Electrochemistry Result 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 K .
$ \left[2.303 \dfrac{R T}{F} \text { at } 298 \mathrm{~K}=0.059 \mathrm{~V}\right] $
(2019 Main, 11 Jan II)
(a) $0.4736 $ $V$
(b) $0.04736$ $ mV$
(c) $0.4736 $ $mV$
(d) $0.04736 $ $V$
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Answer:
Correct Answer: 6. ( a )
Solution:
- According to Nernst equation,
$ E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{2.303 R T}{n F} \log Q $
Given, $\frac{2.303 R T}{F}=0.059 \mathrm{~V}$
$ \therefore \quad E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{0.059}{n} \log Q $
At equilibrium, $E_{\text {cell }}=0$
$ E_{\text {cell }}^{\circ}=\frac{0.059}{n} \log K_C $
For the given reaction, $n=2$
Also,
$ K_C=10 \times 10^{15} \quad $ [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} $
BETA
