Question

Zinc granules are added in excess to $500mL$ of $1.0M$ nickel nitrate solution at ${25}^{\circ }C$ until the equilibrium is reached. If the standard reduction potentials of $Z{n}^{2+}/Zn$ and $N{i}^{2+}/Ni$ are $-0.75$ and $-0.24$ volt respectively, find out the concentration of $N{i}^{2+}$ ions in solution at equilibrium.

Answer 1

The reaction to be considered is,
$Zn\left(s\right)+N{i}^{2+}(aq.)\rightleftharpoons Z{n}^{2+}(aq.)+Ni\left(s\right)$
The cell involving this reaction would be,
$Zn\left(s\right)|Z{n}^{2+}(aq.)\parallel N{i}^{2+}(aq.)|Ni\left(s\right)$
$E_{cell}^{\circ }=-0.24+0.74=0.51volt$
$\log K_{eq}=\frac{nF{E}^{\circ }}{2.303RT}=\frac{n{E}^{\circ }}{0.0591}=\frac{2\times 0.51}{0.0591}=17.25$
So, $K_{eq}=1.78\times {10}^{17}$
Let $x$ be the concentration of $N{i}^{2+}$ that have been reduced to nickel at equilibrium.
$Zn\left(s\right)+\underset{(1.0-x)}{N{i}^{2+}(aq.)}\rightleftharpoons \underset{x}{Z{n}^{2+}(aq.)}+Ni\left(s\right)$
$K_{eq}=\frac{\left[Z{n}^{2+}\right]}{\left[N{i}^{2+}\right]}=\frac{x}{(1-x)}=1.78\times {10}^{17}$
$x\approx 1.0M$
So, $(1-x)=\left[N{i}^{2x}\right]=\frac{1.0}{1.78\times {10}^{17}}=5.6\times {10}^{-18}M$.