Question

how do I find the oxidizing and reducing agents of
1) Cr2O72– + 3SO32– + 8H+ = 2Cr3+ + 3SO42– + 4H2O
2) 4C2O42– + MnO4– + 16H+ = 10CO2 (g) + Mn2+ + 4H2O

Answer 1

Explanation:

Step 1: Plan the problem.

Break the reaction down into a net ionic equation and then into half-reactions. The substance that loses electrons is being oxidized and is the reducing agent. The substance that gains electrons is being reduced and is the oxidizing agent.

Step 2: Solve.

Cl2(g)+2Na+(aq)+2Br−(aq)Cl2(g)+2Br−(aq)→2Na+(aq)+2Cl−(aq)+Br2(l)→2Cl−(aq)+Br2(l)(net ionic equation)(8.2.5)(8.2.6)

Reduction:Cl2(g)+2e−→2Cl−(aq)Oxidation:2Br−(aq)→Br2(l)+2e−(8.2.7)(8.2.8)

The Cl2 is being reduced and is the oxidizing agent. The Br− is being oxidized and is the reducing agent.

Answer 2

1)oxidizing

2)oxidizing