What products are expected to be obtained at the cathode and the anode in the electrolysis of an aqueous solution of `Na_(2)SO_(4)` by using Pt - electrodes?

Explanation

In aqueous solution `Na_(2)SO_(4)` dissociates completely to form `Na^(+) and SO_(4)^(2-)` ions. Thus the aqueous solution of `Na_(2)SO_(4)` contains `Na^(+), SO_(4)^(2-) and H_(2)O(l)` . When electricity is passed through the solution, probable reactions that might occur at the electrodes as follows:
Cathode reaction :
`Na^(+)(aq)+e rarr Na(s), E_("Red")^(@)=-2.17V" ...(1)"`
`2H_(2)O(l)+2e rarr H_(2)(g)+2OH^(-),E_("Red")^(@)=-0.83V" ...(2)"`
As the standard electrode potential of (2) is higher than (1). Hence `H_(2)O` will be reduced at the cathode and `H_(2)(g)` will be produced.
Anode reaction :
`2SO_(4)^(2-) (aq)rarr S_(2)O_(8)^(2-)(aq)+2e , E_("Red")^(@)=+2.0V" ...(1)"`
`2H_(2)O(l)+2e rarr 4H^(+)(aq)+O_(2)(g)+4e, E_("Red")^(@)=+1.23V" ...(2)"`
As the standard electrode potential of (2) is lower than that of (1), hence `H_(2)O` will be reduced at the anode and `O_(2)(g)` will be evolved.