Question

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

• A. $5.15\times {10}^{-18}M$
• B. $7.59\times {10}^{-18}M$
• C. $3.14\times {10}^{-18}M$
• D. $Noneofthese$

Answer 1

The correct option is A $5.15\times {10}^{-18}M$

The redox change is :

$Zn+N{i}^{2+}\rightleftharpoons Z{n}^{2+}+NI$

Initially $\Rightarrow$ 500 0

At eq. $\Rightarrow$ a (500 - a)

$E_{cell}=E_{oxid\left(Zn1Z{n}^{2+}\right)}+E_{red\left(N{i}^{2+}1Ni\right)}$

$E_{cell}=E_{oxid\left(Zn1Z{n}^{2+}\right)}^{\ominus }+E_{red\left(N{i}^{2+}1Ni\right)}^{\ominus }+\frac{0.059}{2}log\frac{\left[N{i}^{2+}\right]}{\left[Z{n}^{2+}\right]}$

At equilibruim, $E_{cell}$ = 0

$\therefore E_{cell}=E_{oxid\left(Zn1Z{n}^{2+}\right)}^{\ominus }+E_{red\left(N{i}^{2+}1Ni\right)}^{\ominus }=-\frac{0.059}{2}log\frac{\left[N{i}^{2+}\right]}{\left[Z{n}^{2+}\right]}$

Or 0.75 + (-0.24) = $-\frac{0.059}{2}log\frac{\left[N{i}^{2+}\right]}{\left[Z{n}^{2+}\right]}$

Or $\frac{\left[N{i}^{2+}\right]}{\left[Z{n}^{2+}\right]}=antilog(-\frac{0.51\times 2}{0.059})=5.15\times {10}^{-18}$

$\therefore \frac{a}{500-a}=5.15\times {10}^{-18}$
($\because 500-a\approx 500)$

$\therefore a=500\times 5.15\times {10}^{-18}$

$\therefore \left[N{i}^{2+}\right]=\frac{a}{V}=\frac{500\times 5.15\times {10}^{-18}}{500}$

= $5.15\times {10}^{-18}M$