For a cell involving one electron $E^\ominus_{ ext{cell}} = 0.59 ext{ V}$ at $298 ext{ K}$. The equilibri

Question

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

Accepted Answer

The relationship between the standard emf of a cell ($E^\ominus_{	ext{cell}}$) and the equilibrium constant ($K_c$) for the cell reaction is given by the Nernst equation at equilibrium:

$E^\ominus_{	ext{cell}} = \frac{2.303 RT}{nF} \log K_c$

Given:
$E^\ominus_{	ext{cell}} = 0.59 	ext{ V}$
$n = 1$ (since the cell involves one electron)
$\frac{2.303 RT}{F} = 0.059 	ext{ V}$

Substituting these values into the equation:
$0.59 = \frac{0.059}{1} \log K_c$
$\log K_c = \frac{0.59}{0.059} = 10$

Taking the antilog of both sides:
$K_c = 10^{10} = 1.0 	imes 10^{10}$

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