Redox Reactions — Oxidation State, Balancing & Electrochemistry (Class 11)

What’s Actually Happening in a Redox Reaction

Most students memorise “oxidation is loss, reduction is gain” (OIL RIG) and stop there. But understanding why electrons transfer — and what drives that transfer — is what separates students who crack JEE from those who get tripped up in the inorganic section.

A redox reaction is any reaction where electrons shift from one species to another. The species that loses electrons gets oxidised; the species that gains electrons gets reduced. These two half-processes always happen together — you can’t have oxidation without a simultaneous reduction.

The key insight: atoms don’t lose electrons into thin air. They lose them to something. That “something” is the oxidising agent — which itself gets reduced in the process. This coupled nature is why we call them oxidation-reduction reactions, not just oxidation reactions.

Why does this matter for JEE and boards? Redox underlies half of inorganic chemistry. Corrosion, bleaching, respiration, combustion, electrochemistry — all redox. The chapter has consistent 2-3 mark weightage in JEE Main and always features in CBSE Class 11 and 12 practicals.

Key Terms and Definitions

Oxidation state (OS) — a hypothetical charge assigned to an atom in a molecule, assuming all bonds are ionic. It’s a bookkeeping tool, not a real charge, but it lets us track electron transfer.

Oxidising agent — the species that accepts electrons, causing another species to be oxidised. It gets reduced itself. Example: $\text{MnO}_4^-$ in acidic medium.

Reducing agent — the species that donates electrons, causing another species to be reduced. It gets oxidised itself. Example: $\text{Fe}^{2+}$ , oxalic acid.

Disproportionation — a special redox reaction where the same element simultaneously gets oxidised and reduced. Classic example: $\text{Cl}_2$ in water, or $\text{H}_2\text{O}_2$ decomposition.

Half-reaction — either just the oxidation part or just the reduction part of a redox reaction, written separately with electrons shown explicitly.

Oxidation Number Rules

These rules are applied in order — earlier rules override later ones.

Free element (uncombined): OS = 0. $\text{Fe}$ , $\text{O}_2$ , $\text{S}_8$ all have OS = 0.
Monatomic ion: OS = ionic charge. $\text{Fe}^{3+}$ has OS = +3.
Fluorine: always −1 (most electronegative element, no exceptions).
Oxygen: usually −2. Exceptions: peroxides (−1), superoxides (−½), $\text{OF}_2$ (+2).
Hydrogen: +1 when bonded to non-metals, −1 in metal hydrides ( $\text{NaH}$ , $\text{CaH}_2$ ).
Sum of OS in a neutral compound = 0.
Sum of OS in a polyatomic ion = charge of the ion.

Worked example — finding OS of Cr in $\text{K}_2\text{Cr}_2\text{O}_7$ :

Let OS of Cr = $x$ .

2(+1) + 2x + 7(-2) = 0

2 + 2x - 14 = 0

2x = 12 \implies x = +6

Cr is in +6 state — the classic dichromate.

Students often forget the subscripts when setting up the equation. In $\text{Cr}_2\text{O}_7^{2-}$ , there are two Cr atoms and seven O atoms. Write $2x$ not $x$ .

Balancing Redox Reactions

There are two methods: oxidation number method and half-reaction (ion-electron) method. JEE tests the half-reaction method more rigorously. CBSE boards accept either.

Method 1: Oxidation Number Method

The core idea: increase in oxidation state × number of atoms = decrease in oxidation state × number of atoms.

Steps:

Identify which atoms change OS.
Calculate the increase and decrease in OS.
Multiply formulas to equalise the total increase and decrease.
Balance remaining atoms (H, O) last.

Example: Balance $\text{MnO}_4^- + \text{Cl}^- \to \text{Mn}^{2+} + \text{Cl}_2$ (acidic medium)

Mn: +7 → +2, decrease of 5.
Cl: −1 → 0, increase of 1 per atom; but $\text{Cl}_2$ has 2 Cl atoms, so increase of 2 per molecule. To equalise: use 2 Cl $^-$ for 5 increase? We need LCM of 5 and 2 — multiply Mn equation by 2, Cl equation by 5:

2\text{MnO}_4^- + 10\text{Cl}^- \to 2\text{Mn}^{2+} + 5\text{Cl}_2

Now balance O (8 O on left, need 8 H $_2$ O on right) and then H:

2\text{MnO}_4^- + 10\text{Cl}^- + 16\text{H}^+ \to 2\text{Mn}^{2+} + 5\text{Cl}_2 + 8\text{H}_2\text{O}

Method 2: Half-Reaction (Ion-Electron) Method

This is the preferred method for electrochemistry problems and JEE Advanced.

