KCET · Chemistry · Electrochemistry
\(\mathrm{H}_{2}(g)+2 \mathrm{AgCl}(s) \rightleftharpoons 2 \mathrm{Ag}(s)+2 \mathrm{HCl}(a q)\) \(E_{\text {cell }}^{\circ}\) at \(25^{\circ} \mathrm{C}\) for the cell is \(0.22 \mathrm{~V}\). The equilibrium constant at \(25^{\circ} \mathrm{C}\) is
- A \(2.8 \times 10^{7}\)
- B \(5.2 \times 10^{8}\)
- C \(2.8 \times 10^{5}\)
- D \(5.2 \times 10^{4}\)
Answer & Solution
Correct Answer
(A) \(2.8 \times 10^{7}\)
Step-by-step Solution
Detailed explanation
For given reaction, \(E_{\text {cell }}^{\circ}\) at \(25^{\circ} \mathrm{C}=0.22 \mathrm{~V}\)
At equilibrium,
\(\begin{aligned}
E_{\text {cell }}^{\circ} &=\frac{0.059}{n} \log K_{C} \\
\Rightarrow \quad \log K_{C} &=\frac{E_{\text {cell }}^{\circ} \times n}{0.059}=\frac{0.22 \times 2}{0.059}=7.45 \\
K_{C} &=\text { antilog } 7.45 \\
&=2.8 \times 10^{7}
\end{aligned}\)
At equilibrium,
\(\begin{aligned}
E_{\text {cell }}^{\circ} &=\frac{0.059}{n} \log K_{C} \\
\Rightarrow \quad \log K_{C} &=\frac{E_{\text {cell }}^{\circ} \times n}{0.059}=\frac{0.22 \times 2}{0.059}=7.45 \\
K_{C} &=\text { antilog } 7.45 \\
&=2.8 \times 10^{7}
\end{aligned}\)
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