Balancing Redox Reactions: Acidic and Basic Practice

Sn²⁺ + Fe³⁺ → Sn⁴⁺ + Fe²⁺ has one tin atom and one iron atom on each side. It still isn't balanced: the total charge is +5 on the left and +6 on the right. Counting atoms gets you only halfway through checking a redox equation.

Start with the two worked examples, then try the seven problems that ask you to solve or repair an equation. Keep a separate line for net charge, even when every atom already matches.

The problems give the reactants, products, and aqueous medium. Change coefficients, never subscripts or ion charges. Use H₂O and H⁺ to balance in acid; for base, neutralize every introduced H⁺ with OH⁻. Balancing conserves atoms and charge for a supplied skeleton. It doesn't identify unknown products or establish whether a reaction occurs under actual conditions.

The half-reaction method, with a check at each stop

Oxidation loses electrons; reduction gains them. Split the skeleton into those two half-reactions using changes in oxidation number. An oxidation half-reaction has electrons on the right; a reduction half-reaction has them on the left. OpenStax explains oxidation, reduction, and oxidation-number rules.

For each half-reaction in acidic solution:

1. Balance elements other than H and O.
2. Balance O with H₂O.
3. Balance H with H⁺.
4. Add electrons to the side with the more positive total charge until both sides match.

Find the smallest common electron count. Multiply every term in each half-reaction by its required factor, add the equations, and cancel equal electrons and identical species on opposite sides. Reduce coefficients to the smallest whole-number ratio. Check every element and net charge again. This follows the University of Texas half-reaction method.

For basic solution, use that same acidic balancing stage, then add as many OH⁻ to both sides as there are remaining H⁺. Replace each H⁺ + OH⁻ pair with H₂O and cancel common water. Recheck the equation. UT Austin's balancing guide describes this neutralization step.

Charge is a signed sum: coefficient × species charge. Neutral water contributes zero; each electron contributes −1. OpenStax's worked checks include both ion and electron charges. Use actual ion charges, rather than the oxidation numbers of atoms within polyatomic ions. If those two quantities blur together, work through formal charge vs oxidation number first.

Acidic example: account for both chromium atoms

Balance this supplied skeleton in acidic solution:

Cr₂O₇²⁻ + Sn²⁺ → Cr³⁺ + Sn⁴⁺

Oxygen is −2 in dichromate. If chromium's oxidation number is x, 2x + 7(−2) = −2, giving x = +6. Chromium changes from +6 to +3: each Cr gains three electrons, and the formula contains two Cr atoms, so one dichromate ion gains six. Tin changes from +2 to +4, losing two electrons per atom.

Build the chromium half-reaction

Balance Cr first, then O with water, then H with H⁺:

Cr₂O₇²⁻ → 2Cr³⁺

Cr₂O₇²⁻ → 2Cr³⁺ + 7H₂O

Cr₂O₇²⁻ + 14H⁺ → 2Cr³⁺ + 7H₂O

The last line conserves atoms. Its charge still needs work:

Check before electrons	Left	Right
Cr atoms	2	2
O atoms	7	7
H atoms	14	14
Net charge	−2 + 14 = +12	2 × (+3) = +6

Add six electrons on the left to lower +12 to +6:

Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O

The left charge is now −2 + 14 − 6 = +6, matching the right. The six-electron result also agrees with the oxidation-number change.

Scale the tin half-reaction and combine

Tin needs no H or O adjustment:

Sn²⁺ → Sn⁴⁺ + 2e⁻

Its charge is +2 on each side because +4 − 2 = +2. To supply six electrons, multiply the whole equation by three:

3Sn²⁺ → 3Sn⁴⁺ + 6e⁻

Add it to the chromium half-reaction and cancel six electrons from each side:

Cr₂O₇²⁻ + 3Sn²⁺ + 14H⁺ → 2Cr³⁺ + 3Sn⁴⁺ + 7H₂O

Final check	Left	Right
Cr atoms	2	2
Sn atoms	3	3
O atoms	7	7
H atoms	14	14
Net charge	−2 + 3 × (+2) + 14 = +18	2 × (+3) + 3 × (+4) = +18

The total charge doesn't have to be zero. It has to be equal on both sides. Spectator ions are omitted from this net ionic equation.

