Question

An acidic solution of copper (II) sulphate containing some contaminations of zinc and iron (II) ions was electrolysed till all the copper is deposited. If electrolysis is further continued for sometime, the product liberated at cathode would be:
(Given : $E_{F{e}^{2+}/Fe}^o=-0.44V$ and $E_{Z{n}^{2+}/Zn}^o=-0.76V$ )

• A. $Fe$
• B. $Zn$
• C. $H_2$
• D. Alloy of $Zn$ and $Fe$

Answer 1

The correct option is C $H_2$

After deposition of $C{u}^{2+}$ ions, $H^{+}$ ion from the $H_{2}O$ molecule will be reduced at cathode as its Standard Reduction Potential is higher than of $F{e}^{2+}$ and $Z{n}^{2+}$.
As, $E_{H^{+}/H_2}^0=0V$
$E_{H^{+}/H_2}^0>E_{F{e}^{2+}/Fe}^0,E_{Z{n}^{2+}/Zn}^0$
So, on reduction of $H^{+}$, $H_2$ gas is produced at cathode.
$H^{+}\left(aq\right)+e^{-}\to \frac{1}{2}{H}_2\left(g\right)$