For the galvanic cell in which the reaction 3Cu + 2Au3+ → 3Cu2+ + 2Au takes place, the value of n to be used in the Nernst equation is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 6
In the Nernst equation, n is the number of electrons transferred in the cell reaction as it is written. The same n is used for both electrodes, so it is found by splitting the balanced reaction into its two half-reactions:
Anode (oxidation): \( \displaystyle \ce{3Cu(s) -> 3Cu^2+(aq) + 6e^-} \)
Cathode (reduction): \( \displaystyle \ce{2Au^3+(aq) + 6e^- -> 2Au(s)} \)Each Cu atom loses 2 electrons (3 × 2 = 6) and each \( \ce{Au^3+} \) ion gains 3 electrons (2 × 3 = 6). The electrons lost and gained balance at 6, so n = 6, and at 298 K:
\( \displaystyle E_{\text{cell}}=E^{\circ}_{\text{cell}}-\frac{0.059}{6}\log\frac{[\ce{Cu^2+}]^3}{[\ce{Au^3+}]^2} \)