In electrolysis of dilute \[{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}\] using platinum electrodes
A. \[{{\rm{H}}_{\rm{2}}}\] is evolved at cathode
B. \[{\rm{N}}{{\rm{H}}_{\rm{3}}}\] is produced at anode
C. \[{\rm{C}}{{\rm{l}}_{\rm{2}}}\] is obtained at cathode
D. \[{{\rm{O}}_{\rm{2}}}\] is produced

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 decide which ion is attracted to which electrode.

Complete Step by Step Solution:
In an 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,
\[2{{\rm{H}}^ + }\left( {aq} \right) + 2{e^ - } \to {{\rm{H}}_{\rm{2}}}\left( g \right)\]
Therefore, liberation of hydrogen gas (\[{{\rm{H}}_{\rm{2}}}\] ) takes place at cathode.
Hence, option A is right.

Note: The advantages of using a platinum electrode in a hydrogen cell is because of its inert nature that is resistance to corrosion,the ability of the platinum to catalyse reduction of proton and requirement of a high intrinsic exchange current density for reduction of a proton in platinum.