A solution of sodium sulphate was electrolyzed using some inert electrode. The product at the electrodes are

Options

A. `O_(2),H_(2)`
B. `O_(2),Na`
C. `O_(2),SO_(2)`
D. `O_(2),S_(2)O_(8)^(2-)`

Answer

a

Explanation

Since the reduction potential of `H_(2)O` is greater than reduction potential of `Na^(o+), ` so `H_(2)O` undergoes reduction to give `H_(2)(g)` at cathode.
`H_(2)O+e^(-) rarr overset(c-)(O)H+(1)/(2)H_(2)(g)`
Similarly , the oxidation potential of `H_(2)O` is greater than the oxidation potential of `SO_(4)^(2-)` ion. So, `H_(2)O` undergoes oxidation to give `O_(2)(g)` at anode.
`H_(2)O rarr 2H^(o+)+2e^(-)+(1)/(2)O_(2)(g)`