On balancing the given redox reaction, $aCr_2O_7^{2-}+bSO_3^{2-}(aq)+cH^+(aq)\rightarrow2aCr^{3+}(aq)+bSO_4^{2-}(aq)+\frac{c}{2}H_2O(l)$ the coefficients a, b and c are found to be, respectively:

A 8, 1, 3

B 1, 3, 8

C 3, 8, 1

D 1, 8, 3

The answer is B.

Solution

Combining half-reactions gives $Cr_2O_7^{2-}+3SO_3^{2-}+8H^+\rightarrow2Cr^{3+}+3SO_4^{2-}+4H_2O$.

Reduction half reaction: $Cr_2O_7^{2-}+14H^++6e^-\rightarrow2Cr^{3+}+7H_2O$ ... (i) Oxidation half reaction: $[SO_3^{2-}+H_2O\rightarrow SO_4^{2-}+2H^++2e^-]\times3$ ... (ii) Oxygen is balanced by adding water and hydrogen is balanced by adding $H^+$ and the charge is balanced by electrons. Add (eq. (i)) + (3 × eq. (ii)): $Cr_2O_7^{2-}+3SO_3^{2-}+8H^+\rightarrow2Cr^{3+}+3SO_4^{2-}+4H_2O$ a = 1, b = 3, c = 8

Concept dichromate–sulphite balancing

Redox Reactions in past papers

30 questions from this chapter have appeared across 19 exam years.

On balancing the given redox reaction, $aCr_2O_7^{2-}+bSO_3^{2-}(aq)+cH^+(aq)\rightarrow2aCr^{3+}(aq)+bSO_4^{2-}(aq)+\frac{c}{2}H_2O(l)$ the coefficients a, b and c are found to be, respectively:

Solution

Redox Reactions in past papers

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