Balancing Redox in Acid

8 min read Updated Mar 26, 2026

Balancing redox reactions requires more than just matching atoms - you also need to balance the electrons transferred. The half-reaction method is the systematic way to do this, and it works every time. In acidic solution, you use H2O to balance oxygen atoms and H+ to balance hydrogen atoms.

The Half-Reaction Method in Acidic Solution

Step 1: Split into half-reactions. Identify which species is oxidized and which is reduced. Write each as a separate half-reaction.

Step 2: Balance atoms other than O and H. Make sure the element being oxidized or reduced has the same number of atoms on both sides.

Step 3: Balance oxygen by adding H2O. For every O atom you need, add one H2O to the opposite side.

Step 4: Balance hydrogen by adding H+. For every H atom you need, add one H+ to the opposite side. (This is why the method works in acidic solution - H+ is available.)

Step 5: Balance charge by adding electrons. Add e- to the more positive side of each half-reaction so the total charge is equal on both sides.

Step 6: Equalize electrons. Multiply each half-reaction by the appropriate integer so both half-reactions transfer the same number of electrons.

Step 7: Add the half-reactions. Cancel electrons (they should cancel completely) and cancel any H2O or H+ that appear on both sides.

Step 8: Verify. Check that atoms balance AND charge balances.

Worked Example

Balance in acidic solution: MnO4-(aq) + Fe2+(aq) —> Mn2+(aq) + Fe3+(aq)

Step 1: Split.

Reduction: MnO4- —> Mn2+

Oxidation: Fe2+ —> Fe3+

Step 2: Balance Mn and Fe. Already balanced (1 each).

Step 3: Balance O with H2O.

MnO4- —> Mn2+ + 4H2O

Step 4: Balance H with H+.

8H+ + MnO4- —> Mn2+ + 4H2O

Step 5: Balance charge with electrons.

Reduction half: 5e- + 8H+ + MnO4- —> Mn2+ + 4H2O

Check: Left side = 5(-1) + 8(+1) + (-1) = +2. Right side = +2 + 0 = +2. Balanced.

Oxidation half: Fe2+ —> Fe3+ + e-

Check: Left = +2. Right = +3 + (-1) = +2. Balanced.

Step 6: Equalize electrons. Reduction transfers 5e-; oxidation transfers 1e-. Multiply oxidation by 5:

5Fe2+ —> 5Fe3+ + 5e-

Step 7: Add and cancel electrons.

5e- + 8H+ + MnO4- + 5Fe2+ —> Mn2+ + 4H2O + 5Fe3+ + 5e-

Cancel the 5e- from both sides:

8H+ + MnO4- + 5Fe2+ —> Mn2+ + 4H2O + 5Fe3+

Step 8: Verify. Mn: 1 = 1. O: 4 = 4. H: 8 = 8. Fe: 5 = 5. Charge: 8 + (-1) + 10 = +17 on the left. +2 + 0 + 15 = +17 on the right. Balanced.

When balancing a redox reaction in acidic solution, what do you add to balance oxygen atoms? What do you add to balance hydrogen atoms?

Click to reveal answer

Add H2O to balance oxygen atoms. Add H+ to balance hydrogen atoms. For each oxygen atom needed, add one H2O to the opposite side. Each H2O introduces 2 H atoms, so you then add H+ to balance those. Electrons are added last to balance charge.

In the reaction Cr2O7^2- + 14H+ + 6e- --> 2Cr3+ + 7H2O, how many electrons does each chromium atom gain? What was its starting oxidation state?

Click to reveal answer

Each Cr gains 3 electrons, going from +6 to +3. In Cr2O7^2-, each Cr is +6 (since 2Cr + 7(-2) = -2, so Cr = +6). In the product Cr3+, each Cr is +3. That is a decrease of 3 per Cr atom. With 2 Cr atoms, the total is 6 electrons gained, matching the 6e- shown in the balanced half-reaction.