Question 13.EP.3: Suppose that we wish to study the possible galvanic corrosio...
Suppose that we wish to study the possible galvanic corrosion between zinc and chromium, so we set up the following cell:
Cr(s)\mid Cr^{2+}(aq)\parallel Zn^{2+}(aq)\mid Zn(s)
What is the chemical reaction that takes place, and what is the standard free energy change for that reaction?
Strategy
To calculate the free energy change, we must know two things: the cell potential and the number of electrons transferred in the reaction. Then we can simply use these values in Equation 13.5
ΔG° = −nFE° (13.5)
to obtain the free energy change.
First we need the balanced chemical equation, which in this case can be written immediately because two electrons are transferred in each half reaction (n = 2):
Zn^{2+}(aq)+Cr(s)\longrightarrow Cr^{2+}(aq)+Zn(s)
Now if we look up the standard reduction potentials, we find
Zn^{2+}(aq) + 2 \ e^{−} \longrightarrow Zn(s) E° = −0.763 V
Cr^{2+}(aq) + 2 \ e^{−} \longrightarrow Cr(s) E° = −0.910 V
According to Equation 13.2, the cell potential is:
E°_{cell} = E°_{red} − E°_{ox} (13.2)
E°_{cell} = −0.763 V − (−0.910 V) = 0.147 V
Inserting this in Equation 13.5,
ΔG° = −nFE° = −2 mol × 96,485 J V^{−1} \ mol^{−1} × 0.147 V = −2.84 × 10^{4} J = −28.4 kJ
Discussion
The most common error students make in this type of problem is forgetting to include n, the number of electrons transferred. Remember that even if you must multiply a chemical equation that represents the reaction given in the standard reduction potential by some value to match the number of electrons between oxidation and reduction, you do not multiply the standard reduction potential. Reduction potentials are not like thermochemical quantities in this regard.