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Application of Electrolysis

A zinc rod is...

Question

A zinc rod is dipped in $0.1 M$ solution of $Z n S O_{4}$ . The salt is $95 %$ dissociated at this dilution at $298 K$ . Calculate the electrode potential. $[E_{Z n^{2 +} / Z n}^{o} = - 0.76 V]$ .

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Solution

As salt $Z n S O_{4}$ is 95% dissociated, $[Z n^{2 +}] = 0.1 \times \frac{95}{100} = 0.095 M$
$Z n^{2 +} + 2 e^{-} \to Z n_{(s)}$
By Nernst Equation,
$E_{Z n^{2 +} / Z n} = E_{Z n^{2 +} / Z n}^{o} - \frac{0.0591}{2} l o g \frac{1}{[Z n^{2 +}]}$

$= - 0.76 V - \frac{0.0591}{2} l o g \frac{1}{0.095}$

$= - 0.76 V - \frac{0.0591}{2} (l o g 1000 - l o g 95)$

$= - 0.76 V - \frac{0.0591}{2} (3.00 - 1.977)$

$= - 0.76 V - \frac{0.0591}{2} \times 1.0223$

$= - 0.76 V - \frac{0.0604 V}{2}$

$= - 0.76 V - 0.0302 V = - 0.7902 V$

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