Question

How can I balance oxidation reduction reactions in basic solution?

Answer 1

WARNING — this is a long answer.

In basic solution, you balance redox equations as if they were in acid. At the end, you use OH⁻ to convert to base.

EXAMPLE:

Balance the following equation in basic solution:
MnO₄⁻ + CN⁻ → MnO₂ + CNO⁻

Solution:

Step 1: Separate the equation into two half-reactions.
MnO₄⁻ → MnO₂
CN⁻ → CNO⁻

Step 2: Balance all atoms other than H and O.
Done

Step 3: Balance O by adding H₂O to the deficient side.
MnO₄⁻ → MnO₂+ 2H₂O
CN⁻+ H₂O → CNO⁻

Step 4: Balance H by adding H⁺ to the deficient side.
MnO₄⁻+ 4H⁺ → MnO₂+ 2H₂O
CN⁻+ H₂O → CNO⁻ + 2H⁺

Step 5: Balance charge by adding electrons to the more positive side.
MnO₄⁻+ 4H⁺ + 3e⁻ → MnO₂+ 2H₂O
CN⁻+ H₂O → CNO⁻ + 2H⁺ + 2e⁻

Step 6: Multiply each half-reaction by a factor that gives the lowest common multiple of the electrons transferred.

In this case, the lowest common multiple of 3 and 2 is 6.
We multiply the first half-reaction by 2 and the second half-reaction by 3.

2 × [MnO₄⁻+ 4H⁺ + 3e⁻ → MnO₂+ 2H₂O]
3 × [CN⁻+ H₂O → CNO⁻ + 2H⁺ + 2e⁻]

Step 7: Add the two half-reactions, cancelling any like terms.
2MnO₄⁻+ 3CN⁻+ 2H⁺ → 2MnO₂ + 3CNO⁻ + H₂O

This is the balanced equation in acid solution. We must now convert to base solution.

Step 8: Add enough multiples of the equations H⁺ + OH⁻ → H₂O or H₂O → H⁺ + OH⁻ to cancel the H⁺ in the redox equation, cancelling like terms.

2MnO₄⁻+ 3CN⁻+ 2H⁺ → 2MnO₂ + 3CNO⁻ + H₂O
2H₂O → 2H⁺ + 2OH⁻

2MnO₄⁻+ 3CN⁻+ H₂O → 2MnO₂ + 3CNO⁻ + 2OH⁻

Step 9: Check that atoms balance.
On the left: 2 Mn; 9 O; 3 C; 3 N; 2 H
On the right: 2 Mn; 9 O; 3 C; 3 N; 2 H

Step 10: Check that charges balance.
On the left: 2- + 3- = 5-
On the right: 3- + 2- = 5-

The balanced equation is
2MnO₄⁻+ 3CN⁻+ H₂O → 2MnO₂ + 3CNO⁻ + 2OH⁻