# Balancing Redox Reactions: Acidic and Basic Practice

*2026-10-13*

`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.

![A woman removes red temporary stitches from a seam joining two aligned striped fabric panels](/blog/balancing-redox-reactions-practice.png)

## 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](https://openstax.org/books/chemistry-2e/pages/4-2-classifying-chemical-reactions).

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](https://ch302.cm.utexas.edu/echem/redox/submodule.php?name=redox-balancing).

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](https://mccord.cm.utexas.edu/courses/help/balanceredox.html) describes this neutralization step.

Charge is a signed sum: coefficient × species charge. Neutral water contributes zero; each electron contributes −1. [OpenStax's worked checks](https://openstax.org/books/chemistry/pages/17-1-balancing-oxidation-reduction-reactions) 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](/blog/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](/blog/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](/docs/getting-started/). 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](/blog/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.*

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