The standard emf of Agalvanic cell involving cell reaction with $n\ = \ 2$ is found to be 0.295 V at 25 ℃. The equilibrium constant of the reaction would be (Given, $F\ = \ 96500\ C\ {mol}^{- 1}\ ,\ R\ = \ 8.314\ {JK}^{- 1}\ {mol}^{- 1}$ )

Question

Dr. Abubacker · Accepted Answer

(d)
By Nernst equation,
$E_{cell} = \ E_{cell}^{{^\circ}}\  - \ \frac{2.303\ RT}{nF}\log_{10}K$
At equilibrium $E_{cell}$ =0
Given that,
$	herefore R\  = 8.315\ JK^{- 1}\ {mol}^{- 1}$
$T\  = \ 25\ ℃\  + \ 273\  = \ 298K$
$F\  = \ 96500\ C\ and\ n\  = \ 2$
$	herefore\ \ \ \ \ \ E_{cell}^{{^\circ}}\  = \ \ \frac{2.303\  	imes 8.314\  	imes 298}{2\  	imes 96500}\ \log_{10}K$
$= \ \frac{0.0591}{2\ }\ \log_{10}K$
∵ Given that $E_{cell}^{{^\circ}}\  = \ 0.295\ V$
∴ $0.295\  = \ \frac{0.0591}{2\ }\ \log_{10}K$
$\log_{10}K\  = \ \ \frac{0.295\  	imes 2}{0.0591}\  = \ 10$
$\log_{10}K\  = \ antilog\ 10$
$K\  = \ 1\  	imes \ 10^{10}$

Q. The standard emf of Agalvanic cell involving cell reaction with $n\ = \ 2$ is found to be 0.295 V at 25 ℃. The equilibrium constant of the reaction would be (Given, $F\ = \ 96500\ C\ {mol}^{- 1}\ ,\ R\ = \ 8.314\ {JK}^{- 1}\ {mol}^{- 1}$ )

Explanation

Reviewed by Dr. Abubacker

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