InterviewSolution
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InterviewSolution
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During an electrolysis of conc `H_(2)SO_(4)` , perdisulphuric acid `(H_(2)S_(2)_(8))` and `O_(2)` are formed in equimolar amount. The moles of `H_(2)` that will be formed simultaneously will be `a. `Thrice that of `O_(2)" "b.` Twice that of `O_(2)` `c.` Equal to that of `O_(2)." "d.` Half of that of `O_(2)` |
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Answer» `a.` This is a special case of electrolysis where two products are being obtained at anode . At anode `:` `4 overset(-)(O)H rarrO_(2)+2H_(2)O+4e^(c-) ,,,,,,,...(i)` `2SO_(4)^(2-) rarr S_(2)O_(8)^(2-)+2e^(-)" "....(ii)` `1 mol O_(2)` requires `4F` electricity and `1 mol S_(2)O_(8)^(2-)(-=H_(2)S_(2)O_(8))` requires `2F` electricity. So, if `x mol `of `O_(2)` are being produced, electricity being passed at anode is `:` `4x( f o r O_(2))+2x(3 f o r S_(2)O_(8)^(2-))=6xF` At cathode `:` `2H^(o+)+2e^(-) rarr H_(2)" "......(iii)` `2F` electricity `-=1 mol H_(2)` is produced `implies 6xF` electricity `-=1 mol H_(2)` is produced `implies ` Moles of `H_(2)` produced at cathode `=3 mol `of `O_(2)` produced at anode. |
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