For a cell reaction involving a two-electron change, the standard Emf of the cell is found to be $0.295 ext{

Question

For a cell reaction involving a two-electron change, the standard Emf of the cell is found to be $0.295	ext{ V}$ at $25^{\circ}	ext{C}$. The equilibrium constant of the reaction at $25^{\circ}	ext{C}$ will be:

Accepted Answer

The relationship between the 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:
$0.295 = \frac{0.059}{2} \log K_c$
$0.295 = 0.0295 \log K_c$
$\log K_c = \frac{0.295}{0.0295} = 10$
Therefore, $K_c = 10^{10} = 1 	imes 10^{10}$.

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