
negative one power and you get this expression to the same expression.
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