Question

For a cell reaction involving a two-electron change, the standard emf of the cell is found to be $0.295V$ at ${25}^{o}C$. The equilibrium constant of the reaction at ${25}^{o}C$ will be :

• A. $1\times {10}^{-10}$
• B. $29.5\times {10}^{-2}$
• C. $10$
• D. $1\times {10}^{10}$

Answer 1

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

The relationship between equilibrium constant for a reaction and its equilibrium constant is $E_{cell}^0=\frac{0.0592}{n}log{K}_c.$
Substitute values in the above equation.
$0.295V=\frac{0.0592}{2}log{K}_c$
$log{K}_c=9.96$
$K_c=9.25\times {10}^9\cong {10}^{10}.$