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The standard emf of a galvanic cell involving 2 moles of electrons in its redox reaction is $0.59 \mathrm{~V}$. The equilibrium constant for the redox reaction of the cell is
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The correct answer is:
$10^{20}$
In a galvanic cell, $\Delta G^{\circ}=-n F E^{\circ}$
Relationship between $\Delta G^{\circ}$ and equilibrium constant
$$
\Delta G^{\circ}=-2.303 R T \log K_{p}
$$
From Eqs. (i) and (ii), we get
$$
\begin{gathered}
-2.303 R T \log K_{p}=-n F E^{\circ} \\
\log K_{p}=\frac{n F E^{\circ}}{2.303 R T}=\frac{n E^{\circ}}{\frac{2.303 R T}{F}} \\
=\frac{2 \times 0.59}{0.059}=20 \\
K_{p}=\mathrm{AL}(20) \quad(\mathrm{AL}=\text { Antilog }) \\
K_{p}=10^{20}
\end{gathered}
$$
Relationship between $\Delta G^{\circ}$ and equilibrium constant
$$
\Delta G^{\circ}=-2.303 R T \log K_{p}
$$
From Eqs. (i) and (ii), we get
$$
\begin{gathered}
-2.303 R T \log K_{p}=-n F E^{\circ} \\
\log K_{p}=\frac{n F E^{\circ}}{2.303 R T}=\frac{n E^{\circ}}{\frac{2.303 R T}{F}} \\
=\frac{2 \times 0.59}{0.059}=20 \\
K_{p}=\mathrm{AL}(20) \quad(\mathrm{AL}=\text { Antilog }) \\
K_{p}=10^{20}
\end{gathered}
$$
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