Question

For a cell reaction involving two electron change, the standard $EMF$ of the cell is $0.295V$ at ${25}^{\circ }C$. The equilibrium constant of the reaction at ${25}^{\circ }C$ will be :

• A. $29.5\times {10}^{-2}$
• B. $10$
• C. $1\times {10}^{10}$
• D. $2.95\times {10}^{-10}$

Answer 1

Answer: (C)

$1\times {10}^{10}$

Using the relation,

$E_{cell}^{\circ }=\frac{2.303RT}{nF}\log K_c=\frac{0.0591}{n}\log K_c$

$\therefore 0.295V=\frac{0.0591}{2}\log K_c$

or $\log K_c=\frac{2\times 0.295}{0.0591}=10$

or $K_c=1\times {10}^{10}$

Hence, option C is correct.