Question

The following reaction takes place at $298K$ in an electrochemical cell involving two metals $A$ and $B$, $A^{2+}(aq.)+B\left(s\right)\to B^{2+}(aq.)+A\left(s\right)$. With $\left[A^{2+}\right]=4\times {10}^{-3}M$ and $\left[B^{2+}\right]=2\times {10}^{-3}M$ in the respective half-cells, the cell EMF is $1.091V$. The equilibrium constant of the reaction is closest to:

• A. $4\times {10}^{36}$
• B. $2\times {10}^{37}$
• C. $2\times {10}^{34}$
• D. $4\times {10}^{37}$

Answer 1

The correct option is A $4\times {10}^{36}$

$E=E_{cell}^o-\frac{RT}{nF}\log \frac{\left[B^{2+}\right]}{\left[A^{2+}\right]}$

$1.091=E_{cell}^o-\frac{0.0591}{2}\log \frac{2\times {10}^{-3}}{{4\times 10}^{-3}}$

$E_{cell}^o=1.082V$

At equlibirium,

$E_{cell}^0=\frac{0.0591}{2}\log K$

On solving we get,

$K=4.16\times {10}^{36}\approx 4\times {10}^{36}$

Option A is correct.