Question

Consider the cell $Zn|Zn^{2+} (a = 0.01) || Fe^{2+} (a = 0.001), Fe^{3+} (a = 0.01) | Pt, E_{cell} = 1.71 V$ at $25^{\circ}C$ for the above cell. The equilibrium constant for the reaction:
$Zn + 2Fe^{3+} \rightleftharpoons Zn^{2+} + 2Fe^{2+}$ at $25^{\circ}C$ would be close to

• A. $10^{27}$
• B. $10^{54}$
• C. $10^{81}$
• D. $10^{40}$

Answer 1

Answer: (B)

$Zn + 2Fe^{3+} \rightleftharpoons Zn^{2+} + 2Fe^{2+}$
$E_{cell}=E^{\circ}-\frac{0.0591}{2}$ log $\frac{\left[Zn^{2+}\right]\left[Fe^{2+}\right]^{2}}{\left[Fe^{3+}\right]^{2}}$
$1.71 =E^{\circ} -\frac{0.0591}{2}$ log $\frac{\left[0.01\right]\left[0.001\right]^{2}}{\left[0.01\right]^{2}}$
$E^{\circ} = 1.71 +\frac{0.0591}{2} log 10^{-4} = 1.5918$
Now $\Delta G = - RT$ In $K_{eq} = -nFE^{\circ}$
$2 \times 96500 \times 1.5918$
$ = 2.303 \times 8.314 \times 298 \times log K_{eq}$
log $K_{eq} = 53.84$
$\Rightarrow K_{eq} \approx10^{54}$