Question

During electrolysis of conc. \( H _{2} SO _{4} \), perdisulphuric acid \( \left( H _{2} S _{2} O _{8}\right) \), and \( O _{2} \) form in equimolar amount. The amount of \( H _{2} \) that will form simultaneously will be : 

(A) Thrice that of \( O _{2} \) in moles. 

(B). Twice that of \( O _{2} \) in moles. 

(C). Equal to that of \( O _{2} \) in moles. 

(D). Half of that of \( O _{2} \) in moles.

Answer 1

Correct option is (A) Thrice that of O₂ in moles

At anode - 2H₂SO₄ \(\longrightarrow\) H₂S₂O₈ + 2H⁺+ 2e^-

                     2H₂O \(\longrightarrow\) O₂ + 4H⁺ + 4e^-

At Cathode- (2H₂O \(\longrightarrow\) H₂ + 2OH^- - 2e^- x 3

Net reaction- 2H₂SO₄ + 8H₂O \(\longrightarrow\) H₂S₂O₈ + O₂ + 3H₂ + 6H⁺ + 6OH^-

Hence, ratio of moles of O₂ and H₂ is 1 : 3

Hence, The amount of H₂ that will form simultaneously will be Thrice that of O₂ in moles.