ElectrochemistryMedium
Question
The EMF of the cell: Zn|Zn2+ (0.01 M)||Fe2+ (0.001 M)|Fe at 298 K is 0.2905 V, then the value of equilibrium constant for the cell reaction is
Options
A.$e^{\frac{0.32}{0.0295}}$
B.$10^{\frac{0.32}{0.0295}}$
C.$10^{\frac{0.26}{0.0295}}$
D.$10^{\frac{0.32}{0.0591}}$
Solution
Cell reaction: $Zn + Fe^{2 +} \rightleftharpoons Zn^{2 +} + Fe$
$E_{cell} = E_{cell}^{o} - \frac{0.059}{2}.\log\frac{\left\lbrack Zn^{2 +} \right\rbrack}{\left\lbrack Fe^{2 +} \right\rbrack} $$${\text{Or, 0.2905 = }\frac{0.059}{2}.\log K_{eq} - \frac{0.059}{2}.\log\frac{0.01}{0.001} }{\therefore K_{eq} = 10^{\left( \frac{0.32}{0.0295} \right)}}$$
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