Question

The equilibrium constant for the reaction $C{d}^{2+}\left(aq\right)+4N{H}_3\left(aq\right)\rightleftharpoons Cd\left(N{H}_3{)}_4^{2+}\right(aq)$ is $2\times {10}^x$. Find the value of $\frac{(x+1)}{5}$.
(Given: $E_{C{d}^{2+}|Cd}^{\circ }=-0.04V;E_{Cd(N{H}_3{)}_4^{2+}|Cd}^{\circ }=-0.61V$)

Answer 1

$E_{cell}^0=E_{C{d}^{2+}|Cd}^{\circ }-E_{Cd(N{H}_3{)}_4^{2+}|Cd}^{\circ }=-0.04V-(-0.61V)=0.57V$

$E_{cell}^0=\frac{0.059}{n}logK$

Substitute values in the above expression.

$0.57V=\frac{0.059}{2}logK$

$logK=19.32$

$K=2\times {10}^{19}$

But $K={10}^x$.

Hence, $x=19$.

$\therefore \frac{x+1}{5}=4$.