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 ]

The emf of the cell increases when small amount of concentrated $N{H}_3$ is added to:

• A. $ZnS{O}_4$ solution
• B. $CuS{O}_4$ solution
• C. Both $ZnS{O}_4$ and $CuS{O}_4$
• D. Can't say

Answer 1

The correct option is A $ZnS{O}_4$ solution

When $N{H}_3$ is added to $ZnS{O}_4$ solution, $N{H}_3$ reacts with $Z{n}^{2+}$ in the following manner:
$Z{n}^{2+}+4{\mathrm{NH}}_3\rightleftharpoons {\left[\mathrm{Zn}\right({\mathrm{NH}}_3{)}_4]}^{2+}$
i.e., [$Z{n}^{2+}$] decreases
In the equation
$E=E°-\frac{0.0591}{2}\log \frac{\left[{\mathrm{Zn}}^{2+}\right]}{\left[{\mathrm{Cu}}^{2+}\right]}$
If [$Z{n}^{2+}$] decreases then $\log \frac{\left[{\mathrm{Zn}}^{2+}\right]}{\left[{\mathrm{Cu}}^{2+}\right]}$ decreases .Hence, EMF of cell increases.