$\begin{array}{1 1}0.184V\\0.827V\\0.521V\\0.490V\end{array} $

Answer : 0.521V

Since $\Delta G^{\large\circ}=-nFE^{\large\circ}$,we have

$Cu^{2+}+2e^-\rightarrow Cu$

$Cu^{2+}+e^-\rightarrow Cu^+$

____________________

$Cu^{+}+e^-\rightarrow Cu$

$\Delta G_1^{\large\circ}=-(2)(F)(0.337V)$

$\Delta G_2^{\large\circ}=-(1)(F)(0.153V)$

$\Delta G^{\large\circ}=\Delta G_1^{\large\circ}-\Delta G_2^{\large\circ}=-(4)F\{(2\times 0.337-0.153)V\}$

$E_{Cu^+|Cu}=2\times 0.337V-0.153V=0.521V$

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