Statement-I: In the redox reaction 8 H ( a q ) + + 4 N O 3 − + 6 C l − + S n ( s ) → S n C l 6 2 − + 4 N O 2 + 4 H 2 O the reducing agent is Sn(s). Statement-II: In balancing the half-reac

Question

Statement-I: In the redox reaction 8 H ( a q ) + ​ + 4 N O 3 − ​ + 6 C l − + S n ( s ) ​ → S n C l 6 2 − ​ + 4 N O 2 ​ + 4 H 2 ​ O the reducing agent is Sn(s). Statement-II: In balancing the half-reaction S 2 ​ O 3 2 − ​ → S ( s ) ​ , the number of electrons added on the left is 4.

TestHub · Accepted Answer

It is intermolecular redox reaction.

Statement-I is True: The oxidation state of $S n$ increases from $0$ in $S n_{(s)}$ to $+ 4$ in $S n C l_{6}^{2 -}$ , meaning it is oxidized and acts as the reducing agent.

Statement-II is True: Balancing the half-reaction gives $S_{2} O_{3}^{2 -} + 6 H^{+} + 4 e^{-} \to 2 S + 3 H_{2} O$ , which correctly shows $4$ electrons added on the left side.

Explanation: Both statements are facts regarding entirely different redox processes and share no logical or causal connection.

Correct Option: (B) Statement-I is true, Statement-II is true; Statement-II is NOT a correct explanation for Statement-I.

Chemistry - Redox Reactions Question with Solution | TestHub

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