Balancing equations

Question

Balance the following equations for reactions in acidic solution by the half-reaction method. Identify the oxidizing agent and the reducing agent in each.

(a) Cr2O7 2−(aq) + HS− (aq) → Cr3+(aq) + S(s)

(b) MnO2(s) + Cl− (aq) → Mn2+(aq) + Cl2(g)

(c) Br− (aq) + HSO4 − (aq) → SO2(g) + Br2(l)

I was able to to (b) but I got really stuck on (a) and for (c) I couldn't figure out how to balance the charges

Answer 1

If you did (b), it's surprising that you couldn't do the others, but ... here goes!

(a). Cr₂O₇^2- ==> Cr³⁺

Cr₂O₇^2- ==> 2Cr³⁺ + 7H₂O .. balanced for Cr and O

Cr₂O₇^2- + 14H⁺ ==> 2Cr³⁺ + 7H₂O .. balanced for Cr, O and H (using acid)

Cr₂O₇^2- + 14H⁺ + 6e- ==> 2Cr³⁺ + 7H₂O .. balanced for Cr, O, H and charge = balanced reduction rxn

HS^- ==> S

HS^- ==> S + H⁺ .. balanced for S and H

HS^- ==> S + H⁺ + 2e- .. balanced for S, H and charge = balanced oxidation reaction

Cr₂O₇^2- + 14H⁺ + 6e- ==> 2Cr³⁺ + 7H₂O

3HS^- ==> 3S + 3H⁺ + 6e-

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Cr₂O₇^2- + 11H⁺ + 3HS^- ==> 2Cr³⁺ + 7H₂O + 3S balanced redox equation

(c). 2Br^- ==> Br₂ .. balanced for Br

2Br^- ==> Br₂ + 2e- .. balanced for Br and charge = balanced oxidation reaction

HSO₄^- ==> SO₂

HSO₄^- ==> SO₂ + 2H₂O .. balanced for O

HSO₄^- + 3H⁺ ==> SO₂ + 2H₂O .. balanced for O and H (using acid)

HSO₄^- + 3H⁺ +2e- ==> SO₂ + 2H₂O .. balanced for O, H and charge = balanced reduction reaction

Adding oxidation and reduction half reactions and combining/canceling like terms, we get:

2Br^- + HSO₄^- + 3H⁺ ==> Br₂ + SO₂ + 2H₂O .. balanced redox equation