1.

How many coulombs are required for the oxidation of 0.01 mol of H_(2)O to O_(2) ?

Answer»

`1.93 xx 10^(3) C`
`9.65 xx 10^(2) C`
`3.86 xx 10^(5) C`
`4.285 xx 10^(2) C`

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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