Balance Redox Reaction in Basic Solution Example Problem

Half-Reaction Method in a Basic Solution

Redox solutions occur in both acidic and basic solutions.
Redox solutions occur in both acidic and basic solutions. Siede Preis, Getty Images

Redox reactions commonly take place in acidic solutions. The could just as easily take place in basic solutions. This example problem shows how to balance a redox reaction in a basic solution.

Redox reactions are balanced in basic solutions using the same half-reaction method demonstrated in the example problem "Balance Redox Reaction Example". In summary:

  1. Identify the oxidation and reduction components of the reaction.
  1. Separate the reaction into the oxidation half-reaction and reduction half-reaction.
  2. Balance each half-reaction both atomically and electronically.
  3. Equalize the electron transfer between oxidation and reduction half-equations.
  4. Recombine the half-reactions to form the complete redox reaction.

This will balance the reaction in an acidic solution, where there is an excess of H+ ions. In basic solutions, there is an excess of OH- ions. The balanced reaction needs to be modified to remove the H+ ions and include OH- ions.

Problem:

Balance the following reaction in a basic solution:

Cu(s) + HNO3(aq) → Cu2+(aq) + NO(g)

Solution:

Balance the equation using the half-reaction method outlined in the Balance Redox Reaction Example. This reaction is the same one used in the example but was balanced in an acidic environment. The example showed the balanced equation in the acidic solution was:

3 Cu + 2 HNO3 + 6 H+→ 3 Cu2+ + 2 NO + 4 H2O

There are six H+ ions to remove.

This is accomplished by adding the same number of OH- ions to both sides of the equation. In this case, add 6 OH- to both sides. 3 Cu + 2 HNO3 + 6 H+ + 6 OH- → 3 Cu2+ + 2 NO + 4 H2O + 6 OH-

The H+ ions and OH- combine to form a water molecule (HOH or H2O). In this case, 6 H2O are formed on the reactant side.



3 Cu + 2 HNO3 + 6 H2O → 3 Cu2+ + 2 NO + 4 H2O + 6 OH-

Cancel out the extraneous water molecules on both sides of the reaction. In this case, remove 4 H2O from both sides.

3 Cu + 2 HNO3 + 2 H2O → 3 Cu2+ + 2 NO + 6 OH-

The reaction is now balanced in a basic solution.