Basic example: cancel water twice

Balance this supplied skeleton in basic solution:

MnO₄⁻ + NO₂⁻ → MnO₂ + NO₃⁻

The stated manganese product is MnO₂. With O = −2, MnO₄⁻ gives x + 4(−2) = −1, so Mn starts at +7. In neutral MnO₂, x + 2(−2) = 0, so it ends at +4 and gains three electrons. The same calculation gives N = +3 in NO₂⁻ and +5 in NO₃⁻: nitrogen loses two electrons.

Balance both halves using the acidic stage

For manganese, add two waters on the right to balance O, then four H⁺ on the left to balance H. Before electrons, left charge is −1 + 4 = +3, while right charge is zero. Add three electrons on the left:

MnO₄⁻ + 4H⁺ + 3e⁻ → MnO₂ + 2H₂O

For nitrogen, put one water on the left and two H⁺ on the right. Left charge is −1 and right charge is −1 + 2 = +1, so add two electrons on the right:

NO₂⁻ + H₂O → NO₃⁻ + 2H⁺ + 2e⁻

Balanced half-reaction	Atom totals on each side	Net charge on each side
Manganese reduction	Mn 1, O 4, H 4	0: left −1 + 4 − 3; right 0
Nitrogen oxidation	N 1, O 3, H 2	−1: left −1; right −1 + 2 − 2

Six is the smallest common electron count. Multiply the manganese half by two and the nitrogen half by three:

2MnO₄⁻ + 8H⁺ + 6e⁻ → 2MnO₂ + 4H₂O

3NO₂⁻ + 3H₂O → 3NO₃⁻ + 6H⁺ + 6e⁻

Cancel six electrons, six H⁺, and three H₂O from opposite sides. The result is:

2MnO₄⁻ + 3NO₂⁻ + 2H⁺ → 2MnO₂ + 3NO₃⁻ + H₂O

This intermediate has Mn 2, N 3, O 14, and H 2 on each side. Charge is −2 − 3 + 2 = −3 on the left and −3 on the right. It's balanced, but it hasn't yet been expressed for the requested basic medium.

Neutralize the remaining H⁺

Two H⁺ remain, so add two OH⁻ to both sides:

2MnO₄⁻ + 3NO₂⁻ + 2H⁺ + 2OH⁻ → 2MnO₂ + 3NO₃⁻ + H₂O + 2OH⁻

Combine the two H⁺ + OH⁻ pairs into two waters:

2MnO₄⁻ + 3NO₂⁻ + 2H₂O → 2MnO₂ + 3NO₃⁻ + H₂O + 2OH⁻

There is still water on both sides. Cancel one H₂O from each:

2MnO₄⁻ + 3NO₂⁻ + H₂O → 2MnO₂ + 3NO₃⁻ + 2OH⁻

Final check	Left	Right
Mn atoms	2	2
N atoms	3	3
O atoms	8 + 6 + 1 = 15	4 + 9 + 2 = 15
H atoms	2	2
Net charge	−2 − 3 = −5	−3 − 2 = −5

Replacing H⁺ directly with OH⁻ would change both atoms and charge. Adding the same amount to both sides preserves the equation; forming and canceling water completes the conversion. For the proton-transfer step itself, see conjugate acid–base pairs.

Seven problems to solve or repair

Work on paper before reading the answer key. Balance the supplied net ionic skeletons in the stated medium. Ions are aqueous, H₂O is liquid, and MnO₂ is a solid. O₂, Br₂, and I₂ denote neutral elemental species; H₂O₂ is neutral too. These are paper exercises rather than instructions for mixing chemicals.

For full balances, show both half-reactions, the electron multipliers, and final atom and charge totals. For repair tasks, name the failed step before fixing it.

