Ncert Exercises & Solutions

$E_{Cell}^{°}$ of some half cell reactions are given below.

I: $\mathrm{H}^{+} (a q)+e^{-} \rightarrow \frac{1}{2} \mathrm{H}_2 (g )\quad ; \quad E_{\text {cell }}^0=0.00 \mathrm{~V} $
II: $2 \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: $2 \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, SO₄^2- 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

E_{Cell}^{°}

will undergo faster oxidation.

Hence, oxidation of water occur preferentially reduction half cell reaction occurs at cathode

H^{+} (aq) + e^{-} ⟶ \underset{At Cathode}{\frac{1}{2} H_{2} (g)}

Hence, options (a) and (c) are correct

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