Question

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

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

Answer 1

Answer: (B)

$Zn+2F{e}^{3+}\rightleftharpoons Z{n}^{2+}+2F{e}^{2+}$
$E_{cell}=E^{\circ }-\frac{0.0591}{2}$ log $\frac{\left[Z{n}^{2+}\right]{\left[F{e}^{2+}\right]}^2}{{\left[F{e}^{3+}\right]}^2}$
$1.71=E^{\circ }-\frac{0.0591}{2}$ log $\frac{\left[0.01\right]{\left[0.001\right]}^2}{{\left[0.01\right]}^2}$
$E^{\circ }=1.71+\frac{0.0591}{2}log10^{-4}=1.5918$
Now $\Delta G=-RT$ In $K_{eq}=-nF{E}^{\circ }$
$2\times 96500\times 1.5918$
$=2.303\times 8.314\times 298\times log{K}_{eq}$
log $K_{eq}=53.84$
$\Rightarrow K_{eq}\approx 10^{54}$