Potential for some half cell reactions are given below. On the basis of these mark the correct answer.
(i) `H^(+)(aq) + e^(-) rightarrow (1/2)H^(2)(g)` `E_(cell)^(@) = 0.00V`
(ii) `2H_(2)O(l) rightarrow O_(2)(g) + 4H^(+)(aq) + 4e^(-)`, `E_(cell) = 1.23V`
(iii) `2SO_(4)^(2-)(aq) rightarrow S_(2)O_(8)^(2-)(aq) + 2e^(-), E_(cell)^(@) = 1.96V`

Options

A. In dilute sulphuric acid solution, hydrogen will be reduced at cathode
B. In concentrated sulphuric acid solution, water will be oxidised at anode
C. In dilute sulphuric acid solution water will be oxidised at anode
D. In dilute sulphuric acid solution `SO_(4)^(2-)` ion will be oxidised to tetrahionate ion at anode.

Explanation

During the electrolysis of dilute sulphuric acid above given three reaction occurs each of which represents particular reaction either oxidation half cell reaction or reduction half cell reaction.
Oxidation half cell reactions occur at anode are as follows
`2SO_(4)^(2-)(aq)rarrS_(2)O_(8)^(2-)+2e^(-) E_("cell")^(@)=1.96V`
`2H_(2)O^(+)(l)rarrO_(2)(g)+4H^(+)(aq)+4e^(-) E_("cell")^(@)=1.23V`
Reaction having lower value of `E_("cell")^(@)` will undergo faster oxidation.
Hence, oxidation of water occur preferentially reduction half cell reaction occurs at cathode
Hence, oxidation of water occur preterentially reduction half cell reaction occurs at cathode
`H^(+)(aq)+e^(-)rarr(1)/(2) H_(2 )(g)`
at cathode
Hence option (a) aned (b) are correct.