The standard Emf of a galvanic cell involving cell reaction with $n = 2$ is found to be $0.295 ext{ V}$ at $2

Question

The standard Emf of a galvanic cell involving cell reaction with $n = 2$ is found to be $0.295	ext{ V}$ at $25^{\circ}	ext{C}$. The equilibrium constant of the reaction would be:
(Given $F = 96500	ext{ C mol}^{-1}$; $R = 8.314	ext{ J K}^{-1}	ext{ mol}^{-1}$)

Accepted Answer

The relationship between standard cell potential ($E^{\circ}_{cell}$) and the equilibrium constant ($K_c$) at $298	ext{ K}$ ($25^{\circ}	ext{C}$) is given by the Nernst equation at equilibrium:
$E^{\circ}_{cell} = \frac{2.303RT}{nF} \log K_c = \frac{0.059}{n} \log K_c$
Given:
$E^{\circ}_{cell} = 0.295	ext{ V}$
$n = 2$
Substituting the values into the equation:
$0.295 = \frac{0.059}{2} \log K_c$
$\log K_c = \frac{0.295 	imes 2}{0.059} = \frac{0.590}{0.059} = 10$
Therefore, $K_c = 10^{10} = 1.0 	imes 10^{10}$ .

Step-by-Step Solution

Practice All CHEMISTRY Questions