Question

The standard emf of the cell $ (E^{\circ}_{cell} $ and equilibrium constant $ (K_{eq}) $ of the following reaction of $ 298\, K $
$ Cd^{2+} + 4NH_{3} {\rightleftharpoons} Cd (NH_{3})^{2+}_{4} $

• A. $ E^{\circ}_{cell}=1.0V, K_{eq}=126\times10^{7} $
• B. $ E^{\circ}_{cell}=0.21 V, K_{eq}=126\times10^{7} $
• C. $ E^{\circ}_{cell}=1.0 V, K_{eq}=6.60\times10^{33} $
• D. $ E^{\circ}_{cell}=0.21 V, K_{eq}=6.60\times10^{33} $

Answer 1

Answer: (B)

Key point (we assume that standard emf of the cell $E^{\circ}_{cell}$ is known)
For the given equilibrium,
$Cd^{2+}+4NH_{3} \rightleftharpoons Cd(NH_{3})^{2+}_{4}$
At equilibrium,
$E^{\circ}_{cell}=0$
Hence, we can calculate the equilibrium constant for reaction as follows :
$E^{\circ}_{cell}=\frac{0.0591\,V}{n} log \,K_{c}$
Where, $ K_{c}$ is unknown
$E^{\circ}_{cell}=0.21\,V $
$0.21\,V=\frac{0.0591}{2}log K_{c}$
$log\,K_{c}=\frac{0.21\times2}{0.0591}$
$log\,K_{c}=\frac{0.42}{0.0591}=7.1065$
$K_{c}=1.27\times10^{7}$