Electrochemistry 2 Question 6
6. Given the equilibrium constant $\left(K_{C}\right)$ of the reaction :
$$ \mathrm{Cu}(s)+2 \mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Cu}^{2+}(a q)+2 \mathrm{Ag}(s) $$
is $10 \times 10^{15}$, calculate the $E_{\text {cell }}^{\circ}$ of this reaction at $298 \mathrm{~K}$.
$\left[2.303 \frac{R T}{F}\right.$ at $\left.298 \mathrm{~K}=0.059 \mathrm{~V}\right]$
(a) $0.4736 \mathrm{~V}$
(b) $0.04736 \mathrm{mV}$
(c) $0.4736 \mathrm{mV}$
(d) $0.04736 \mathrm{~V}$
(2019 Main, 11 Jan II)
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Solution:
- According to Nernst equation,
$$ \begin{aligned} E_{\text {cell }} & =E_{\text {cell }}^{\circ}-\frac{2.303 R T}{n F} \log Q \ \text { Given, } \quad \frac{2.303 R T}{F} & =0.059 \mathrm{~V} \ \therefore \quad E_{\text {cell }} & =E_{\text {cell }}^{\circ}-\frac{0.059}{n} \log Q \ \text { At equilibrium, } E_{\text {cell }} & =0 \ E_{\text {cell }}^{\circ} & =\frac{0.059}{n} \log K_{C} \end{aligned} $$
For the given reaction, $n=2$
$$ \begin{aligned} & \text { Also, } \quad K_{C}=10 \times 10^{15} \quad \text { [given] } \ & \therefore \quad E_{\text {cell }}^{\circ}=\frac{0.059}{2} \log \left(10 \times 10^{15}\right)=0.472 \mathrm{~V} \approx 0.473 \mathrm{~V} \end{aligned} $$
