Question

If you pass current through copper sulphate solution, copper gets deposited on the plate connected to the ___________terminal of the battery.

Answer 1

Electrolysis

• Electrolysis is the process of dissociation of salt into ionic forms with the help of electricity.
• In the electrolysis of copper sulphate solution, the electrolyte copper sulphate produces a high concentration of copper ions and sulphate ions.
• The electrode used in the electrolysis is a platinum electrode.
• The Copper sulphate dissociates into ions as:

$\underset{\mathrm{Copper}\mathrm{sulphate}}{{\mathrm{CuSO}}_4\left(\mathrm{aq}\right)}\to \underset{\mathrm{Copper}\mathrm{ion}}{{\mathrm{Cu}}^{2+}}\left(\mathrm{aq}\right)+\underset{\mathrm{Sulphate}\mathrm{ion}}{{{\mathrm{SO}}_4}^{2-}\left(\mathrm{aq}\right)}$

• On passing a current through copper sulphate solution different reactions occurs at the cathode and anode.
• There are tiny concentrations of hydrogen ions${\mathrm{H}}^{+}$ and hydroxide ions $\left({\mathrm{OH}}^{–}\right)$ from the self-ionization of water, as the electrolytic solution is aqueous in nature.

$\underset{\mathrm{Water}}{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}\rightleftharpoons \underset{\mathrm{Hydrogen}\mathrm{ion}}{{\mathrm{H}}^{+}\left(\mathrm{aq}\right)}+\underset{\mathrm{Hydroxyl}\mathrm{ion}}{{\mathrm{OH}}^{–}\left(\mathrm{aq}\right)}$

• Reaction at the cathode( Negative terminal)

${\mathrm{Cu}}^{2+}\left(\mathrm{aq}\right)+2{\mathrm{e}}^{-}\to \mathrm{Cu}\left(\mathrm{s}\right)$

Copper ions accept 2 electrons to form copper which gets deposited at the cathode.

• Reaction at the anode( Positive terminal)

$$\begin{gathered} 4\underset{\mathrm{Hydroxyl}\mathrm{ion}}{{\mathrm{OH}}^{–}\left(\mathrm{aq}\right)}–4{\mathrm{e}}^{–}\to \underset{\mathrm{Water}}{2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}+\underset{\mathrm{Oxygen}}{{\mathrm{O}}_2(\mathrm{g}}) \\ \underset{\mathrm{Water}}{2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}–4\mathrm{e}{}^{-}\to \underset{\mathrm{Hydrogen}\mathrm{ion}}{4{\mathrm{H}}^{+}\left(\mathrm{aq}\right)}+\underset{\mathrm{Oxygen}}{{\mathrm{O}}_2\left(\mathrm{g}\right)} \end{gathered}$$

Oxygen gas is formed at the positive electrode, an oxidation reaction (electron loss).

The negative sulfate ions $\left({{\mathrm{SO}}_4}^{2-}\right)$ are attracted to the positive electrode, but due high oxidation potential of the hydroxyl ion, it gets oxidized and the sulphate ion being stable remains in the solution.

• On passing current in copper sulphate solution using a copper electrolyte, copper metal gets deposited on the negative terminal of the battery as mentioned above.

Therefore, if you pass current through copper sulphate solution, copper gets deposited on the plate connected to the negative terminal of the battery.