Calculate the equailibrium constant for the reaction:
${C u}_{(s)} + {2 A g}_{(a q)}^{+} ⟶ {C u}_{(a q)}^{2 +} + {2 A g}_{(s)}$
Given that : $E_{A g^{+} / A g}^{0} = 0.80 V, E_{C u^{2 +} / C u}^{0} = 0.34 V .$

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Solution

$E_{A g^{+} / A g}^{o} = 0.80 V$ , $E_{C u^{2 +} / C u}^{o} = 0.34 V$

$E^{o}$ cell $= E^{o}$ reduction $+ E^{o}$ oxidation
$= (0.80 - 0.34) V$
$= 0.46 V$

$E^{o}$ cell $= (\frac{0.0591}{n}) log K$

$\frac{0.46 \times 2}{0.0591} = log K$

$log K = 15.5$

$K = 10^{15.5}$

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