How do you balance redox equations in acidic solutions?

1 Answer
Jul 16, 2014

Answer:

A Socratic answer here shows how to balance redox equations.

Explanation:

EXAMPLE:

Balance the following equation in acidic solution:
Cr₂O₇²⁻ + NO₂⁻ → Cr³⁺ + NO₃⁻

Solution:

1: The two half-reactions are.
Cr₂O₇²⁻ → Cr³⁺
NO₂⁻ → NO₃⁻

2: Balance all atoms other than H and O.
Cr₂O₇²⁻ → 2 Cr³⁺
NO₂⁻ → NO₃⁻

3: Balance O.
Cr₂O₇²⁻ → 2 Cr³⁺ + 7 H₂O
NO₂⁻ + 1 H₂O → NO₃⁻

4: Balance H.
Cr₂O₇²⁻ + 14 H⁺→ 2 Cr³⁺ + 7 H₂O
NO₂⁻ + 1 H₂O → NO₃⁻ + 2 H⁺

5: Balance charge.
Cr₂O₇²⁻ + 14 H⁺ + 6 e⁻→ 2 Cr³⁺ + 7 H₂O
NO₂⁻ + 1 H₂O → NO₃⁻ + 2 H⁺ + 2 e⁻

6: Equalize electrons transferred.
1 × [Cr₂O₇²⁻ + 14 H⁺ + 6 e⁻→ 2 Cr³⁺ + 7 H₂O]
3 × [NO₂⁻ + 1 H₂O → NO₃⁻ + 2 H⁺ + 2 e⁻]

7: Add the two half-reactions.
1 Cr₂O₇²⁻ +3 NO₂⁻ + 8 H⁺ → 2 Cr³⁺ + 3 NO₃⁻ + 4 H₂O

8: Check mass balance.
On the left: 2 Cr; 13 O; 3 N; 8 H
On the right: 2 Cr; 3 N; 13 O; 8 H

9: Check charge balance.
On the left: 2- + 3- + 8+ = 3+
On the right: 6+ + 3- = 3+

The balanced equation is
Cr₂O₇²⁻ +3NO₂⁻ + 8H⁺ → 2Cr³⁺ + 3NO₃⁻ + 4H₂O