Looking for classes? Ksquare Career Institute, Bengaluru →
The standard e.m.f. of a galvanic cell involving cell reaction with n = 2 is found to be 0.295 V at $25^\circ C$. The equilibrium constant of the reaction would be :- (Given F = 96500 C mol$^{-1}$; R = 8.314 J K$^{-1}$ mol$^{-1}$)
A
$4.0 \times 10^{12}$
B
$1.0 \times 10^{2}$
C
$1.0 \times 10^{10}$
D
$2.0 \times 10^{11}$
Detailed Solution
At equilibrium, $\Delta G^\circ = -nFE^\circ_{cell} = -RT\ln K$
$\Rightarrow \log K = \dfrac{nFE^\circ_{cell}}{2.303RT}$
At 298 K, $\dfrac{2.303RT}{F} = \dfrac{2.303 \times 8.314 \times 298}{96500} = 0.0591$ V
$\log K = \dfrac{nE^\circ_{cell}}{0.0591} = \dfrac{2 \times 0.295}{0.0591} = \dfrac{0.590}{0.0591} \approx 10$
$K = 1.0 \times 10^{10}$
$\Rightarrow \log K = \dfrac{nFE^\circ_{cell}}{2.303RT}$
At 298 K, $\dfrac{2.303RT}{F} = \dfrac{2.303 \times 8.314 \times 298}{96500} = 0.0591$ V
$\log K = \dfrac{nE^\circ_{cell}}{0.0591} = \dfrac{2 \times 0.295}{0.0591} = \dfrac{0.590}{0.0591} \approx 10$
$K = 1.0 \times 10^{10}$