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Current time:0:00Total duration:11:27

AP.Chem:

ENE‑6 (EU)

, ENE‑6.C (LO)

, ENE‑6.C.1 (EK)

, ENE‑6.C.3 (EK)

, ENE‑6.C.4 (EK)

- [Instructor] We already know how to calculate cell potential when the reactants and products are in their standard states. However, what if that's not the case? We can find cell potential
when reactants and products are not in their standard states by using the Nernst equation,
which is shown here. And we're gonna call that cell potential the instantaneous cell potential. E cell is the instantaneous
cell potential, the cell potential or voltage
at a specific moment in time. E naught of the cell is the
standard cell potential. This is the cell potential when the reactants and products are in their standard states. R is the ideal gas constant. T is the temperature. N is the number of electrons transferred in the redox reaction. F is Faraday's constant. And Q is the reaction quotient. So, this is one way to
write the Nernst equation. And let's compare this
form of the Nernst equation to a simplified form. In the simplified form
of the Nernst equation, E of the cell is equal
to E naught of the cell, minus 0.0592 volts, divided by n, times the log of the reaction quotient Q. So, going back to the first
form of the Nernst equation, if we assume the temperature
is 25 degrees Celsius, or 298 Kelvin, we multiply that by
the ideal gas constant, divide that by Faraday's constant, and for changing natural log to log, then we end up with this
form for the bottom equation, 0.0592 volts divided
by n times the log of Q for this second term
of the Nernst equation. You can use the Nernst equation
to quantitatively calculate the exact voltage at a
specific moment in time. However, we're gonna try to
think about the Nernst equation from a qualitative point of view. Let's use the Nernst equation to think about the instantaneous cell potential for a zinc copper cell. And we're gonna look at a
few different situations. In a zinc copper cell, solid zinc is oxidized
to zinc two plus ions and copper two plus ions are
reduced to form solid copper. The standard cell potential
for the zinc copper cell at 25 degrees Celsius is
equal to positive 1.10 volts. And our goal is to find the cell potential at the moment of time when the concentration
of copper two plus ions is equal to 1.0 Mueller and the concentration of zinc
two plus ions in solution is also equal to 1.0 Mueller. To find the instantaneous cell potential, we're gonna use the simplified
form of the Nernst equation. E naught of the cell is
equal to positive 1.10 volts. Next, let's think about n, the number of electrons
transferred in the redox reaction. So, looking at our equation here is a solid zinc going to zinc two plus, is a loss of two electrons, and copper two plus going to solid copper is a gain of two electrons. So, two electrons were
transferred and n is equal to two. So, I've gone ahead and made n
equal to two in our equation. Next, we need to think about
the reaction quotient Q. Q has the same form as the equilibrium constant expression. So, if we think about
the balanced equation, remember, pure solids are left out of equilibrium constant expressions. So, it would be the
concentration of products over the concentration of reactants. And therefore, Q is equal
to the concentration of zinc two plus ions,
divided by the concentration of copper two plus ions. Because our goal is to find the instantaneous cell potential when the concentrations are
both equal to one Mueller, Q is equal to one divided by one. So, Q is equal to one. And the log of one is equal to zero. Therefore this entire second term in the Nernst equation is equal to zero, which means the
instantaneous cell potential is equal to the standard cell potential. So, the instantaneous cell potential is equal to positive 1.10 volts. And this answer makes a lot of sense because standard state for solutions is a one Mueller concentration. Therefore, the
instantaneous cell potential is equal to the standard cell potential at this moment in time. Next, let's think about the
instantaneous cell potential when the concentration
of copper two plus ions is equal to 10.0 Mueller and the concentration
of zinc two plus ions is equal to 1.0 Mueller. The standard cell potential
at 25 degrees Celsius is still equal to positive 1.10 volts and n is still equal to two. So, we need to think about
the reaction quotient Q at this moment in time. The concentration of zinc two plus ions is equal to 1.0 Mueller so we plug that in and the concentration
of copper two plus ions is equal to 10.0 Mueller. So, at this moment in
time, Q is less than one. The log of a number less
than one is negative. Therefore, this second
term would be negative or less than zero. And since we are subtracting
a negative number, we would actually be adding that value to the standard cell potential, which means that the
instantaneous cell potential will be greater than the
standard cell potential. And if you went ahead and calculated the instantaneous cell potential
with the Nernst equation, you would find it's equal
to positive 1.13 volts, which is greater than positive 1.10 volts. Next, let's think about the
