For a cell involving one electron $E^\circ_{cell} = 0.59 ext{ V}$ at $298 ext{ K}$, the equilibrium constant

Question

For a cell involving one electron $E^\circ_{cell} = 0.59	ext{ V}$ at $298	ext{ K}$, the equilibrium constant for the cell reaction is : [Given that $\frac{2.303 RT}{F} = 0.059	ext{ V at } T = 298	ext{ K}$]

Accepted Answer

$E_{cell} = E^\circ_{cell} - \frac{0.059}{n} \log Q$ ...(i). At equilibrium, $Q = K_{eq}$ and $E_{cell} = 0$. $0 = E^\circ_{cell} - \frac{0.059}{1} \log K_{eq}$ (from equation (i)). $\log K_{eq} = \frac{E^\circ_{cell}}{0.059} = \frac{0.59}{0.059} = 10$. $K_{eq} = 10^{10} = 1 	imes 10^{10}$.

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