Question

During the electrolysis of copper [II] sulphate solution using platinum as cathode & carbon as anode, the reactions at the cathode is :

• A. $C{u}^{2+}\left(aq\right)+2e^{-}⟶Cu\left(s\right)$
• B. $Cu\left(aq\right)\to Cu\left(s\right)+2e^{-}$
• C. $4OH\left(aq\right)+4e^{-}⟶2H_{2}O\left(l\right)+O_2\left(g\right)$
• D. $4O{H}^{-}\left(aq\right)⟶2H_{2}O\left(l\right)+O_2\left(g\right)+4e^{-}$

Answer 1

The correct option is A $C{u}^{2+}\left(aq\right)+2e^{-}⟶Cu\left(s\right)$

The half-equations for the electrolysis of copper(II) sulfate solution

The negative cathode reaction with graphite electrodes

The negative cathode electrode attracts $C{u}^{2+}$ ions (from copper sulfate) and $H^{+}$ ions (from water). Only the copper ion is discharged, being reduced to copper metal. The less reactive a metal, the more readily its ion is reduced on the electrode surface.

A copper deposit forms as the positive copper ions are attracted to the negative electrode (cathode)

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

positive ion reduction by electron gain

The blue colour of the copper ion will fade as the copper ions are converted to the copper deposit on the cathode.

The positive anode reaction with graphite electrodes

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

The negative sulphate ions ($S{O}_4^{2-}$) or the traces of hydroxide ions ($O{H}^{-}$) are attracted to the positive electrode. But the sulfate ion is too stable and nothing happens. Instead, either hydroxide ions or water molecules are discharged and oxidized to form oxygen.

$4O{H}^{-}\left(aq\right)+4e^{-}⟶2H_{2}O\left(l\right)+O_2\left(g\right)$