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

Explanation

To solve the question regarding the half-cell reactions and their potentials, we will analyze each option based on the provided half-cell reactions and their standard reduction potentials.

### Step-by-Step Solution:

1. **Understanding Half-Cell Reactions**:
- The half-cell reactions provided are:
- (i) \( H^+(aq) + e^- \rightarrow \frac{1}{2} H_2(g) \) with \( E^\circ_{cell} = 0.00V \)
- (ii) \( 2H_2O(l) \rightarrow O_2(g) + 4H^+(aq) + 4e^- \) with \( E^\circ_{cell} = 1.23V \)
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