Please help me with this question's solution

Copper is purified upto 99.95% purity using electrolytic refining method. An electrolytic cell is prepared to do this refining .Diagram shown below is an electrolytic cell used for copper refining.

In such a cell a thin sheet of high-purity Cu serves as the cathode, and the anode is the impure Cu which is to be refined. The electrolyte is a solution of copper(II)sulfate. When current is passed in system the Cu+ moves towards the cathode where it gains electron and turn as metallic Cu. At anode impure copper release electrons go into solution as Cu2+.

At the cathode,  $C{u^{2+}}_{\left(aq\right)} +\hspace{0.17em}2e^{-} \to Cu$
 At the anode, $C{u}_{\left(S\right)} \to C{u^{2+}}_{\left(aq\right)} +\hspace{0.17em}2e^{-}$
 In this way pure copper is obtained as cathod rod at the end of the process.

Hence, the correct option is (c).

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