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E^{\Theta }_{cell} for some half cell reactions are given below. Based on these mark the correct answer.

(i) H^{+}(aq) + e^{-} \rightarrow \frac{1}{2}H_{2}(g) ; E^{o}_{cell} = 0.00V

(ii) 2H_{2}O (l) \rightarrow O_{2} (g) + 4H^{+}(aq) + 4e^{-}; E^{o}_{cell} = 1.23V

(iii) 2SO_{4}^{2-} (aq) \rightarrow S_{2}O_{8}^{2-} (aq) + 2e^{-} ; E^{o}_{cell} = 1.96V

(i) In dilute sulphuric acid solution, hydrogen will be reduced at the cathode.

(ii) In concentrated sulphuric acid solution, water will be oxidised at the anode.

(iii) In dilute sulphuric acid solution, water will be oxidised at the anode.

(iv) In dilute sulphuric acid solution, SO_{4}^{2-} ion will be oxidised to tetrathionate ion at the anode.

Answers (1)

The answer is the option (ii, iii) In dilute sulphuric acid solution -
Reduction at cathode: H^++e^-\rightarrow \frac{1}{2} H_2

Oxidation at anode : 2H_2 O \rightarrow O_2+4H^++4e^-

In concentrated sulphuric acid solution, sulphate (SO{_{4}}^{2-}) ions oxidize to form tetrathionate (S_{2}O{_{8}}^{2-}) ions.

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