If the \(E_{\text {cell }}^{\circ}\) of an equilibrium reaction \(A(s)+2 B^{2+}(a q) \rightleftharpoons…
If the \(E_{\text {cell }}^{\circ}\) of an equilibrium reaction \(A(s)+2 B^{2+}(a q) \rightleftharpoons A^{2+}(a q)+2 B(s)\)
at \(298 \mathrm{~K}\) is \(0.59 \mathrm{~V}\), the equilibrium constant \(K_c\) is
\(1.0 \times 10^{10}\)
\(1.0 \times 10^2\)
\(1.0 \times 10^{-20}\)
\(1.0 \times 10^{20}\)
Solution
For the reaction,
\(\begin{gathered}
A+2 B^{2+}(a q) \rightleftharpoons A^{2+}(a q)+2 B(s) \\
\Delta G^{\circ}=-R T \ln K_c=-n F E_{\text {cell }}^{\circ} \quad \ldots (i)
\end{gathered}\)
where,
\(\begin{aligned}
\Delta G^{\circ} & =\text { Gibbs free energy } \\
n & =\text { number of electrons involved }(=2) \\
F & =\text { Faraday's constant }=96500 \mathrm{C} \text { charge. } \\
E^{\circ} & =\text { standard electrode cell potential }(=0.59 \mathrm{~V}) \\
R & =\text { gas constant } \\
T & =\text { temperature }(=298 \mathrm{~K})
\end{aligned}\)
\(\therefore\) From Eq. (i), we have
\(\begin{aligned}
\text {or } \log K_c & =\frac{n F E^{\circ}}{2.303 R T} \\
& =\frac{2 \times 96500 \times 0.59}{2303 \times 8.314 \times 298}=-19.96 \approx 20 \\
\therefore \quad K_c & =1.0 \times 10^{20}
\end{aligned}\)
Hence, option (c) is the correct answer.
Alternatively
At, \(T=298 \mathrm{~K}\)
\(\ln \quad K_c=\frac{n E_{\text {ccll }}^{\circ}}{0.059}=\frac{2 \times 0.59}{0.059}=20\)
\(\therefore \quad K_c=1 \times 10^{20}\)