Ecellofor some $$half-cell$$ reactions are given below.(i) $$H+(aq)+e$$−⟶$$12H2(g)$$;Ecell ∘$$=0.00V(ii)$$ $$2H2O(l)$$⟶$$O2(g)+4H+(aq)+4e$$−;Ecell∘$$=1.23V(iii)$$ $$2SO42$$−(aq)⟶$$S2O82$$−$$(aq)+2e$$−; Ecell∘$$=1.96V$$ On the basis of these mark the correct answer.

Question

Ecellofor some $$half-cell$$ reactions are given below.(i) $$H+(aq)+e$$−⟶$$12H2(g)$$;Ecell ∘$$=0.00V(ii)$$ $$2H2O(l)$$⟶$$O2(g)+4H+(aq)+4e$$−;Ecell∘$$=1.23V(iii)$$ $$2SO42$$−(aq)⟶$$S2O82$$−$$(aq)+2e$$−; Ecell∘$$=1.96V$$    On the basis of these mark the correct answer.

Infinity Learn · Accepted Answer

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                 $$2SO42$$−(aq)⟶$$S2O82$$−$$+2e$$−Ecell∘$$=1.96V2H2O(l)$$⟶$$O2(g)+4H+(aq)+4e$$−;Ecell ∘$$=1.23VReaction$$ having lower value of Ecello will undergo faster oxidation. Hence, oxidation of water occur preferentially for dilute sulphuric acid but in case of concentrated $$H2SO4$$ former reaction (oxidation$$ of $$SO42-) is preferred. Reduction $$half-cell$$ reaction occurs at cathode is $$H+(aq)+e$$−⟶$$12H2(g)$$     at cathode

Ecellofor some half-cell reactions are given below.(i) H+(aq)+e−⟶12H2(g);Ecell ∘=0.00V(ii) 2H2O(l)⟶O2(g)+4H+(aq)+4e−;Ecell∘=1.23V(iii) 2SO42−(aq)⟶S2O82−(aq)+2e−; Ecell∘=1.96V On the basis of these mark the correct answer.

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