Q30 · CSIR-NET Chemistry, June 2015
Paper: CSIR-NET June 2015 · Subject: Physical Chemistry · Chapter: Electrochemistry · Topic: Nernst Cells · Marks: 2 · Difficulty: Medium
The correct $\Delta G$ for the cell reaction involving steps \[ \begin{array}{l} \mathrm{Zn(s) \rightarrow Zn^{2+}(aq)+2e} \\ \mathrm{Cu^{2+}(aq)+2e \rightarrow Cu(s)} \end{array} \]
is (a)$\Delta G^{o}-RT\ln\dfrac{a_{\mathrm{Zn}^{2+}}}{a_{\mathrm{Cu}^{2+}}}$
(b)$\Delta G^{o}+RT\ln\dfrac{a_{\mathrm{Zn}^{2+}}}{a_{\mathrm{Cu(s)}}}$
(c)$\Delta G^{o}-RT\ln\dfrac{a_{\mathrm{Zn(s)}}}{a_{\mathrm{Cu}^{2+}}}$
(d)$\Delta G^{o}+RT\ln\dfrac{a_{\mathrm{Zn}^{2+}}}{a_{\mathrm{Cu}^{2+}}}$
Answer
Answer: D ✓ checked by 4AB · confidence high
The source book printed no answer; this one was worked out and checked — see the explanation.
Explanation
Zn + Cu²⁺ → Zn²⁺ + Cu: ΔG = ΔG° + RT ln(a(Zn²⁺)/a(Cu²⁺)). The pure solids have unit activity (d).
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