Question

E.M.F of $Ni\left(s\right)|N{i}^{2+}\left(aq\right)||C{u}^{2+}\left(aq\right)|Cu\left(s\right)$ cell can be increased by:

• A. adding $N{H}_3$ in the right half-cell
• B. increasing the conc. of $N{i}^{2+}$ ions
• C. adding dimethyl glyoxime into the left half- cell
• D. changing the electrolyte present in salt bridge.

Answer 1

Answer: (C)

adding dimethyl glyoxime into the left half- cell

$Ni|N{i}^{2+}\parallel C{u}^{+2}|Cu$

Cathode: $C{u}^{2+}+2e^{-}⟶Cu$

Anode: $Ni⟶N{i}^{2+}+2e^{-}$

$C{u}^{2+}+Ni⟶N{i}^{2+}+Cu$

$E_{cell}=E_{cell}^o-\frac{0.0591}{2}\log \frac{\left[N{i}^{2+}\right]}{\left[C{u}^{2+}\right]}$

If we add $DMG$ in left-half cell, then $DMG$ will bind to $N{i}^{2+}$ and decrease the concentration of free $N{i}^{2+}$ in solution.

Thus $\frac{0.059}{2}\log \frac{\left[N{i}^{2+}\right]}{\left[C{u}^{2+}\right]}$ term also decrease; result in increase in $E_{cell}$ .