Q86 · 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 \[ \begin{array}{l} \mathrm{Zn(s) \rightarrow Zn^{2+}(aq) + 2e} \\ \mathrm{Cu^{2+}(aq) + 2e \rightarrow Cu(s)} \end{array} \]
(a)$\Delta G^{o}-R T \ln \frac{a_{Z n^{2+}}}{a_{C u^{2+}}}$
(b)$\Delta G^{o}-R T \ln \frac{a_{Z n^{2+}}}{a_{C u(s)}}$
(c)$\Delta G^{o}-R T \ln \frac{a_{Z n(s)}}{\mathrm{a}_{\mathrm{Cu}^{2+}}}$
(d)$\Delta G^{o}+R T \ln \frac{a_{Z n^{2+}}}{\mathrm{a}_{\mathrm{Cu}^{2+}}}$
Answer
Answer: D ✓ checked by 4AB · confidence high
Explanation
Zn + Cu²⁺ → Zn²⁺ + Cu: ΔG = ΔG° + RT ln(a_Zn²⁺/a_Cu²⁺), with solids at unit activity (d).
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