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On electrolysis of dil. sulphuric acid using Platinum electrode, the product obtained at anode will be
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Oxygen gas
Dissociation of sulfuric acid,
\(\mathrm{H}_2 \mathrm{SO}_4 \rightarrow 2 \mathrm{H}^{+}(\mathrm{aq})+\mathrm{SO}_4^{-2}(\mathrm{aq})\)
Disssociation of water,
\(\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq})\)
At Anode: \(4 \mathrm{OH}^{-} \rightarrow 2 \mathrm{H}_2 \mathrm{O}^{+}(\mathrm{l})+\mathrm{O}_2(\mathrm{~g})+4 \mathrm{e}^{-}\)
At Cathode: \(2 \mathrm{H}^{+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_2(\mathrm{~g})\)
\(\mathrm{H}_2 \mathrm{SO}_4 \rightarrow 2 \mathrm{H}^{+}(\mathrm{aq})+\mathrm{SO}_4^{-2}(\mathrm{aq})\)
Disssociation of water,
\(\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq})\)
At Anode: \(4 \mathrm{OH}^{-} \rightarrow 2 \mathrm{H}_2 \mathrm{O}^{+}(\mathrm{l})+\mathrm{O}_2(\mathrm{~g})+4 \mathrm{e}^{-}\)
At Cathode: \(2 \mathrm{H}^{+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_2(\mathrm{~g})\)
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