Answer to Question #180382 in General Chemistry for Chaihcai

Question

Balancing Redox Reaction

Balance the reactions below using the change in oxidation number method.

1. C₂O₄^2- = CO₂

2. KMnO₄ + Na₂C₂O₄ + H₂SO₄ = K₂SO₄ + MnSO₄ + Na₂SO₄ + SO + CO₂ + H₂O

Answer 1

1. (balancing atoms and charge)

C2O4^2- —> 2CO2 + 2e^-

C. Rewrite and reconcile the two half reactions, then add them and simplify:

Cr2O7^-2(aq) + 14H^+(aq) + 6e^- —> 2Cr^+3 + 7H2O

3C2O4^2- —> 6CO2 + 6e^-

(I multiplied the oxidation by 3 so that the total number of electrons transferred will be 6 as it is in the reduction)

Cr2O7^-2(aq) + 14H^+(aq) + 6e^- + 3C2O4^2- —> 2Cr^+3 + 7H2O + 6CO2 + 6e^-

(added left and right sides ... then we simplify through cancellations)

Cr2O7^-2(aq) + 14H^+(aq) + 3C2O4^2- —> 2Cr^+3 + 7H2O + 6CO2

2. 2KMnO₄(aq) + 5 Na₂C₂O4₄(aq) + 8 H₂SO₄(aq) → 8 H₂O (l) + 10 CO₂ (g) + K₂SO₄ (aq) + 2 MnSO₄ (aq) + 5 Na₂CO₄ (aq)

This is an oxidation-reduction (redox) reaction:

2 Mn^VII + 10 e^- → 2 Mn^II

(reduction)

10 C^III - 10 e^- → 10 C^IV

(oxidation)