Electrochemistry Ques 21
21. For a cell involving one electron $E _{\text{cell }}^{!}=0.59 V$ at $298 K$, the equilibrium constant for the cell reaction is :
[2019]
$[.$ Given that $\frac{2.303 R T}{F}=0.059 V$ at $.T=298 K]$
(a) $1.0 \times 10^{2}$
(b) $1.0 \times 10^{5}$
(c) $1.0 \times 10^{10}$
(d) $1.0 \times 10^{30}$
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Solution:
(c) $E _{\text{cell }}^{\circ}=\frac{2.303 RT}{nF} \log K$
Given : $E _{\text{cell }}^{\circ}=0.59 V, n=1$
$0.59=\frac{0.059}{1} \log K$
$\frac{0.59}{0.059}=\log K$
$10=\log k$
$K=10^{10}$
BETA
