Question

Balance the following redox reaction if it occurs in H₂SO₄. What are the coefficients in front of H₂O and Cr₂(SO₄)₃ in the balanced reaction?

C₃H₈O₂(aq) + K₂Cr₂O₇(aq) → C₃H₄O₄(aq) + Cr₂(SO₄)₃(aq)

Can someone please explain everything step by step since how to determine oxidation numbers and half reactions?

Answer 1

3 C₃H₈O₂ + 4 K₂Cr₂O₇ + 16 H₂SO₄ 3 C₃H₄O₄ + 4 Cr₂(SO₄)₃ + 4 K₂SO₄ + 22 H₂O

Explanation is as below:

Oxidation : C₃H₈O₂ C₃H₄O₄ + 8e^-
Reduction : K₂Cr₂O₇ + 6e^- Cr₂(SO₄)₃

While balancing the number of electrons, the total number of electron transferred is 24.

So the balanced oxidation reaction is 3C₃H₈O₂   3C₃H₄O₄ + 24 e^-
and the reduction reaction is 4K₂Cr₂O₇ + 24 e^- 4Cr₂(SO₄)₃  

In order to balance the equation, we have to add the amount of oxygen missing in the oxidation reaction and and the amount of sulfuric acid (H2SO4) in the reduction reaction.
6H₂O + 3C₃H₈O₂   3C₃H₄O₄ + 24 e^- + 24H⁺ -------------------------------------------------(1)

16H₂SO₄ + 4K₂Cr₂O₇ + 24 e^- + 24H⁺ 4Cr₂(SO₄)₃ +  28H₂O +4K₂SO₄ -----------------------(2)

Adding equation (1) and (2), the electrons and the protons (H⁺) will cancel out, the water coefficient will subtract and become 22.

Thus, the correct balanced reaction is
3 C₃H₈O₂ + 4 K₂Cr₂O₇ + 16 H₂SO₄ 3 C₃H₄O₄ + 4 Cr₂(SO₄)₃ + 4 K₂SO₄ + 22 H₂O