The standard reduction potential for the reduction half-cell; Cu2+ (aq) + 2e− → Cu(s) is 0.34 V.
This half cell is connected to the standard hydrogen electrode (oxidation half-cell). The change in Gibbs free energy for the cell is:
(Given: F = 96487 C mol−1)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → − 65.6 kJ mol−1
In a galvanic cell, the decrease in Gibbs energy equals the electrical work the cell can do. That work is the charge that flows (nF) multiplied by the cell potential (E):
\( \displaystyle \Delta_rG^{\circ}=-nFE^{\circ}_{cell} \)
Here n is the number of electrons transferred in the balanced cell reaction.
The standard hydrogen electrode is the anode (oxidation) and the copper half-cell is the cathode (reduction):
\( \displaystyle \text{Anode: }\ce{H2(g) -> 2H^+(aq) + 2e^-} \)
\( \displaystyle \text{Cathode: }\ce{Cu^2+(aq) + 2e^- -> Cu(s)} \)
\( \displaystyle \text{Cell: }\ce{H2(g) + Cu^2+(aq) -> 2H^+(aq) + Cu(s)} \)
So n = 2, and
\( \displaystyle E^{\circ}_{cell}=E^{\circ}_{cathode}-E^{\circ}_{anode}=0.34-0=0.34\ \mathrm{V} \)
\( \displaystyle \Delta_rG^{\circ}=-2\times96487\ \mathrm{C\,mol^{-1}}\times0.34\ \mathrm{V} \)
\( \displaystyle \Delta_rG^{\circ}=-65611\ \mathrm{J\,mol^{-1}} \)
\( \displaystyle \Delta_rG^{\circ}\approx-65.6\ \mathrm{kJ\,mol^{-1}} \)
Shortcut: 2 × 96.5 × 0.34 = 193 × 0.34 ≈ 65.6, so the answer is −65.6 kJ mol−1.