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Question: During the electrolysis of dilute sulphuric acid, the following process is possible at anode....

During the electrolysis of dilute sulphuric acid, the following process is possible at anode.

A

2H2O(l)→O2(g)+4H+(aq)+4e−2H_{2}O_{(l)} \rightarrow O_{2(g)} + 4{H^{+}}_{(aq)} + 4e^{-}

B

2SO42−(aq)→S2O82−(aq)+2e−2S{O_{4}^{2 -}}_{(aq)} \rightarrow S_{2}{O_{8}^{2 -}}_{(aq)} + 2e^{-}

C

H2O(l)→H+(aq)+OH−(aq)H_{2}O_{(l)} \rightarrow {H^{+}}_{(aq)} + O{H^{-}}_{(aq)}

D

H2O(l)+e−→12H2(g)+OH−(aq)H_{2}O_{(l)} + e^{-} \rightarrow \frac{1}{2}H_{2(g)} + O{H^{-}}_{(aq)}

Answer

2H2O(l)→O2(g)+4H+(aq)+4e−2H_{2}O_{(l)} \rightarrow O_{2(g)} + 4{H^{+}}_{(aq)} + 4e^{-}

Explanation

Solution

During the electrolysis of dilute sulphuric acid, the following process is possible at anode:

2H2O(l)→O2(g)+4H+(aq)+4e−(dil.H2SO4)2H_{2}O_{(l)} \rightarrow O_{2(g)} + 4{H^{+}}_{(aq)} + 4e^{-}(dil.H_{2}SO_{4})