1.

Electrolysis is an important technique for extraction of metals and each low of the solution needs a minimum voltage to get discharged and this value is expressed in terms of discharge potential. For some metal ions the discharge potentials follow the order give below : Li^(+) gt K^(+) gt Ca^(2+) gt Na^(+) gt Mg^(2+) gt Al^(3+) gt Zn^(2+) gt Fe^(2+) gt Ni^(2+) gt H_(3)O^(+) gt Cu^(2+) gt Hg_(2)^(2+) gt Ag^(+) gt Au^(3+) For some anions the discharge potentials are in the order: SO_(4)^(2-) gt NO_(3)^(-) gt OH^(-) gt Br^(-) gt I^(-) The product formed at anode and cathode, when dilute H2.So4 is electrolysed are

Answer»

`SO_(2),H_(2)`
`SO_(3),H_(2)`
`H_(2)S_(2)O_(8)`
`O_(2),H_(2)`

Solution :`H_(2)SO_(4) to 2H^(+)+SO_(4)^(2-)`
At anode `2H_(2)O to O_(2)+4H^(+)+4e^(-), "At Cathode" , 2H^(+)+2e^(-) to H_(2)`


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