Question

The electrolytic decomposition of dilute sulphuric acid with platinum electrode in cathodic reaction is,
A. Oxidation
B. Reduction
C. Oxidation and reduction
D. Neutralisation

Answer 1

Hint: A process in which the passing of electricity takes place through the electrolyte and a chemical reaction occurs is termed electrolysis. There is the presence of different types of ions in the solution. Here, we have to determine the ions present in the solution and then we have to decide which ion is attracted to which electrode.

Complete Step by Step Solution:
In aqueous solution, the dissociation of dilute sulphuric acid to sulphate ions and hydrogen ions takes place. Also water gives hydroxide ions and protons. The chemical reactions are:
\[{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}\left( {aq} \right) \to 2{{\rm{H}}^ + }\left( {aq} \right) + {\rm{S}}{{\rm{O}}_{\rm{4}}}^{2 - }\left( {aq} \right)\]
\[{{\rm{H}}_{\rm{2}}}{\rm{O}}\left( l \right) \to {{\rm{H}}^ + }\left( {aq} \right) + {\rm{O}}{{\rm{H}}^ - }\left( {aq} \right)\]

As anode is a positive electrode, it attracts sulphate or hydroxide ions towards itself. The discharge of hydroxide ions at anode liberates oxygen gas and the discharge of sulphate ions at anode forms sulphur dioxide gas. But electrode potential for discharge of hydroxide ion is less than the sulphate ions. So, preferential discharge of hydroxide ion occurs. Therefore, liberation of oxygen gas takes place at anode.
\[4{\rm{O}}{{\rm{H}}^ - }\left( {aq} \right) \to 2{{\rm{H}}_{\rm{2}}}{\rm{O}}\left( l \right) + {{\rm{O}}_{\rm{2}}}\left( g \right) + 4{e^ - }\]

As we know, cathode is a negative electrode. Therefore, protons are attracted towards the cathode. So, the discharge of protons at the cathode forms hydrogen gas. The chemical reaction is,
\[2{{\rm{H}}^ + }\left( {aq} \right) + 2{e^ - } \to {{\rm{H}}_{\rm{2}}}\left( g \right)\]
As the cathodic reaction accepts electrons, the reaction is a reduction reaction.
Hence, option B is right.

Note: In the electrochemical cells, platinum is used because of its nature to resist oxidation. It does not undergo chemical reactions easily. So, it will not become part of the redox reactions occurring at the electrochemical cells.