Question 18.PS.8: Determining E°cell and ΔG° Consider the redox reaction Zn^2+......

E^\circ_{cell} = – 0.763  V; ΔG° = 147  kJ. The reaction as written is not product-favored.

Strategy and Explanation   The first step is to write the half-reactions for the oxidation and reduction that occur and to use their standard reduction potentials from Table 18.1.

Reduction:              Zn^{2+}(aq) + 2 e^-→ Zn(s)                 E^\circ_{cathode} = – 0.763  V

Oxidation:             H_2(g) + 2 H_2O(\ell) → 2 H_3O^+(aq) + 2 e^-                  E^\circ_{node} = 0  V

We obtain the E^\circ_{cell} for the reaction from the standard reduction potentials for the two halfreactions.

E^\circ_{cell} = E^\circ_{cathode} –  E^\circ_{anode} = –  0.763  V  –  0  V = –  0.763  V

Since E^\circ_{cell} is negative, the reaction is not product-favored in the direction written. Zn^{2+} will not oxidize H_2O. The reverse reaction is product-favored; that is, zinc metal reacts with acid.

From the calculated E^\circ_{cell} we can calculate ΔG°.

ΔG° =- nFE^\circ_{cell}
= – (2  mol  e^-) × (\frac{9.65 × 10^4  C}{1  mol  e^-})(\frac{1  J}{1  V × 1  C})(- 0.763  V)
= 1.47 × 10^5  J = 147  kJ

The positive value for ΔG° also shows that the reaction as written is not product-favored.

Reasonable Answer Check   The given oxidation is the half-reaction at the standard hydrogen electrode with a potential of zero, so the overall reaction is governed by the Zn reduction. The negative cell potential and the positive ΔG° are consistent with the reaction being reactant-favored as written.