Question

On electrolysis of dilute sulphuric acid using platinum electrodes, the product obtained at the anode will be:A ) HydrogenB ) OxygenC ) Hydrogen sulphideD ) Sulphur dioxide

Hint: During electrolysis, an electric current is passed through electrolytic solution, and a chemical reaction is carried out. Determine the different types of ions that are present in the solution. Then determine, which types of ions are attracted to anode. Then using electrode potential values, determine which ion gets preferentially discharged.

In aqueous solution, dilute sulphuric acid dissociates to form protons and sulphate ions. Also, the autoionisation of water gives protons and hydroxide ions. Thus, the dilute aqueous sulphuric acid solution contains protons, hydroxide ions and sulphate ions.

${{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4}\left( {aq} \right){\rm{ }} \to {\rm{ 2}}{{\rm{H}}^ + }\left( {aq} \right){\rm{ + SO}}_4^{2 - }\left( {aq} \right)$
${{\rm{H}}_2}{\rm{O}}\left( l \right){\rm{ }} \to {\rm{ }}{{\rm{H}}^ + }\left( {aq} \right){\rm{ + O}}{{\rm{H}}^ - }\left( {aq} \right){\rm{ }}$

Anode is a positive electrode. It attracts hydroxide ions and sulphate ions. At anode, hydroxide ions get discharged (in preference to sulphate ions) to form oxygen gas.

${\rm{4 O}}{{\rm{H}}^ - }\left( {aq} \right){\rm{ }} \to {\rm{ 2 }}{{\rm{H}}_2}{\rm{O}}\left( l \right){\rm{ + }}{{\rm{O}}_2}\left( g \right){\rm{ + 4 }}{{\rm{e}}^ - }$

The electrode potential for the above reaction is 0.0 V.
The electrode potential for the discharge of sulphate ions is 1.23 V

${\rm{SO}}_4^{2 - }\left( {aq} \right){\rm{ + 4 }}{{\rm{H}}^ + }\left( {aq} \right){\rm{ }} \to {\rm{ 2 }}{{\rm{H}}_2}{\rm{O}}\left( l \right){\rm{ + S}}{{\rm{O}}_2}\left( g \right){\rm{ + 4 }}{{\rm{e}}^ - }$

When hydroxide ions are discharged, oxygen gas is liberated at anode. However if sulphate ions are discharged, sulphur dioxide gas is liberated. Since the value of the electrode potential for the discharge of hydroxide ions is less than that of sulphate ions, the hydroxide ions will get discharged preferentially at anode. Due to this, oxygen gas will liberate in preference to sulphur dioxide gas at anode.

During the electrolysis of dilute aqueous sulphuric acid, using platinum electrodes, oxygen gas is liberated at anode.

Hence, the option B ) oxygen is the correct answer.

Note: Cathode is a negative electrode. It attracts protons. At cathode, protons get discharged to form hydrogen gas.
${\rm{2 }}{{\rm{H}}^ + }\left( {aq} \right){\rm{ + 2 }}{{\rm{e}}^ - }{\rm{ }} \to {\rm{ }}{{\rm{H}}_2}\left( g \right)$
When hydroxide ions are discharged, oxygen gas is liberated at anode. However if sulphate ions are discharged, sulphur dioxide gas is liberated. Since the value of the electrode potential for the discharge of hydroxide ions is less than that of sulphate ions, the hydroxide ions will get discharged preferentially at anode. Due to this, oxygen gas will liberate in preference to sulphur dioxide gas at anode.