Steps:

Split into oxidation half-reaction and reduction half-reaction.
Balance atoms other than O and H first.
Balance O by adding $\text{H}_2\text{O}$ .
Balance H by adding $\text{H}^+$ (acidic) or $\text{OH}^-$ (basic medium).
Balance charge by adding electrons ( $e^-$ ).
Multiply half-reactions so electrons cancel, then add.

Example: Balance $\text{MnO}_4^- + \text{Fe}^{2+} \to \text{Mn}^{2+} + \text{Fe}^{3+}$ (acidic medium)

Reduction half-reaction:

\text{MnO}_4^- \to \text{Mn}^{2+}

Balance O: $\text{MnO}_4^- \to \text{Mn}^{2+} + 4\text{H}_2\text{O}$ Balance H: $\text{MnO}_4^- + 8\text{H}^+ \to \text{Mn}^{2+} + 4\text{H}_2\text{O}$ Balance charge: Left = $-1 + 8 = +7$ ; Right = $+2$ . Add 5 $e^-$ to left:

\text{MnO}_4^- + 8\text{H}^+ + 5e^- \to \text{Mn}^{2+} + 4\text{H}_2\text{O}

Oxidation half-reaction:

\text{Fe}^{2+} \to \text{Fe}^{3+} + e^-

Multiply oxidation by 5 and add:

\text{MnO}_4^- + 8\text{H}^+ + 5\text{Fe}^{2+} \to \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}

In basic medium, after balancing in acidic medium, add equal $\text{OH}^-$ to both sides to neutralise $\text{H}^+$ , forming water. Simplify the water molecules. This is faster than starting from scratch in basic conditions.

Electrochemical Cells

Redox reactions can be made to do electrical work in an electrochemical cell. Two types you must know:

Galvanic (Voltaic) Cells

Spontaneous redox reactions produce electrical energy. The classic example: Daniell cell.

Anode (negative terminal): oxidation occurs. $\text{Zn} \to \text{Zn}^{2+} + 2e^-$
Cathode (positive terminal): reduction occurs. $\text{Cu}^{2+} + 2e^- \to \text{Cu}$
Salt bridge: maintains electrical neutrality by allowing ion flow between the two half-cells.

\text{Zn}(s) \mid \text{Zn}^{2+}(aq) \parallel \text{Cu}^{2+}(aq) \mid \text{Cu}(s)

Single vertical line = phase boundary. Double vertical line = salt bridge. Anode is always written on the left.

Electrolytic Cells

Non-spontaneous redox reactions are driven by external electrical energy. Used in electroplating, electrolysis of water, chlor-alkali process.

CBSE boards frequently ask: “What is the role of a salt bridge?” Answer in two points — (1) completes the electrical circuit by allowing ion migration, (2) maintains electrical neutrality in both half-cells. One-line answers lose marks.

Solved Examples

Example 1 — CBSE Level

Q: Find the oxidation state of S in $\text{H}_2\text{SO}_4$ .

Let OS of S = $x$ .

2(+1) + x + 4(-2) = 0 \implies 2 + x - 8 = 0 \implies x = +6

S is in the +6 oxidation state.

Example 2 — JEE Main Level

Q: Identify the oxidising and reducing agents in:

\text{Br}_2 + 2\text{KOH} \to \text{KBr} + \text{KBrO} + \text{H}_2\text{O}

Br in $\text{Br}_2$ : OS = 0 Br in KBr: OS = −1 (reduction — gained electrons) Br in KBrO: OS = +1 (oxidation — lost electrons)

This is a disproportionation reaction. $\text{Br}_2$ is both the oxidising agent and the reducing agent.

Example 3 — JEE Main Level

Q: In the reaction $\text{XeF}_4 + \text{H}_2\text{O} \to \text{Xe} + \text{O}_2 + \text{HF}$ , identify if this is a redox reaction.

In $\text{XeF}_4$ : OS of Xe = +4 (F is −1, so $x + 4(-1) = 0 \implies x = +4$ ). In products: Xe → OS = 0 (reduction), $\text{O}_2$ → OS = 0 (H in H₂O was −2, so O is oxidised from −2 to 0).

Yes, this is a redox reaction and also a disproportionation — both Xe is reduced and O is oxidised.

