Question

The standard emf of the cell $(E_{cell}^{\circ }$ and equilibrium constant $(K_{eq})$ of the following reaction of $298K$
$C{d}^{2+}+4N{H}_3\rightleftharpoons Cd(N{H}_3{)}_4^{2+}$

• A. $E_{cell}^{\circ }=1.0V,K_{eq}=126\times 10^7$
• B. $E_{cell}^{\circ }=0.21V,K_{eq}=126\times 10^7$
• C. $E_{cell}^{\circ }=1.0V,K_{eq}=6.60\times 10^{33}$
• D. $E_{cell}^{\circ }=0.21V,K_{eq}=6.60\times 10^{33}$

Answer 1

Answer: (B)

Key point (we assume that standard emf of the cell $E_{cell}^{\circ }$ is known)
For the given equilibrium,
$C{d}^{2+}+4N{H}_3\rightleftharpoons Cd(N{H}_3{)}_4^{2+}$
At equilibrium,
$E_{cell}^{\circ }=0$
Hence, we can calculate the equilibrium constant for reaction as follows :
$E_{cell}^{\circ }=\frac{0.0591V}{n}log{K}_c$
Where, $K_c$ is unknown
$E_{cell}^{\circ }=0.21V$
$0.21V=\frac{0.0591}{2}log{K}_c$
$log{K}_c=\frac{0.21\times 2}{0.0591}$
$log{K}_c=\frac{0.42}{0.0591}=7.1065$
$K_c=1.27\times 10^7$