Question

Electrolysis of aqueous with $CuS{O_4}$ inert electrodes gives:
A) $Cu$ at cathode, anode gets dissolved
B) $Cu$ at cathode, ${O_2}$ at anode.
C) ${O_2}$ at anode, ${H_2}$ at cathode
D) ${O_2}$ at anode, cathode gets dissolved.

Answer 1

Hint: It is a question of product electrolysis for cells, we know that when an electrochemical cell is formed that one part gets to the cathode and other part gets to anode and gives appropriate gases after electrolysis. In this question a cell setup is described where copper sulphate is in aqueous medium and with inert electrodes, which do not participate in reaction.

Complete step-by-step answer:
Let’s try to understand firstly by the diagrammatic representation in which we have a vessel having copper sulphate in it with water that’s why aqueous conditions are given. After that the electrodes used are inert. It means that they didn’t react or involve in reaction so, we have two types of species in the reaction.

In the reaction, we have competition between copper ions and hydrogen ions at cathode because these ions get attracted towards cathode. Similarly we have sulphate ion and hydroxide ion at anode because these ions get attracted towards the anode.
$CuS{O_4} \to C{u^{ + 2}}\, + \,SO_4^{2 - }\,$
${H_2}O\, \to \,{H^ + }\, + \,{}^ - OH$

At cathode, two cations move towards it and these are $C{u^{ + 2}}\,$ and ${H^ + }$ so here reduction potential of $C{u^{ + 2}}\,$ is more so it will get reduced while ${H^ + }$ get evolved as gas. At anode, there are two anions which will get attracted towards anode, $SO_4^{2 - }\,$ and ${}^ - OH$ here oxidation potential of $SO_4^{2 - }\,$ is more but the reaction is very slow hence, ${}^ - OH$ get attracted towards at anode and evolve as gas ${O_2}$ .

Hence the correct answer is option ‘C’.

Note: The ${H^ + }$ and ${}^ - OH$ ions are come from the aqueous medium, it means from water ${H_2}O\, \to \,{H^ + }\, + \,{}^ - OH$ so they will also participate. There are certain examples in which use of electrodes takes place which are not inert in nature. So at that time the electrode also participates in reaction and also gets reduced or oxidized.