Question

Electrode potential of cadmium is $-0.4$V and electrode potential of chromium $-0.74$V. $\left[C{d}^{2+}\right]=0.1$M and $\left[C{r}^{3+}\right]=0.01$M. Calculate the equilibrium constant at $298K$.

Answer 1

Overall cell reaction is:

$3C{d}_{\left(aq\right)}^{2+}+2C{r}_{\left(s\right)}\to 3C{d}_{\left(s\right)}+2C{r}_{\left(aq\right)}^{3+}$

Equilibrium constant, $K_{eq}=\frac{\left[C{r}^{3+}{]}^2\right[Cd\left(s\right){]}^3}{\left[C{d}_{\left(aq\right)}^{2+}{]}^3\right[C{r}_{\left(s\right)}{]}^2}$

$=\frac{(0.01{)}^2\times 1^3}{(0.1{)}^3\times 1^2}$ [As $\left[C{d}_{\left(s\right)}\right]=1$ and $\left[C{r}_{\left(s\right)}\right]=1$]

At 298K, $K_{eqm}=0.1$