Given,the standard potentials E_{(Cu^{2+}/Cu) | vedclass

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(B) Given:$E_{Cu^{2+}/Cu}^{\circ} = 0.340 \ V$$E_{Cu^{+}/Cu}^{\circ} = 0.522 \ V$Step $1$: Write the half-cell reactions:$(i) \ Cu^{2+} + 2e^{-} \long

Vedclass · Accepted Answer

$0.158$

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(B) Given:$E_{Cu^{2+}/Cu}^{\circ} = 0.340 \ V$$E_{Cu^{+}/Cu}^{\circ} = 0.522 \ V$Step $1$: Write the half-cell reactions:$(i) \ Cu^{2+} + 2e^{-} \longrightarrow Cu, \ E_{1}^{\circ} = 0.340 \ V, \ \Delta G_{1}^{\circ} = -2FE_{1}^{\circ}$$(ii) \ Cu^{+} + e^{-} \longrightarrow Cu, \ E_{2}^{\circ} = 0.522 \ V, \ \Delta G_{2}^{\circ} = -1FE_{2}^{\circ}$Step $2$: We need the reaction for $Cu^{2+} + e^{-} \longrightarrow Cu^{+}$:This is obtained by $(i) - (ii)$:$Cu^{2+} + e^{-} \longrightarrow Cu^{+}, \ E_{3}^{\circ} = ?, \ \Delta G_{3}^{\circ} = -1FE_{3}^{\circ}$Step $3$: Use the relation $\Delta G_{3}^{\circ} = \Delta G_{1}^{\circ} - \Delta G_{2}^{\circ}$:$-1FE_{3}^{\circ} = -2FE_{1}^{\circ} - (-1FE_{2}^{\circ})$$E_{3}^{\circ} = 2E_{1}^{\circ} - E_{2}^{\circ}$$E_{3}^{\circ} = 2(0.340) - 0.522$$E_{3}^{\circ} = 0.680 - 0.522 = 0.158 \ V$

Given,the standard potentials $E_{(Cu^{2+}/Cu)}^{\circ}$ and $E_{(Cu^{+}/Cu)}^{\circ}$ as $0.340 \ V$ and $0.522 \ V$ respectively,the value of $E_{(Cu^{2+}/Cu^{+})}^{\circ}$ is $.... \ V$

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