O_2 gas will be evolved as a product of elect | vedclass

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Question

(A) In the electrolysis of an aqueous solution of $AgNO_3$ using silver electrodes $(A)$,the anode dissolves $(Ag \rightarrow Ag^+ + e^-)$,so no $O_2$

Vedclass · Accepted Answer

$B$ and $C$ only

Vedclass · Answer

(A) In the electrolysis of an aqueous solution of $AgNO_3$ using silver electrodes $(A)$,the anode dissolves $(Ag \rightarrow Ag^+ + e^-)$,so no $O_2$ is evolved.In the electrolysis of an aqueous solution of $AgNO_3$ using platinum electrodes $(B)$,the oxidation of water occurs at the anode $(2H_2O \rightarrow O_2 + 4H^+ + 4e^-)$,evolving $O_2$ gas.In the electrolysis of a dilute solution of $H_2SO_4$ using platinum electrodes $(C)$,the oxidation of water occurs at the anode $(2H_2O \rightarrow O_2 + 4H^+ + 4e^-)$,evolving $O_2$ gas.In the electrolysis of a high concentration solution of $H_2SO_4$ using platinum electrodes $(D)$,the oxidation of $SO_4^{2-}$ ions occurs at the anode $(2SO_4^{2-} \rightarrow S_2O_8^{2-} + 2e^-)$,forming peroxodisulphuric acid instead of $O_2$ gas.Therefore,$O_2$ is evolved in cases $(B)$ and $(C)$.

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