Calculate E value for the cell reaction:
$C u | C u^{2 +} (0.001 M) | | C u^{2 +} (0.1 M) | C u$

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Solution

$C u + {C u}_{L}^{2 +} (0.001) \to {C u}_{R}^{2 +} (0.1) + C u$
$E = E^{0} - \frac{0.0591}{2} log \frac{{[{C u}^{2 +}]}_{R}}{{[{C u}^{2 +}]}_{L}} : (n_{f} = 2)$
$= Z e r o - \frac{0.0591}{2} log (\frac{0.1}{0.001}) = - 0.0591 V$

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E

C u | {C u}^{2 +} (0.001 M) | | {C u}^{2 +} (0.1 m) | C u

E

C u | {C u}^{2 +} (0.001 M) | | {C u}^{2 +} (0.1 M) | C u

Z n | Z n^{2 +} (0.001 M) | | C u^{2 +} (0.1 M) | C u

C u / C u^{2 +}

Z n / Z n^{2 +}

C u^{2 +} + e^{-} \to C u^{+}

E^{o} = 0.15 V

C u^{2 +} + 2 e^{-} \to C u

E^{o} = 0.34 V

E^{o}

C u^{+} + e^{-} \to C u

C u^{+}

EMF of the cell:

$C u / C u^{2 +} (0.01 M) | | C u^{2 +} (0.001 M) | C u$

is $(E^{0} C u^{2 +} / C u = + 0.34 V)$

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