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On balancing the given redox reaction, aCr2O72−(aq)+bSO32−(aq)+cH+(aq)→2Cr3+(aq)+bSO42−(aq)+c2H2O(l)aCr_2O_7^{2-} (aq) + bSO_3^{2-} (aq) + cH^+ (aq) \rightarrow 2Cr^{3+} (aq) + bSO_4^{2-} (aq) + \frac{c}{2} H_2O(l), the coefficients a, b and c are found to be, respectively-

1

1, 3, 8

2

3, 8, 1

3

1, 8, 3

4

8, 1, 3

Step-by-Step Solution

The balanced equation is 1Cr2O72−+3SO32−+8H+→2Cr3++3SO42−+4H2O1Cr_2O_7^{2-} + 3SO_3^{2-} + 8H^+ \rightarrow 2Cr^{3+} + 3SO_4^{2-} + 4H_2O. Comparing this with the given equation aCr2O72−+bSO32−+cH+→2Cr3++bSO42−+c2H2OaCr_2O_7^{2-} + bSO_3^{2-} + cH^+ \rightarrow 2Cr^{3+} + bSO_4^{2-} + \frac{c}{2} H_2O, we get a=1a=1, b=3b=3, and c=8c=8.

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