Given $$ \mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\circ}=0.34 \mathrm{~V}, \mathrm{E}_{\mathrm{Cu}^{2+}…

Given $$ \mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\circ}=0.34 \mathrm{~V}, \mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\circ}=0.15 \mathrm{~V} $$ Standard electrode potential for the half cell $\mathrm{Cu}^{+} / \mathrm{Cu}$ is
  1. $0.38 \mathrm{~V}$
  2. $0.53 \mathrm{~V}$
  3. $0.19 \mathrm{~V}$
  4. $0.49 \mathrm{~V}$

Solution

$ \begin{aligned} & \text { } \mathrm{Cu}^{++}+e^{-} \longrightarrow \mathrm{Cu}^{+} \text {; } \\ & E_1^0=0.15 \mathrm{~V} ; \Delta G_1^0=-n_1 E_1^0 F \\ & \mathrm{Cu}^{2+}+2 e \longrightarrow \mathrm{Cu} \quad \text {; } \\ & E_2^0=0.34 \mathrm{~V} ; \Delta G_2^0=-n_2 E_2^0 F \\ & \end{aligned} $ On subracting eq.(i) from eq. (ii) we get $ \begin{aligned} & \mathrm{Cu}^{+}+e^{-} \longrightarrow \mathrm{Cu} ; \Delta G^0=\Delta G_2^0-\Delta G_1^0 \\ &-n E^0 F=-\left(n_2 E^0 F-n_1 E_1^0 F\right) \\ & E^0=\frac{n_2 E_2^0 F-n_1 E_1^0 F}{n F} \\ &=\frac{2 \times 0.34-0.15}{1} \\ &=0.53 \mathrm{~V} \end{aligned} $

Asked in: JEE Main 2012 (26 May Online)

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