1. Acidic balance. BrO₃⁻ + Br⁻ → Br₂. Bromate's Br is +5; bromide's is −1. Both become elemental bromine, oxidation number 0.
2. Basic disproportionation. Br₂ → Br⁻ + BrO₃⁻. One starting species supplies both the oxidized and reduced bromine. Keep both stated products.
3. Peroxide oxidation in acid. MnO₄⁻ + H₂O₂ → Mn²⁺ + O₂. Oxygen in H₂O₂ is −1. Find its electron change and balance the full equation.
4. Atoms match; charge doesn't. In acidic solution, repair Sn²⁺ + Fe³⁺ → Sn⁴⁺ + Fe²⁺. Explain why a matching atom count isn't enough. Does either half need H⁺ or H₂O?
5. Incomplete neutralization. A basic-solution answer reads ClO⁻ + 2I⁻ + H⁺ → Cl⁻ + I₂ + OH⁻. It passes atom and charge checks. Explain what remains unfinished, then express it for basic solution.
6. Scale every term. The balanced acidic half-reaction is CrO₄²⁻ + 8H⁺ + 3e⁻ → Cr³⁺ + 4H₂O. To double it, a student writes 2CrO₄²⁻ + 8H⁺ + 6e⁻ → 2Cr³⁺ + 4H₂O. Correct the doubled equation and check its charge.
7. The product is given. In basic solution, balance MnO₄⁻ + SO₃²⁻ → MnO₄²⁻ + SO₄²⁻. Someone replaces the supplied MnO₄²⁻ with MnO₂ because it appeared in the worked example. Explain why that changes the task. Does the correct balance prove that the reaction happens under any basic conditions?

Answer key with the failed steps exposed

1. Bromate needs five bromides per ion

Use integer coefficients for the bromate reduction:

2BrO₃⁻ + 12H⁺ + 10e⁻ → Br₂ + 6H₂O

The oxidation half is 2Br⁻ → Br₂ + 2e⁻. Multiply it by five, add, cancel ten electrons, and divide all coefficients by two:

BrO₃⁻ + 5Br⁻ + 6H⁺ → 3Br₂ + 3H₂O

Both sides contain Br 6, O 3, and H 6. Left charge is −1 − 5 + 6 = 0; right charge is zero. In the reduction half, charge is −2 + 12 − 10 = 0 on both sides; the oxidation half has charge −2 on each side.

2. The same reactant appears in both halves

Write the reduction as Br₂ + 2e⁻ → 2Br⁻.

For oxidation, first balance in acid:

Br₂ + 6H₂O → 2BrO₃⁻ + 12H⁺ + 10e⁻

Each half has matching atoms. The reduction half has charge −2 on each side; the oxidation half has charge zero on each side, since −2 + 12 − 10 = 0. Multiply the reduction half by five and add. After electron cancellation and division by two, the acidic intermediate is:

3Br₂ + 3H₂O → 5Br⁻ + BrO₃⁻ + 6H⁺

Add six OH⁻ to both sides, form six waters on the right, then cancel three waters:

3Br₂ + 6OH⁻ → 5Br⁻ + BrO₃⁻ + 3H₂O

Br 6, O 6, and H 6 match. Charge is −6 on each side. Don't assign all bromine atoms the same change: some go from 0 to −1, while others go from 0 to +5.

3. Peroxide loses two electrons per molecule

Each of the two O atoms changes from −1 to 0, giving a total loss of two electrons:

H₂O₂ → O₂ + 2H⁺ + 2e⁻

This half has H 2 and O 2 on both sides, with zero net charge on each side. The manganese half is:

MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

It has Mn 1, O 4, H 8, and charge +2 on each side. Multiply that half by two and the peroxide half by five. Cancel ten electrons and ten of the sixteen H⁺ on the left:

2MnO₄⁻ + 5H₂O₂ + 6H⁺ → 2Mn²⁺ + 5O₂ + 8H₂O

Both sides contain Mn 2, O 18, and H 16. Charge is −2 + 6 = +4 on the left and 2 × (+2) = +4 on the right. Using O = −2 in peroxide would give the wrong electron count.

