Question

Question: Electrolysis of aqueous$CuS{O_4}$with inert electrodes gives:
(A) $Cu$at cathode, anode gets dissolved. $$$$
(B) $Cu$at cathode, ${O_2}$at anodes.
(C) ${O_2}$at anode, cathode.
(D) ${O_2}$at anode, cathode gets dissolved.

Answer 1

Electrolysis may be defined as decomposition of electrolyte by passage of electricity. Moreover when one electrode is inert, it does not take part in chemical reaction. It only acts as a sink for electrons.

Complete step by step answer:**
AS we know electrolysis may be defined as a process of decomposition of an electrolyte by the passage of electricity through its aqueous solution or molten (fused) stats. The apparatus used to bring about electrolysis is called an electrolytic cell. In electrolytic cells electrical energy is used to bring about a chemical reaction. It consists of a glass vessel containing the electrolyte and two metal rods are dipped into it. These rods are called electrolyte and two metal rods are dipped into it. These rods are called electrodes and are connected to a source of electricity i.e. battery. The electrode connected to the negative pole of the battery is called cathode and the other connected to the positive pole of battery is called anode. The mechanism of electrolysis is as follows. On passing electric current, the ions of the electrolyte move towards the oppositely charged electrodes and get discharged cation move towards the cathode and undergo reduction while anion move towards the anode and undergo oxidation there become neutral. Product of electrolysis depends on the nature of material being electrolysed and the type of electrodes being used. If an electrode is inert, it does not participate in chemical reaction and only acts only as a source or sink for electrons. It should be noted that ions with lower discharge potential are discharged in preferences to those which have high discharged potentials.
When Electrolysis of aqueous copper sulphate solution using inert electrons, copper sulphate and water ionize as under:
$\begin{gathered} \\\ \\\ \end{gathered}$ $CuS{O_4}(aq) \rightleftharpoons C{u^{2 + }}(aq) + S{O_4}^{2 - }(aq)$
& ${H_2}O(l) \rightleftharpoons {H^ + }(aq) + O{H^ - }(aq)$
On passing electricity, $C{u^{2 + }}(aq)$and ${H^ + }(aq)$move towards the cathode while $S{O_4}^{2 - }$ions and $O{H^ - }$ ions move towards.
At cathode: Since the discharge potentials of $C{u^{2 + }}$ ions is lower than that of ${H^ + }$ ions, therefore $C{u^{2 + }}$ ions deposited on platinum electrode.
$C{u^{2 + }} + 2{e^ - } \to 2Cu$
At anode: The discharge potential of $O{H^ - }$ions is lower than that of $S{O_4}^{2 - }$ions and ${O_2}$(g) is liberated.
I.e. $4O{H^ - } \to 4OH + 4{e^ - }{\text{ (Primary Change)}}$
$4OH \to 2{H_2}O(l) + {O_2}(g){\text{ (Secondary Change)}}$
Hence we get $Cu$ at cathode and ${O_2}$at anode on electrolysis of aqueous $CuS{O_4}$ with inert electrodes.

So, the correct answer is Option B .

Note:
Some electrochemical processes, though feasible., are kinetically so slow that at lower voltage, they do not seem to occur, Hence in such cases extra voltage is required. This extra voltage required is called overvoltage. Overvoltage is defined as the additional voltage required to cause electrolysis.