Question

For a cell reaction involving two electron change, the standard e.m.f. of the cell is found to be 0.295 Vat 298 K. The equilibrium constant of the reaction at the same temperature will be

• A. $2.95 \times 10^2$
• B. 10
• C. $1 \times 10^{10}$
• D. $1 \times 10^{-10}$

Answer 1

The Correct Option is C

$E^{^\circ}_{cell} = 0.295 V$
$ T =298 \,K $
$E^{^{\circ} }_{cell} = \frac{2.303 RT}{nF} \log K_{c}$
$ \log K_{c} = \frac{E^{^{\circ} }_{cell} \times nF}{2.303 RT} = \frac{0.295 \times2\times96500}{2.303 \times 8.314 \times 298} $
$\log K_{c} = 9.978 $
$K_{c} = 9.51 \times 10^{9}$
$ = 0.95 \times 10^{10} \approx 1\times 10^{10}$