instantaneous cell potential when the concentration
of copper two plus ions is equal to 1.0 Mueller
and the concentration of zinc two plus ions is
equal to 10.0 Mueller. Going to the Nernst equation, the standard cell potential
is still positive 1.10 volts and n is still equal to two. So next, we think about
the reaction quotient Q. At this moment in time, the concentration of zinc two plus ions is equal to 10.0 Mueller and the concentration
of copper two plus ions is equal to 1.0 Mueller. So, at this moment in time,
Q is greater than one. The log of a number greater
than one is positive. Therefore the second term in our equation will be positive or greater than zero. And because we are
subtracting a positive number, we would be subtracting a value from the standard cell potential, which means that the
instantaneous cell potential will be less than the
standard cell potential. And if you plug the numbers
into the Nernst equation, you would find that the
instantaneous cell potential is equal to positive 1.07 volts, which is less than the standard
cell voltage of 1.10 volts. We just saw that at the moment in time when the concentration
of copper two plus ions is equal to 1.0 Mueller and the concentration
of zinc two plus ions is equal to 10.0 Mueller, the instantaneous cell potential is equal to positive 1.07 volts. Let's use this moment in
time as a starting point and let's think about what happens to the instantaneous cell potential as the reaction progresses. Remember that the instantaneous
cell potential is related to the instantaneous change in free energy, delta G, by this equation. Delta G is equal to negative nFE. Because we have a positive value for the instantaneous cell potential, if we plug in a positive
value into our equation because of the negative
sign in the equation, we would get a negative sign for delta G. And when delta G is negative, the reaction is thermodynamically favored, which means the reactants
will turn into products. And as the reactions turn into products, the concentration of zinc
two plus ions will increase and the concentration of copper
two plus ions will decrease. And think about what that does
to the reaction quotient Q. There would be an increase
in the concentration of zinc two plus ions and a
decrease in the concentration of copper two plus ions. So, Q would no longer be
equal to 10.0 over 1.0. We would see an increase in Q as the reactants turn into the products. An increased value of Q
means that the second term in the Nernst equation will
be a larger, positive number. So, if we subtract a
larger positive number than we did in the previous example from the standard cell potential, we would get a lower
instantaneous cell potential than positive 1.07 volts. So, what the Nernst equation tells us is as the reaction goes to the right, there is an increase in
Q and as Q increases, there's a decrease in the
instantaneous cell potential. So, we've just seen as the
reaction goes to the right, that's an increase in Q. And according to the Nernst equation, that's a decrease in the
instantaneous cell potential. And therefore, the
instantaneous cell potential would decrease from 1.07 volts. However, the voltage
would still be positive, which means that delta G
would still be negative, which means that the reaction is still thermodynamically favored. And so, this keeps happening. As the reaction goes to the
right, Q keeps increasing and the instantaneous cell
potential keeps decreasing. So, we had started out
at positive 1.07 volts and we saw that that had decreased and it keeps on decreasing,
keeps on decreasing. Eventually the
instantaneous cell potential will go to zero volts. And when the voltage goes to zero, if you plug in zero into our equation, delta G is equal to zero. And when delta G is equal to zero, the reaction is at equilibrium. And the reaction quotient Q is equal to the equilibrium constant K. So, if you were to plug in
the equilibrium constant K in for Q, you would find that
this entire second term would be equal to positive 1.10 volts and cancel out the
standard cell potential, giving an instantaneous
potential of zero volts. So, if we think about our
zinc copper cell as a battery, when the reaction reaches equilibrium, the voltage is equal to zero
and the battery is dead. Let's do a quick summary
of what we've learned from the Nernst equation. So, this is the simplified form, which is valid at 25 degrees Celsius. When Q is equal to one, the instantaneous cell potential is equal to the standard cell potential. When Q is greater than one, the instantaneous cell potential is less than the standard cell potential. When Q is less than one, the instantaneous cell potential is greater than the
standard cell potential. And when Q is equal to the
equilibrium constant K, the instantaneous cell potential
is equal to zero volts. When Q is equal to K and the
potential is equal to zero, the reaction is at equilibrium. However, for these first three situations, Q is not equal to K and the
instantaneous cell voltage is not equal to zero. Therefore, the reaction
is not at equilibrium.

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