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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. |
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Answer» 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. |
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