Balance the redox equation below in acidic solution.

Question

[a] H⁺ + [b] ClO₃^1- + [c] Cl^1- → [d] Cl₂ + [e] ClO₂ + [f] H₂O

the answers are:

4, 2, 2, 1, 2, 2 respectively.

Having trouble balancing the electrons to get to the answer.

Answer 1

Cl- ==> Cl₂ ... unbalanced oxidation half reaction

2Cl^- ==> Cl₂ + 2e- balanced for Cl and charge

ClO₃^- ==> ClO₂ ... unbalanced reduction half reaction

ClO₃^- + 2H⁺ + e- ==> ClO₂ + H₂O ... balanced for Cl, O and H by adding acid (H⁺)

To equalize electrons, multiply reduction half reaction by two, and then add the two together:

2ClO₃^- + 4H⁺ + 2e- ==> 2ClO₂ + 2H₂O

2Cl^- ==> Cl₂ + 2e-

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2ClO₃^- + 4H⁺ + 2e- + 2Cl^- ==> 2ClO₂ + 2H₂O + Cl₂ + 2e-

Combine and/or cancel like terms to obtain the final balanced redox equation:

2ClO₃^- + 4H⁺ + 2Cl^- ==> 2ClO₂ + 2H₂O + Cl₂ ... BALANCED REDOX EQUATION

Answer 2

1) Skeleton of 1/2 reactions and balance the element being ox or red.

ClO₃^- → ClO₂ (Red)

2Cl^- → Cl₂ (Ox)

2) Use H₂O to balance O and H⁺ to balance H and e^- to balance charge in each 1/2 rxn.

ClO₃^- + 2H⁺ + e^- → ClO₂ + H₂O

2Cl^- → Cl₂ + 2e^-

3) add equations so that e^- cancel (2x first equation + second equation:

2ClO₃^- + 4H⁺ + 2Cl^- → 2ClO₂ +Cl₂ + 2H₂O

Please consider a tutor. Take care.