Question

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

• A. $1.0 \times 10^{10}$
• B. $1.0 \times 10^{30}$
• C. $1.0 \times 10^{2}$
• D. $1.0 \times 10^{5}$

Answer 1

Answer: (A)

$1.0 \times 10^{10}$

Answer 2

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