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Question
`E_(cell)^Θ` for some half cell reactions are given below. On the basis of these mark the correct answer.
(a) \[\ce{H^{+} (aq) + e^{-} -> 1/2 H_2 (g); E^Θ_{cell} = 0.00V}\]
(b) \[\ce{2H2O (1) -> O2 (g) + 4H^{+} (aq) + 4e^{-}; E^Θ_{cell} = 1.23V}\]
(c) \[\ce{2SO^{2-}_{4} (aq) -> S2O^{2-}_{8} (aq) + 2e^{-}; E^Θ_{cell} = 1.96V}\]
(i) In dilute sulphuric acid solution, hydrogen will be reduced at cathode.
(ii) In concentrated sulphuric acid solution, water will be oxidised at anode.
(iii) In dilute sulphuric acid solution, water will be oxidised at anode.
(iv) In dilute sulphuric acid solution, \[\ce{SO4^{2-}}\] ion will be oxidised to tetrathionate ion at anode.
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Solution
(i) In dilute sulphuric acid solution, hydrogen will be reduced at cathode.
(iii) In dilute sulphuric acid solution, water will be oxidised at anode.
Explanation:
In dilute sulphuric acid solution, hydrogen will be reduced at cathode.
\[\ce{H^{+} + e^{-} -> 1/2 H2}\]
And \[\ce{H2O}\] is oxidized at node.
\[\ce{2H2O -> O2 + 4H^{+} + 4e^{-}}\]
While in concentrated solution of sulphuric acid, SO-2 ions are oxidized to tetrathionate (SO2) ions.
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