Question

$H _{2(g)}+2 AgCl _{(s)} \rightleftharpoons 2 Ag _{(s)}+2 HCl _{(a q)}$
$E _{\text {cell }}^{\circ}$ at $25^{\circ} C$ for the cell is $0.22\, V$. The equilibrium constant at $25^{\circ} C$ is

• A. $2.8 \times 10^{7}$
• B. $5.2 \times 10^{8}$
• C. $2.8 \times 10^{5}$
• D. $5.2 \times 10^{4}$

Answer 1

Answer: (A)

$E _{\text {cell }}^{0}=\frac{0.059}{ n } \log K _{ C }$
$0.22=\frac{0.059}{2} \log K _{ c }$
$\log K _{ c }=\frac{0.22 \times 2}{0.059}=7.457$
$K _{ C }=2.8 \times 10^{7}$