1.

During an electrolysis of conc. H_(2)SO_(4) perdisulphuric acid (H_(2)S_(2)O_(6)) and O_(2) form in equimolar amount. The amount of H_(2) than will form simultaneously at other electrode will be (2H_(2)SO_(4)toH_(2)S_(2)O_(8)+2H^(+)+2e^(-))

Answer»

thrice that of `O_(2)` in moles
twice that of `O_(2)` in moles
equal to that of `O_(2)` in moles
half of the of `O_(2)` in moles

Solution :Anode `{{:(2H_(2)SO_(4)toH_(2)S_(2)O_(8)+2H^(+)+2e^(-)),(2H_(2)OtoO_(2)+4H^(+)+4e^(-)):}}`
`underline(" Cathode")""{2H_(2)OtoH_(2)+2OH^(-)-2e^(-)}3)`.
Net: `2H_(2)SO_(4)+8H_(2)OtoH_(2)S_(2)O_(8)+O_(2)+3H_(2)+6H^(+)+6OH^(-)`
HENCE ratio of `n_(O_(2))` and `n_(H_(2))` is `1:3`


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