ECell° of some half cell reactions are given below. 

I: \(\small\mathrm{H}^{+} (a q)+e^{-} \rightarrow \frac{1}{2} \mathrm{H}_2 (g )\quad ; \quad E_{\text {cell }}^0=0.00 \mathrm{~V} \)
II: \(\small2 \mathrm{H}_2 \mathrm{O(l)} \rightarrow O_2 (g)+4 \mathrm{H}^{+} (a q)+4 e^{-} ; E_{\text {cell }}^{0}=1.23 \mathrm{~V} \)
III: \(\small2 \mathrm{SO}_4^{2-} (a q) \rightarrow \mathrm{S}_2 \mathrm{O}_8^{2-} (a q)+2 e^{-}, E_{\text {cell }}^{0}=1.96 \mathrm{~V} \)

The correct statements among the following are:

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 a dilute sulphuric acid solution, water will be oxidised at anode.
d. In dilute sulphuric acid solution, SO42- ion will be oxidised to tetrathionate ion at anode.


1. (a, b)
2. (b, c)
3. (c, d)
4. (a, c)

Hint: Higher the reduction potential, the higher will be oxidising power


During the electrolysis of dilute sulphuric acid above given three reactions occur each of which represents a particular reaction either oxidation half cell reaction or reduction half cell reaction.
Oxidation half cell reactions that occur at the anode are as follows
\(2SO^{2-}_4 aq \rightarrow S_2O^{2-}_8 + 2e^- ~~~~~~~~E^\circ _{cell} = 1.96 V\\ 2 \mathrm{H}_2 \mathrm{O(l)} \rightarrow O_2 (g)+4 \mathrm{H}^{+} (a q)+4 e^{-} ; E_{\text {cell }}^{\ominus}=1.23 \mathrm{~V} \)
Reaction having a lower value of ECell° will undergo faster oxidation.
Hence, oxidation of water occur preferentially reduction half cell reaction occurs at cathode
H+aq+e-12H2 gAt Cathode
Hence, options (a) and (c) are correct