1.

In electrolysis of dil. H_(2)SO_(4) the ratio by weight of gases evolved at anode and cathode respectively are

Answer»

`1:16`
`8:1`
`16:1`
`3:1`

SOLUTION :In the electrolysis of DIL. `H_(2)SO_(4)` the electrolysis takes PLACE as
`H_(2)O hArr H^(+)+OH^(-)`
At anode `"" 2OH^(-) to H_(2)O+[O]+2e^(-)`
At cathode `"" 2H^(+)+2e^(-) to 2[H]`
`therefore W+[O]:W+2[H^(+)] therefore 16:2 therefore 8:1`


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