\( E_{\mathrm{cell}}^{\ominus} \) for some half cell reactions are ...
0\( E_{\mathrm{cell}}^{\ominus} \) for some half cell reactions are given below. On the basis of these mark the correct answer.
(a) \( \mathrm{H}^{+}(\mathrm{aq})+\mathrm{e}^{-} \longrightarrow \frac{1}{2} \mathrm{H}_{2}(\mathrm{~g}) \);
\( E_{\text {Cell }}^{\ominus}=0.00 \mathrm{~V} \)
(b) \( 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow \mathrm{O}_{2}(\mathrm{~g})+4 \mathrm{H}^{+}(\mathrm{aq})+4 \mathrm{e}^{-} ; \quad E_{\text {Cell }}^{\ominus}=1.23 \mathrm{~V} \)
(c) \( 2 \mathrm{SO}_{4}^{2-}(\mathrm{aq}) \longrightarrow \mathrm{S}_{2} \mathrm{O}_{8}^{2-}(\mathrm{aq})+2 \mathrm{e}^{-} ; \quad E_{\text {cell }}^{\ominus}=1.96 \mathrm{~V} \)
(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, \( \mathrm{SO}_{4}^{2-} \) ion will be oxidised to tetrathionate ion at anode.
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