Question

The standard emf of a galvanic cell involving cell reaction with n = 2 is found to be 0.295 V at ${25}^{o}C$. The equilibrium constant of the reaction would be
(Given, $F=96500Cmo{l}^{-1},R=8.314J{K}^{-1}mo{l}^{-1}$ )

• A. $2.0\times {10}^{11}$
• B. $4\times {10}^{12}$
• C. $1\times {10}^2$
• D. $1\times {10}^{10}$

Answer 1

The correct option is D $1\times {10}^{10}$

For a cell reaction in equilibrium at 298 K

$E_{Cell}^o=\frac{0.0591}{n}log{K}_c$

$\Rightarrow log{K}_c=\frac{E_{cell}^o\times n}{0.0591}$

Substitute the given values of $E_{cell}^o$ and n in the above equation, we get
$log{K}_C=\frac{0.295\times 2}{0.0591}$
$log{K}_C=10\Rightarrow K_C=1\times {10}^{10}$