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Calculate the electromotive force (→ potential) under the standard conditions of a battery (concentration battery) containing the electrodes: (1) $Cu^{2+}$ // $Cu$ $E^o$= $0.35 \;V$ (2) $Cu^{2+}$ // $Cu$ $E^o$= $0.35 \;V$ knowing that $[Cu^{2+}]_{(1)}=0.10 \frac{mol}{L}$ $[Cu^{2+}]_{(2)}=0.010 \frac{mol}{L}$
$E^o_{battery}$ $ =$ $ 0.35-0.35=0\; V$
$n\; =\; 2$ indeed $Cu^{2+}$ + 2$e^-$ + $Cu$ - 2$e^- $ $\leftrightarrows$ $Cu+Cu^{2+}$
$E$ $ =$ $ E^o_{battery}-\frac{0.059}{2}log\frac{[Cu^{2+}]_{(2)}}{ [Cu^{2+}]_{(1)}}$ $E$ = $ 0-\frac{0.059}{2}log\frac{0.01}{0.1} $ = $0.03\;V$