Electrochemistry - Result Question 46
####48. 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 $F=96500 Cmol^{-1} ; R=8.314 JK^{-1} mol^{-1}$ )
(a) $2.0 \times 10^{11}$
(b) $4.0 \times 10^{12}$
(c) $1.0 \times 10^{2}$
(d) $1.0 \times 10^{10}$
[2004]
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Solution:
(d) $E^{\circ}=\frac{0.0591}{n} \log _{10} K$
Here, $n=2, E^{\circ}=0.295$
$\therefore \quad \log _{10} K=\frac{2 \times 0.295}{0.0591}=10$ or $K=1 \times 10^{10}$