Example 4 — JEE Advanced Level

Q: Balance in acidic medium:

\text{Cr}_2\text{O}_7^{2-} + \text{C}_2\text{O}_4^{2-} \to \text{Cr}^{3+} + \text{CO}_2

Reduction: $\text{Cr}_2\text{O}_7^{2-} \to 2\text{Cr}^{3+}$ Balance O: $\text{Cr}_2\text{O}_7^{2-} \to 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$ Balance H: $\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ \to 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$ Balance charge: Left = $-2 + 14 = +12$ ; Right = $+6$ . Add 6 $e^-$ to left:

\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \to 2\text{Cr}^{3+} + 7\text{H}_2\text{O}

Oxidation: $\text{C}_2\text{O}_4^{2-} \to 2\text{CO}_2 + 2e^-$

Multiply oxidation by 3:

3\text{C}_2\text{O}_4^{2-} \to 6\text{CO}_2 + 6e^-

Add both:

\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 3\text{C}_2\text{O}_4^{2-} \to 2\text{Cr}^{3+} + 6\text{CO}_2 + 7\text{H}_2\text{O}

Exam-Specific Tips

CBSE Class 11 Boards: Balancing by the oxidation number method is sufficient. However, the half-reaction method earns full marks and is faster once you practise it. Questions on identifying oxidising/reducing agents appear almost every year — 2-3 marks.

JEE Main: This chapter has appeared every year in recent sessions. JEE Main 2024 Shift 1 (January) had a question on calculating OS in a complex oxyanion. JEE Main 2023 tested disproportionation. Focus on: (1) OS in unusual compounds like $\text{XeF}_6$ , $\text{KO}_2$ , $\text{Na}_2\text{O}_2$ ; (2) identifying disproportionation; (3) balancing in both acidic and basic medium.

Oxygen exceptions are a high-yield area:

Compound	O Oxidation State
Normal oxides ( $\text{Na}_2\text{O}$ , $\text{CO}_2$ )	−2
Peroxides ( $\text{H}_2\text{O}_2$ , $\text{Na}_2\text{O}_2$ )	−1
Superoxides ( $\text{KO}_2$ )	−½
$\text{OF}_2$	+2
$\text{O}_2\text{F}_2$	+1

Memorise this table. It appears in roughly 30% of redox questions at JEE Main level.

Common Mistakes to Avoid

Mistake 1 — Confusing oxidising agent with the species that gets oxidised. The oxidising agent causes oxidation in something else, so it itself gets reduced. The reducing agent gets oxidised. Many students mix this up under pressure. Trick: the oxidising agent is the one that’s “used up” — it takes electrons.

Mistake 2 — Forgetting subscripts in OS calculations. In $\text{Fe}_3\text{O}_4$ , there are three Fe atoms. Set up: $3x + 4(-2) = 0 \implies x = +8/3$ . This is a real fractional OS (Fe₃O₄ contains both Fe²⁺ and Fe³⁺). Don’t panic at fractions.

Mistake 3 — Treating $\text{Na}_2\text{O}_2$ as a normal oxide. $\text{Na}_2\text{O}_2$ is sodium peroxide, so O has OS = −1, not −2. This is a favourite trick in MCQs. If OS of Na is +1 and O were −2, the formula would be Na₂O. The $\text{O}_2^{2-}$ unit in peroxides always has OS = −1 per oxygen.

Mistake 4 — Not balancing charge in half-reactions before combining. Students sometimes forget to add electrons, combine the half-reactions, and end up with an equation where charge doesn’t balance. Always do a charge check before combining: left side charge = right side charge in each half-reaction.

Mistake 5 — Wrong medium for balancing. If the problem says “acidic medium”, use $\text{H}^+$ and $\text{H}_2\text{O}$ . If it says “basic medium”, use $\text{OH}^-$ and $\text{H}_2\text{O}$ . Never mix them. In CBSE theory questions, the medium is always specified.

Practice Questions

Q1. Find the OS of Mn in $\text{KMnO}_4$ .

$x + (+1) + 4(-2) = 0 \implies x = +7$ . Mn is in the +7 state — the highest common OS of manganese.

Q2. Find the OS of Cl in $\text{NaClO}_3$ .

$(+1) + x + 3(-2) = 0 \implies x = +5$ . Cl is in the +5 state (chlorate ion).

Q3. Is the following a redox reaction? $\text{CaO} + \text{CO}_2 \to \text{CaCO}_3$

No. OS of Ca remains +2, O remains −2 throughout, C remains +4 in CO₂ and +4 in CaCO₃. No change in oxidation states means no electron transfer — this is a combination reaction but not a redox reaction.

