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A zinc rod is...

Question

A zinc rod is dipped in $0.095 M$ solution of $Z n S O_{4}$ at $298 K$ . Calculate the electrode potential of zinc electrode $(E_{Z a^{2 +} / Z a}^{0} = - 0.76 V)$ .

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Solution

Given:
$E_{z n^{21} / z n}^{o} = - 0.76 V$ ; and $[2 n^{21}] = 0.095 M$
Redox reaction: $z n^{21} + 2 e^{(-)} ⟶ Z n^{(s)}$
For electrode potential of zinc electrode $E_{z n^{21} / z n} = E_{z n^{21} / z n}^{o} - \frac{0.0591}{2} log \frac{1}{[2 n^{21}]}$
$= - 0.76 - \frac{0.0591}{2} log \frac{1}{0.095}$
$= - 0.7902 V$

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