23.2. Cell Potential http://www.ck12.org
aA+bB→cC+dD
Q=[C]c[D]d
[A]a[B]bIf the reaction is run at the standard temperature of 25°C, and it is just the concentrations that are non-standard, the
Nernst equation simplifies as follows:
E=E◦−^0 .02569 voltsn ln QOr, if you prefer base-10 logarithms:
E=E◦−^0 .05619 voltsn log QLet’s look at an example problem to see how this equation is used.
Example Problem 23.2
Calculate the cell potential at 25°C for the following reaction:
Cu(s) + 2Fe^3 +(aq)→Cu^2 +(aq) + 2Fe^2 +(aq)The ions are present in the following concentrations: 1.0 x 10−^4 M Fe^3 +, 0.25 M Cu^2 +, and 0.20 M Fe^2 +.
Answer:
The Nernst equation has the following form at 25°C:
E=E◦−^0 .05619 voltsn log QIn order to find the cell potential (E), we need to find values for E°, n, and Q. Let’s start by splitting this reaction into
half-reactions. Because there are only two different metals, we can easily see that the unbalanced half-reactions will
have the following forms:
Cu(s)→Cu^2 +(aq)
2Fe^3 +(aq)→2Fe^2 +(aq)Then, balance each of these half-reactions by adding electrons to the appropriate side (enough to balance the charge
on each side of the equation):
Cu(s)→Cu^2 +(aq) + 2e−
2Fe^3 +(aq) + 2e−→2Fe^2 +(aq)This gives us two pieces of information. First, we know that the value of n in the Nernst equation is equal to 2 (the
coefficient of the electrons in the balanced half-reactions above). Second, we know that copper is being oxidized and
iron is being reduced. Now, look at the table of standard reduction potentials to find the appropriate values. What
we will find are values for the following two half-reactions:
Cu^2 +(aq) + 2e−→Cu(s)
Fe^3 +(aq) + e−→Fe^2 +(aq)The reduction of Cu^2 +(aq) to Cu(s) has a standard potential of +0.34 V, and the reduction of Fe^3 +(aq) to Fe^2 +(aq)
has a standard potential of +0.77 V. To find E° for this complete cell, we subtract the reduction potential of the
species being oxidized from the reduction potential of the ion actually being reduced:
E°cell= E°red- E°oxid