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Chemistry Question on Electrolysis

In electrolysis of dilute H2SO4 H_2SO_4, what is liberated at anode?

A

H2H_{2}

B

SO42−SO_{4}^{2-}

C

SO2SO_2

D

O2O_2

Answer

O2O_2

Explanation

Solution

Pure water does not conduct electricity. But when small amount of acid say H2SO4H _{2} SO _{4} is added to it, water ionises. On passing electricity, it decomposes. H2SO4H _{2} SO _{4} being a strong electrolyte ionise completely whereas water is feebly ionised H2SO4⟶2H++SO42−H _{2} SO _{4} \longrightarrow 2 H ^{+}+ SO _{4}^{2-} H2O⇌H++OH−H _{2} O \rightleftharpoons H ^{+}+ OH ^{-} During electrolysis, the hydrogen ions migrate towards the cathode and discharge here in the form of hydrogen gas 2H++2e−⟶H2↑2 H ^{+}+2 e^{-} \longrightarrow H _{2} \uparrow At anode, the concentration of OH−OH ^{-}ions is too low to maintain a reaction and sulphate ions are not oxidised but remain in solution. Thus water molecules must be the species reacting at anode 2H2O⟶O2+4H++4e−2 H _{2} O \longrightarrow O _{2}+4 H ^{+}+4 e^{-} The overall reactions are : At cathode 2H++2e−⟶H22 H ^{+}+2 e^{-} \longrightarrow H _{2} 4H++4e−⟶2H24 H ^{+}+4 e^{-} \longrightarrow 2 H _{2} At anode 2H2O⟶O2+4H++4e−2 H _{2} O \longrightarrow O _{2}+4 H ^{+}+4 e^{-} Overall cell reaction : 4H++2H2O⇌2H2+O2+4H+4 H ^{+}+2 H _{2} O \rightleftharpoons 2 H _{2}+ O _{2}+4 H ^{+} as we see that acid used is regenerated at the end. Therefore, the whole electrolysis reaction is the dissociation of water to give oxygen at anode and hydrogen at cathode catalysed by H2SO4H _{2} SO _{4}.