# Balance Redox Reaction Example Problem

When balancing redox reactions, the overall electronic charge must be balanced in addition to the usual molar ratios of the component reactants and products. This example problem illustrates how to use the half-reaction method to balance a redox reaction in a solution.

## Question

Balance the following redox reaction in an acidic solution:

Cu(s) + HNO3(aq) → Cu2+(aq) + NO(g)

## Solution

Step 1: Identify what is being oxidized and what is being reduced.

To identify which atoms are being reduced or oxidized, assign oxidation states to each atom of the reaction.

For review:

• Cu(s): Cu = 0
• HNO3: H = +1, N = +5, O = -6
• Cu2+: Cu = +2
• NO(g): N = +2, O = -2

Cu went from oxidation state 0 to +2, losing two electrons. Copper is oxidized by this reaction.
N went from oxidation state +5 to +2, gaining three electrons. Nitrogen is reduced by this reaction.

Step 2: Break the reaction into two half-reactions: oxidation and reduction.

Oxidation: Cu → Cu2+

Reduction: HNO3 → NO

Step 3: Balance each half-reaction by both stoichiometry and electronic charge.

This is accomplished by adding substances to the reaction. The only rule is that the only substances you can add must already be in the solution. These include water (H2O), H+ ions (in acidic solutions), OH- ions (in basic solutions) and electrons.

The half-reaction is already balanced atomically. To balance electronically, two electrons must be added to the product side.

Cu → Cu2+ + 2 e-

Now, balance the reduction reaction.

This reaction requires more work. The first step is to balance all atoms except oxygen and hydrogen.

HNO3 → NO

There is only one nitrogen atom on both sides, so nitrogen is already balanced.

The second step is to balance the oxygen atoms. This is done by adding water to the side that needs more oxygen. In this case, the reactant side has three oxygens and the product side has only one oxygen. Add two water molecules to the product side.

HNO3 → NO + 2 H2O

The third step is to balance the hydrogen atoms. This is accomplished by adding H+ ions to the side that needs more hydrogen. The reactant side has one hydrogen atom while the product side has four. Add 3 H+ ions to the reactant side.

HNO3 + 3 H+ → NO + 2 H2O

The equation is balanced atomically, but not electrically. The final step is to balance the charge by adding electrons to the more positive side of the reaction. One the reactant side, the overall charge is +3, while the product side is neutral. To counteract the +3 charge, add three electrons to the reactant side.

HNO3 + 3 H+ + 3 e- → NO + 2 H2O

Now the reduction half-equation is balanced.

Step 4: Equalize the electron transfer.

In redox reactions, the number of electrons gained must equal the number of electrons lost. To accomplish this, each reaction is multiplied by whole numbers to contain the same number of electrons.

The oxidation half-reaction has two electrons while the reduction half-reaction has three electrons. The lowest common denominator between them is six electrons. Multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2.

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

Step 5: Recombine the half-reactions.

This is accomplished by adding the two reactions together. Once they are added, cancel out anything that appears on both sides of the reaction.

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

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

Both sides have six electrons that can be canceled.

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

The complete redox reaction is now balanced.