The electrode potentials for $\mathrm{Cu}^{2+}(a q)+e^{-} \longrightarrow \mathrm{Cu}^{+}(a q) \text { and }…

The electrode potentials for $\mathrm{Cu}^{2+}(a q)+e^{-} \longrightarrow \mathrm{Cu}^{+}(a q) \text { and } \mathrm{Cu}^{+}(a q)+e^{-} \longrightarrow \mathrm{Cu}(s) \text { are }+0.15 \mathrm{~V} \text { and }+0.50 \mathrm{~V} \text { respectively. The value of } E_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\circ} \text { will be}$
  1. $0.325 \mathrm{~V}$
  2. $0.650 \mathrm{~V}$
  3. $0.150 \mathrm{~V}$
  4. $0.500 \mathrm{~V}$

Solution

$\begin{aligned} & \mathrm{Cu}^{2+}+e^{-} \rightarrow \mathrm{Cu}^{+} ; \\ & E_{1}^{\circ}=0.15 \mathrm{~V} ; \Delta G_{1}^{\circ}=-n_{1} E_{1}^{\circ} F \\ & \mathrm{Cu}^{+}+e^{-} \rightarrow \mathrm{Cu} ; \\ & E_{2}^{\circ}=0.50 \mathrm{~V} ; \Delta G_{2}^{\circ}=-n_{2} E_{2}^{\circ} F \\ & \mathrm{Cu}^{2+}+2 e^{-} \rightarrow \mathrm{Cu} ; E^{\circ}=? ; \Delta G^{\circ}=-n E^{\circ} F \\ & \Delta G^{\circ}=\Delta G_{1}^{\circ}+\Delta G_{2}^{\circ} \\ & \text{or } -2 E^{\circ} F=-1 F \times 0.15+(-1 F \times 0.50) \\ & \text{or } -2 E^{\circ} F=-0.15 F-0.50 F \\ & \text{or } -2 F E^{\circ}=-F(0.15+0.50) \\ & \therefore \quad E^{\circ}= \frac{0.65}{2}=0.325 \mathrm{~V} \end{aligned}$

Asked in: NEET 2011 (Screening)

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