$E_{cell}^{o}$ for some half-cell reactions are given below.
(i) $H^{+} (aq) + e^{-} ⟶ \frac{1}{2} H_{2} (g); E_{cell}^{\circ} = 0 .00 V$
(ii) $2 H_{2} O (l) ⟶ O_{2} (g) + 4 H^{+} (aq) + 4 e^{-}; E_{cell}^{\circ} = 1 .23 V$
(iii) $2 {SO}_{4}^{2 -} (aq) ⟶ S_{2} O_{8}^{2 -} (aq) + 2 e^{-}; E_{cell}^{\circ} = 1 .96 V$
On the basis of these mark the correct answer.

Moderate

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 tetrathionate ion at anode

Solution

During the electrolysis of dilute sulphuric acid solution all the given three reactions occur 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
$\begin{array}{l} 2 {SO}_{4}^{2 -} (aq) ⟶ S_{2} O_{8}^{2 -} + 2 e^{-} E_{cell}^{\circ} = 1 .96 V \\ 2 H_{2} O (l) ⟶ O_{2} (g) + 4 H^{+} (aq) + 4 e^{-}; E_{cell}^{\circ} = 1 .23 V \end{array}$
Reaction having lower value of $E_{cell}^{o}$ will undergo faster oxidation. Hence, oxidation of water occur preferentially for dilute sulphuric acid but in case of concentrated H2SO4 former reaction (oxidation of ${SO}_{4}^{2 -}$ ) is preferred.
Reduction half-cell reaction occurs at cathode is $H^{+} (aq) + e^{-} ⟶ \frac{1}{2} H_{2} (g)$ at cathode

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