Question

A beaker contains an acidified copper sulphate solution. A copper plate and a carbon rod are kept in this copper sulphate solution. The copper plate is connected to the positive terminal of a battery whereas the carbon rod is connected to the negative terminal of the battery. What will you observe when an electric current is passed through this set-up for a considerable time?

Answer 1

Electrolysis of copper sulphate using carbon electrodes:
The negative cathode electrode which is connected to the carbon rod attracts $C{u}^{2+}$ ions.

(-) $C{u}^{2+}$ (aq) + 2 $e^{-}$ $\to$ $C{u}_{\left(s\right)}$ (copper deposit)

The negative sulphate ions $(S{O}_4{)}^{2-}$ (from copper sulfate) or the traces of hydroxide ions $O{H}^{-}$ (from water) are attracted to the positive electrode. But both the sulfate ion and hydroxide ion are too stable and nothing happens to them because the copper anode is preferentially oxidized to discharge $C{u}^{2+}$ copper ions.

An oxidation electrode reaction at the positive anode

(+) $C{u}_{\left(s\right)}$ – 2 $e^{-}$ $\to$ $C{u}^{2+}$ _(aq) (copper dissolves, oxidation - 2 electrons lost)

So copper is deposited on a carbon rod and some copper is dissolved from the copper plate