Tricks to Balance Redox Reactions in 3 Minutes
A Shortcut Guide for JEE & NEET
Introduction
When I teach Redox Reactions in class, I notice one common problem: most students understand the concept, but they get stuck when it comes to balancing the reaction. Not because it is difficult, but because they don’t know the right method to approach it.
So in this guide, I’m going to explain the simplest and fastest way to balance any redox reaction. No confusion, no extra theory — just a clear step-by-step shortcut with examples you can follow immediately.
By the time you finish this guide, you will be able to balance:
- Reactions in acidic medium
- Reactions in basic medium
- Simple redox reactions
- Complex ion-based reactions
Let’s start with the easiest and most reliable method: the Ion–Electron (Half-Reaction) Method.
Step-by-Step Shortcut Method to Balance Redox Reactions
We always follow these steps:
1 Identify oxidation and reduction
2 Write half-reactions separately
3 Balance atoms
4 Balance oxygen using H₂O
5 Balance hydrogen using H⁺ (acidic medium)
6 Balance charges using electrons
7 Equalize electrons
8 Add and simplify
Let’s understand each step clearly with examples.
Example Reaction:
Step 1: Identify Oxidation and Reduction
Look at oxidation numbers.
Check oxidation numbers:
- Fe²⁺ → Fe³⁺ (from +2 to +3) → oxidation
- MnO₄⁻ (Mn = +7) → Mn²⁺ (Mn = +2) → reduction
Step 2: Write Half-Reactions
Fe²⁺ → Fe³⁺
Reduction:
MnO₄⁻ → Mn²⁺
Step 3: Balance Atoms Other Than H & O
In this example:
- Fe is already balanced.
- Mn is already balanced.
Move on.
Step 4: Balance Oxygen Using H₂O
MnO₄⁻ has 4 oxygen atoms.
So add 4 water molecules to the right:
Step 5: Balance Hydrogen Using H⁺ (Acidic Medium)
We added 4 water molecules → 8 H atoms on the right.
So add 8 H⁺ to the left:
Step 6: Balance Charge Using Electrons
Now check charges.
In oxidation reaction:
Charge increases by +1
So add 1 electron on the right:
In reduction reaction:
Left = 8H⁺ (+8) + MnO₄⁻ (−1) = +7 total
Right = Mn²⁺ (+2)
Difference = 5
So add 5 electrons to the left:
Step 7: Equalize Electrons
Oxidation has 1 electron, reduction has 5 electrons.
Multiply the oxidation half-reaction by 5:
Step 8: Add the Two Reactions and Cancel Electrons
Final balanced reaction:
Done! This entire balancing takes only 2–3 minutes once you practice it 3–4 times.
Balancing Redox in Basic Medium (Shortcut)
For basic medium, balance the reaction normally (as acidic), then:
1 Add OH⁻ to both sides to neutralize H⁺
2 Convert H⁺ + OH⁻ → H₂O
3 Simplify
Example:
Balanced in acidic:
Now convert to basic:
Add 8OH⁻ on both sides:
Left: 8H⁺ + 8OH⁻ → 8H₂O
Right already has water → simplify
Result after simplification:
This is the balanced form in basic medium.
Quick Practice Examples
Try these using the same steps:
If you want, I can solve these step-by-step as well.
Step-by-Step Solutions
Example 1 — Cr₂O₇²⁻ + Fe²⁺ → Cr³⁺ + Fe³⁺ (Acidic Medium)
Step A — Write the two half-reactions
Reduction (Cr): Cr₂O₇²⁻ → Cr³⁺
Step B — Balance atoms except H and O
Fe and Cr counts are already fine (Fe:1, Cr:2 on left give 2 Cr³⁺ needed on right).
Step C — Balance O by adding H₂O (in reduction half)
Cr₂O₇²⁻ has 7 O, so add 7 H₂O on the right:
Step D — Balance H by adding H⁺ (acidic medium)
7 H₂O → 14 H on right, so add 14 H⁺ to the left:
Step E — Balance charge by adding electrons
Left charge: 14(+1) + (−2) = +12. Right charge: 2(+3) = +6. To reduce +12 → +6, add 6 e⁻ to the left:
Oxidation half (balance charge):
Step F — Equalize electrons
Reduction uses 6 e⁻, oxidation gives 1 e⁻. Multiply oxidation by 6:
Step G — Add halves and cancel electrons
(14H⁺ + Cr₂O₇²⁻ + 6e⁻) + (6Fe²⁺ → 6Fe³⁺ + 6e⁻)
Cancel 6e⁻:
This is the balanced equation.
Example 2 — H₂O₂ → O₂ (Basic Medium)
Step A — Split into half-reactions
Oxidation: H₂O₂ → O₂
Reduction: H₂O₂ → OH⁻ (in basic medium)
Step B — Balance in acidic form first
Oxidation: H₂O₂ → O₂ + 2H⁺ + 2e⁻
Reduction: H₂O₂ + 2e⁻ → 2OH⁻
Step C — Add the two half-reactions
H₂O₂ + 2e⁻ → 2OH⁻
————————-
2H₂O₂ → O₂ + 2H⁺ + 2OH⁻
Step D — Convert H⁺ + OH⁻ to H₂O
2H⁺ + 2OH⁻ → 2H₂O, so we get:
This is the balanced equation for H₂O₂ → O₂ in basic medium.
Example 3 — SO₂ + MnO₄⁻ → SO₄²⁻ + Mn²⁺ (Acidic Medium)
Step A — Half-reactions
Reduction (Mn): MnO₄⁻ → Mn²⁺
Step B — Balance O by adding H₂O
Oxidation half: SO₂ → SO₄²⁻ needs 2 extra oxygens, add 2 H₂O on left:
Step C — Balance H by adding H⁺
Left has 4 H (from 2 H₂O), so add 4 H⁺ to the right:
Step D — Balance charge with electrons (oxidation)
Left charge 0; right: SO₄²⁻ (−2) + 4H⁺ (+4) = +2. Add 2 e⁻ to the right:
Step E — Reduction half (balance O and H)
Add 4 H₂O on the right to balance oxygen:
Add 8 H⁺ on the left to balance hydrogen:
Balance charge with electrons: left charge = +8 −1 = +7; right charge = +2. Add 5 e⁻ to the left:
Step F — Equalize electrons
Oxidation gave 2 e⁻. Reduction uses 5 e⁻. LCM = 10.
Multiply oxidation half by 5 and reduction half by 2:
Reduction ×2: 16H⁺ + 2MnO₄⁻ + 10e⁻ → 2Mn²⁺ + 8H₂O
Step G — Add and cancel
Add both:
This is the balanced equation in acidic medium.
Summary (What You Should Remember)
- Break into oxidation and reduction
- Balance atoms
- Balance O with H₂O
- Balance H with H⁺
- Balance charge with e⁻
- Make electrons equal
- Add, cancel, and simplify
Master this, and redox becomes one of the easiest scoring chapters in Chemistry.
If you enjoyed learning this way, you can learn many more topics on the Mohit Tyagi YouTube channel. And if you want full preparation support, feel free to explore our courses as well.
