Topic: Balancing redox reactions· Redox Reactions and Electrochemistry

On balancing the given redox reaction,a Cr_2O_7^2- + b SO_3^2-(aq) + c H^+(aq) → 2a Cr^3+(aq) + b SO_4^2-(aq) + c2H_2O()the coefficients a, b and c are found to be, respectively -

1. 8, 1, 3

2. 1, 3, 8

3. 3, 8, 1

4. 1, 8, 3

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Answer: 2. 1, 3, 8

Key idea: Balance by the ion-electron (or oxidation-number) method.

Cr goes from +6 to +3: gain of 3 e\(^-\) per Cr, 6 e\(^-\) per \(\mathrm{Cr_2O_7^{2-}}\). S goes from +4 to +6: loses 2 e\(^-\) per \(\mathrm{SO_3^{2-}}\). To equalise electrons: 1 \(\mathrm{Cr_2O_7^{2-}}\) to 3 \(\mathrm{SO_3^{2-}}\).

\(\mathrm{Cr_2O_7^{2-} + 3SO_3^{2-} + 8H^+ \rightarrow 2Cr^{3+} + 3SO_4^{2-} + 4H_2O}\)

So a = 1, b = 3, c = 8 (check: H\(^+\) = 8, and \(c/2 = 4\) water). Option 2.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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On balancing the given redox reaction,a Cr_2O_7^2- + b SO_3^2-(aq) + c H^+(aq) → 2a Cr^3+(aq) + b SO_4^2-(aq) + c2H_2O()the coefficients a, b and c are found to be, respectively -

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