Question

Which substance is obtained in the solution on electrolysis aqueous $CuSO_4$ solution using graphite electrode?
(A) ${H_2}O$
(B) ${H_2}S{O_4}$
(C) $N{a_2}S{O_4}$
(D) $Cu{(OH)_2}$

Answer 1

Electrolysis, process by which electric current is passed through a substance to effect a chemical change. The process is carried out in an electrolytic cell, an apparatus consisting of positive and negative electrodes held apart and dipped into a solution containing positively and negatively charged ions.

Complete answer:
Electrolysis is a technique that uses direct electric current to drive an otherwise non-spontaneous chemical reaction. Electrolysis is commercially important as a stage in the separation of elements from naturally occurring sources such as ores using an electrolytic cell. The voltage that is needed for electrolysis to occur is called the decomposition potential.
The key process of electrolysis is the interchange of atoms and ions by the removal or addition of electrons due to the applied current. The desired products of electrolysis are often in a different physical state from the electrolyte and can be removed by physical processes
At anode:
$2{H_2}O(I) \to {O_2}(g) + 4H + (aq) + 4{e^ - }$
At cathode:
$2C{u_2} + (aq) + 4e \to 2Cu(s)$
Net electrolysis reaction is
$2C{u_2}(aq) + 2{H_2}O(l) \to 2Cu(s) + 4H + (aq) + {O_2}(g)$
So, the correct answer is (A) ${H_2}O$ .

Note:
In electrolysis, Each electrode attracts ions that are of the opposite charge. Positively charged ions (cations) move towards the electron-providing (negative) cathode. Negatively charged ions (anions) move towards the electron-extracting (positive) anode. In this process electrons are effectively introduced at the cathode as a reactant and removed at the anode as a product. In chemistry, the loss of electrons is called oxidation, while electron gain is called reduction.