Question

An electrochemical cell is constructed by immersing a piece of copper wire in 50 ml of 0.1 M $CuS{O}_4$ solution and zinc strip in 50 ml of 0.1 M $ZnS{O}_4$ solution
[ $E_{C{u}^{+2}/Cu}^0$= 0.34 V, $E_{Z{n}^{+2}/Zn}^0$= -0.76 V ]

In a separate experiment, 50 ml of 1.4 M $N{H}_3$ is added to $CuS{O}_4$ solution. EMF of the cell is:
$K_f$($\left[Cu\right(N{H}_3{)}_4]$)=$6\times {10}^{13}$

• A. 0.993 V
• B. 1.327 V
• C. 1.467 V
• D. 0.720 V

Answer 1

The correct option is D 0.720 V

Due to complex formation, $\left[C{u}^{2+}\right]$ decreases and it can be calculated by the reaction,
$\underset{50\times 0.1}{{\mathrm{Cu}}^{2+}}+4\underset{50\times 1.4}{{\mathrm{NH}}_3}\rightleftharpoons {\left[\mathrm{Cu}\right({\mathrm{NH}}_3{)}_4]}^{2+}$
This reaction goes to completion, and then moves backwards slightly. When this happens, the copper ion concentration will be:
${\mathrm{Cu}}^{2+}=\frac{{\left[\mathrm{Cu}\right({\mathrm{NH}}_3{)}_{{}_4}]}^{2+}}{{\left[{\mathrm{NH}}_3\right]}^4\times K_f}=\frac{\frac{50\times 0.1}{100}}{{(0.50)}^4\times 6\times {10}^{13}}=13.33\times {10}^{-15}M$
$E_{\mathrm{cell}}=E°-\frac{0.0591}{2}\log \frac{0.1}{13.3\times {10}^{-15}}$
$E°=E{°}_{{\mathrm{Cu}}^{2+}/\mathrm{Cu}}-E{°}_{{\mathrm{Zn}}^{2+}/\mathrm{Zn}}=0.34-(-0.76)=1.1V$
$E_{\mathrm{cell}}=E{°}_{\mathrm{cell}}-\frac{0.0591}{2}\log \frac{0.1}{13.3\times {10}^{-15}}=0.720V$