Given :(i) , Cu^{2+} + 2e^- rightarrow Cu ,, | vedclass

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(D) The standard Gibbs free energy change is given by $\Delta G^{o} = -nFE^{o}$.For reaction $(i): Cu^{2+} + 2e^{-} \rightarrow Cu$,$\Delta G_{1}^{o}

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$0.52$

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(D) The standard Gibbs free energy change is given by $\Delta G^{o} = -nFE^{o}$.For reaction $(i): Cu^{2+} + 2e^{-} ightarrow Cu$,$\Delta G_{1}^{o} = -2 	imes F 	imes 0.337 = -0.674F$.For reaction $(ii): Cu^{2+} + e^{-} ightarrow Cu^{+}$,$E^{o} = 0.153 \, V$. We need the reverse reaction: $Cu^{+} ightarrow Cu^{2+} + e^{-}$,where $E^{o} = -0.153 \, V$.Thus,for $Cu^{+} ightarrow Cu^{2+} + e^{-}$,$\Delta G_{2}^{o} = -1 	imes F 	imes (-0.153) = 0.153F$.Adding the two reactions:$(Cu^{2+} + 2e^{-}$ $ightarrow Cu) + (Cu^{+}$ $ightarrow Cu^{2+} + e^{-})$ $ightarrow (Cu^{+} + e^{-}$ $ightarrow Cu)$.$\Delta G_{total}^{o} = \Delta G_{1}^{o} + \Delta G_{2}^{o} = -0.674F + 0.153F = -0.521F$.For the target reaction $Cu^{+} + e^{-} ightarrow Cu$,$n=1$,so $\Delta G^{o} = -1 	imes F 	imes E^{o}$.$-0.521F = -1 	imes F 	imes E^{o} \implies E^{o} = 0.52 \, V$.

Given $:$
$(i) \, Cu^{2+} + 2e^- \rightarrow Cu \,, \, E^o = 0.337 \, V$
$(ii) \, Cu^{2+} + e^- \rightarrow Cu^{+} \,, \, E^o = 0.153 \, V$
Electrode potential,$E^o$ for the reaction,
$Cu^{+} + e^- \rightarrow Cu \,,$ will be $............$ $V$.

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Given $:$$(i) \, Cu^{2+} + 2e^- \rightarrow Cu \,, \, E^o = 0.337 \, V$$(ii) \, Cu^{2+} + e^- \rightarrow Cu^{+} \,, \, E^o = 0.153 \, V$Electrode potential,$E^o$ for the reaction,$Cu^{+} + e^- \rightarrow Cu \,,$ will be $............$ $V$.