4. Two irons accept the electrons from one tin

The original atom count matches, but its +5 left charge differs from +6 on the right. The halves are:

Sn²⁺ → Sn⁴⁺ + 2e⁻

Fe³⁺ + e⁻ → Fe²⁺

Both halves have charge +2 on each side. Neither needs H⁺ or H₂O because neither contains H or O. Double the iron half:

Sn²⁺ + 2Fe³⁺ → Sn⁴⁺ + 2Fe²⁺

Sn 1 and Fe 2 match. Charge is 2 + 2 × 3 = +8 on the left and 4 + 2 × 2 = +8 on the right. Changing Fe²⁺ to a different ion to fix the charge would change the supplied product.

5. A valid intermediate still contains H⁺

The proposed line has Cl 1, I 2, O 1, H 1, and charge −2 on each side. Those checks are correct. It still contains H⁺ that must be neutralized for the requested basic form.

Add one more OH⁻ to both sides, combine H⁺ + OH⁻ on the left into H₂O, and keep the resulting two OH⁻ on the right:

ClO⁻ + 2I⁻ + H₂O → Cl⁻ + I₂ + 2OH⁻

Now Cl 1, I 2, O 2, and H 2 match, with charge −3 on each side. Deleting the H⁺ alone would break both conservation checks.

6. H⁺ and H₂O need the multiplier too

The correct doubled half-reaction is:

2CrO₄²⁻ + 16H⁺ + 6e⁻ → 2Cr³⁺ + 8H₂O

Cr 2, O 8, and H 16 match. Charge is 2 × (−2) + 16 − 6 = +6 on the left and +6 on the right.

The student's line has O 8 on the left but O 4 on the right; its charges are −2 and +6. Its H count happens to match at 8, but doubling the original equation requires 16 on each side. Matching electron counts alone can't rescue a half-reaction whose other terms weren't scaled.

7. Manganate changes the electron requirement

In MnO₄²⁻, manganese is +6, so each Mn gains one electron from its +7 starting state:

MnO₄⁻ + e⁻ → MnO₄²⁻

Each side has Mn 1, O 4, and charge −2. Sulfur changes from +4 to +6. Start its half-reaction in acid:

SO₃²⁻ + H₂O → SO₄²⁻ + 2H⁺ + 2e⁻

Each side has S 1, O 4, H 2, and charge −2. Add two OH⁻ to both sides, form two waters on the right, then cancel one water. The basic half-reaction is:

SO₃²⁻ + 2OH⁻ → SO₄²⁻ + H₂O + 2e⁻

Each side has S 1, O 5, H 2, and charge −4. Double the manganese half and add:

2MnO₄⁻ + SO₃²⁻ + 2OH⁻ → 2MnO₄²⁻ + SO₄²⁻ + H₂O

Both sides contain Mn 2, S 1, O 13, and H 2. Charge is −2 − 2 − 2 = −6 on the left and −4 − 2 = −6 on the right.

Replacing manganate with MnO₂ changes manganese's final oxidation state from +6 to +4, and therefore changes the electron count and equation. The balanced result establishes conservation for the given products. Whether those products form under particular conditions requires chemical information beyond balancing, including concentrations and reaction energetics.

Review the decision, then solve a fresh equation

Use full equations for paper practice. Short review prompts work better for a recurring decision you missed:

Front	Back
Every element matches. What check remains for a redox equation?	Sum coefficient × species charge on each side. Include electrons in half-reactions.
A half-reaction needs multiplying by three. Which terms change?	Every coefficient on both sides, including electrons, H⁺, and H₂O.
Two H⁺ remain in a basic-solution intermediate. What do I add?	Two OH⁻ to each side; form H₂O, then cancel common water.
Why does one Cr₂O₇²⁻ need six electrons to produce 2Cr³⁺?	Two Cr atoms each change from +6 to +3: 2 × 3 = 6 electrons gained.
Can I change the supplied manganese product because another example used MnO₂?	No. Product identity determines the electron change; balance the stated skeleton.

Keep these on paper or add the ones you need to Nibomo using the getting-started guide. Cover the answer, explain the missed step, and then balance a different skeleton without help. Recognizing a finished equation doesn't show that you can produce it.

Once the equation passes both checks, its coefficients can support mole calculations. Our limiting reactant practice works through that next step.

AI assistance was used to draft this guide and check its atom and charge arithmetic. The exercises are constructed paper problems; the linked chemistry sources support the balancing method.