Question

Zinc granules are added in excess to 50 mL of 1 M $Ni(N{O}_3{)}_2$ solution at ${25}^o$C until the equilibrium is reached. If$E_{Z{n}^{2+}/Zn}^o$ and $E_{N{i}^{2+}/Ni}^o$ are $-0.75V$ and $-0.24V$ respectively.Find out the $\left[N{i}^{2+}\right]$ at equilibrium:

• A. $5.15\times {10}^{-18}M$
• B. $6.65\times {10}^{-18}M$
• C. $6.75\times {10}^{-18}M$
• D. None of these

Answer 1

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

The redox change is:
$Zn+N{i}^{2+}\rightleftharpoons Z{n}^{2+}+Ni$
mM before equilibrium 500 0
mM at equilibrium a (500 - a)
$\therefore E_{cell}=E_{O{P}_{Zn/Z{n}^{2+}}}=E_{R{P}_{N{i}^{2+}/Ni}}$
$\therefore E_{cell}=E_{O{P}_{Zn/Z{n}^{2+}}}=E_{R{P}_{N{i}^{2+}/Ni}}+\frac{0.059}{2}log\frac{\left[N{i}^{2+}\right]}{\left[Z{n}^{2+}\right]}$
At equilibrium $E_{cell}=0$
$\therefore E_{O{P}_{Zn/Z{n}^{2+}}}+E_{R{P}_{N{i}^{2+}/Ni}}=-\frac{0.059}{2}log\frac{\left[N{i}^{2+}\right]}{\left[Z{n}^{2+}\right]}$
or $0.75+(-0.24)==-\frac{0.059}{2}log\frac{\left[N{i}^{2+}\right]}{\left[Z{n}^{2+}\right]}$
$\frac{\left[N{i}^{2+}\right]}{\left[Z{n}^{2+}\right]}=antilog(-\frac{0.51\times 2}{0.059})=5.15\times {10}^{-18}$
$\therefore \frac{a}{500-a}=5.15\times {10}^{-18}$
$\therefore a=500\times 5.15\times {10}^{-18}$
$\therefore \left[N{i}^{2+}\right]=\frac{mM}{V}=\frac{500\times 5.15\times {10}^{-18}}{500}$
$=5.15\times {10}^{18}M$