Q4. Identify the disproportionation reaction: (a) $\text{Cl}_2 + 2\text{KOH} \to \text{KCl} + \text{KClO} + \text{H}_2\text{O}$ (b) $2\text{H}_2\text{O}_2 \to 2\text{H}_2\text{O} + \text{O}_2$ (c) Both

Both are disproportionation reactions. In (a), Cl goes from 0 to −1 (reduced in KCl) and 0 to +1 (oxidised in KClO). In (b), O goes from −1 to −2 (reduced in H₂O) and −1 to 0 (oxidised in O₂). The same element plays both roles.

Q5. Balance in basic medium: $\text{MnO}_4^- + \text{I}^- \to \text{MnO}_2 + \text{I}_2$

Reduction: $\text{MnO}_4^- + 2\text{H}_2\text{O} + 3e^- \to \text{MnO}_2 + 4\text{OH}^-$

Oxidation: $2\text{I}^- \to \text{I}_2 + 2e^-$

Multiply reduction by 2 and oxidation by 3:

2\text{MnO}_4^- + 4\text{H}_2\text{O} + 6e^- \to 2\text{MnO}_2 + 8\text{OH}^-

6\text{I}^- \to 3\text{I}_2 + 6e^-

Combined: $2\text{MnO}_4^- + 6\text{I}^- + 4\text{H}_2\text{O} \to 2\text{MnO}_2 + 3\text{I}_2 + 8\text{OH}^-$

Q6. In the Daniell cell, which electrode is the anode and why?

The zinc electrode is the anode because oxidation occurs there: $\text{Zn} \to \text{Zn}^{2+} + 2e^-$ . Zinc has a lower reduction potential (−0.76 V vs SHE) than copper (+0.34 V vs SHE), so Zn spontaneously loses electrons to Cu²⁺ ions.

Q7. What is the OS of Fe in $\text{Fe}_3\text{O}_4$ ?

$3x + 4(-2) = 0 \implies x = +8/3$ . This is a fractional average. Fe₃O₄ is a mixed oxide containing one Fe²⁺ and two Fe³⁺ ions (average = 8/3). Both OS coexist in the same compound.

Q8. Which is the stronger oxidising agent: $\text{F}_2$ or $\text{Cl}_2$ ? Give reason.

$\text{F}_2$ is the stronger oxidising agent. It has the highest reduction potential (+2.87 V) of all halogens. Its small size and weak F−F bond make it easy to accept electrons. $\text{Cl}_2$ (+1.36 V) is a weaker oxidising agent in comparison, though still strong enough to oxidise Br⁻ and I⁻.

FAQs

What is the difference between oxidation state and valency?

Valency is the combining capacity of an element — always a positive whole number. Oxidation state is an assigned charge (can be negative, zero, or fractional) used to track electron shifts in redox reactions. Carbon has a valency of 4, but its OS ranges from −4 (in CH₄) to +4 (in CO₂).

How do I know whether a reaction is redox or not?

Check if any element changes its oxidation state between reactants and products. If at least one element’s OS changes, it’s a redox reaction. If all OS values stay the same, it’s not (e.g., precipitation, acid-base, and most complex formation reactions are non-redox).

Why does $\text{H}_2\text{O}_2$ act as both oxidising and reducing agent?

O in $\text{H}_2\text{O}_2$ is in the −1 state, which is intermediate between 0 and −2. With a stronger oxidising agent, $\text{H}_2\text{O}_2$ donates electrons (O goes from −1 to 0, acting as reducing agent). With a stronger reducing agent, $\text{H}_2\text{O}_2$ accepts electrons (O goes from −1 to −2, acting as oxidising agent).

What exactly does a salt bridge do?

It serves two functions: (1) completes the internal circuit by allowing ions to migrate between the two half-cells, and (2) maintains electrical neutrality by supplying or absorbing ions as charges build up. Without it, the reaction stops almost immediately as one half-cell becomes positively charged and the other negatively charged.

Is the OS of O always −2?

No — this is one of the most common misconceptions in Class 11. O is −2 in normal oxides, −1 in peroxides ( $\text{H}_2\text{O}_2$ , $\text{Na}_2\text{O}_2$ ), −½ in superoxides ( $\text{KO}_2$ ), and +2 in $\text{OF}_2$ (because F is more electronegative than O and always takes priority as −1).

Can oxidation state be a fraction?

Yes. $\text{Fe}_3\text{O}_4$ has Fe at +8/3, $\text{KO}_2$ has O at −½. These are averages — the actual atoms in the compound have integer OS values, but the average across all atoms of that element works out to a fraction.

How is the half-reaction method better than the oxidation number method?

The half-reaction method is more systematic for complex reactions in solution, especially when the medium matters (acidic vs basic). It’s also the foundation for electrochemistry calculations. For simple reactions, both methods are equally fast; for ionic reactions in aqueous solution, half-reaction is less error-prone.