When a copper plate is kept in a $0.1 \ M$ solution of $CuSO_4$ at $298 \ K$ temperature and if $70 \ \%$ dissociation has occurred,then calculate the potential of the copper electrode.
(N/A) The reduction reaction is: $Cu^{2+} + 2e^{-} \rightarrow Cu_{(s)}$
The effective concentration of $Cu^{2+}$ ions is calculated as: $[Cu^{2+}] = \text{Molarity} \times \text{Degree of dissociation} = 0.1 \times 0.70 = 0.07 \ M$
Using the Nernst equation for the electrode potential: $E_{Cu^{2+}/Cu} = E^{\circ}_{Cu^{2+}/Cu} - \frac{0.059}{n} \log \frac{1}{[Cu^{2+}]}$
Given $E^{\circ}_{Cu^{2+}/Cu} = +0.34 \ V$ and $n = 2$:
$E_{Cu^{2+}/Cu} = 0.34 - \frac{0.059}{2} \log \frac{1}{0.07}$
$E_{Cu^{2+}/Cu} = 0.34 - 0.0295 \times \log(14.286)$
$E_{Cu^{2+}/Cu} = 0.34 - 0.0295 \times 1.1549$
$E_{Cu^{2+}/Cu} = 0.34 - 0.03407 = 0.3059 \ V$
The effective concentration of $Cu^{2+}$ ions is calculated as: $[Cu^{2+}] = \text{Molarity} \times \text{Degree of dissociation} = 0.1 \times 0.70 = 0.07 \ M$
Using the Nernst equation for the electrode potential: $E_{Cu^{2+}/Cu} = E^{\circ}_{Cu^{2+}/Cu} - \frac{0.059}{n} \log \frac{1}{[Cu^{2+}]}$
Given $E^{\circ}_{Cu^{2+}/Cu} = +0.34 \ V$ and $n = 2$:
$E_{Cu^{2+}/Cu} = 0.34 - \frac{0.059}{2} \log \frac{1}{0.07}$
$E_{Cu^{2+}/Cu} = 0.34 - 0.0295 \times \log(14.286)$
$E_{Cu^{2+}/Cu} = 0.34 - 0.0295 \times 1.1549$
$E_{Cu^{2+}/Cu} = 0.34 - 0.03407 = 0.3059 \ V$
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