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For a cell involving one electron $E^\ominus_{cell}=0.59\ V$ at 298 K, the equilibrium constant for the cell reaction is: [Given that $\frac{2.303RT}{F}=0.059\ V$ at $T=298\ K$]
A
$1.0\times10^2$
B
$1.0\times10^5$
C
$1.0\times10^{10}$
D
$1.0\times10^{30}$
Detailed Solution
$E_{cell}=E^\ominus_{cell}-\frac{0.059}{n}\log Q$
At equilibrium, $Q=K_{eq}$ and $E_{cell}=0$:
$0=E^\ominus_{cell}-\frac{0.059}{1}\log K_{eq}$
$\log K_{eq}=\frac{0.59}{0.059}=10$
$K_{eq}=1.0\times10^{10}$