Redox Reactions

Written and reviewed by Akanksha Tiwari, Physics & Chemistry faculty at UniDrill (M.Sc. Physics, Banaras Hindu University · 8 years teaching Class XI–XII science). Based on NCERT Chemistry Class XI, Chapter 7.

Why does a cut apple turn brown?

Bite an apple and leave it. Within minutes the white flesh browns. An iron gate in the monsoon turns to flaky rust. A torch battery lights a bulb, and bleach whitens a shirt. These look unrelated, but all of them are the same event at the atomic level: electrons moving from one species to another. Chemists call these changes redox reactions. Master the bookkeeping of those electrons and you can predict products, balance any equation and understand how a battery works.

From oxygen to electrons: three ways to define it

The definition of oxidation grew as chemists saw more reactions.

Oxidation and reduction always happen together, which is why the word is redox. Take 2Na + Cl₂ → 2NaCl. Sodium loses electrons (2Na → 2Na⁺ + 2e⁻), so it is oxidised. Chlorine gains them (Cl₂ + 2e⁻ → 2Cl⁻), so it is reduced. Each of these is called a half reaction.

The agent and the process are opposite. The oxidising agent causes oxidation of the other species, but it is reduced. This is a favourite trap.

Competition for electrons. Put a zinc strip in blue copper sulphate and the strip is coated with red copper while the blue fades: Zn + Cu²⁺ → Zn²⁺ + Cu. Zinc wins the competition and the reaction goes almost to completion. A copper strip in silver nitrate turns the solution blue and grows a silver coat: Cu + 2Ag⁺ → Cu²⁺ + 2Ag. But Cu in zinc sulphate does nothing. The order of reducing power is Zn > Cu > Ag, and Co + Ni²⁺ ⇌ Co²⁺ + Ni stops partway because the two metals are close in strength. This ranking becomes the electrochemical series later in the chapter.

Oxidation number: electron bookkeeping

The oxidation number (ON) of an atom is the charge it would have if every bonding pair were handed to the more electronegative atom. It is a counting device, so it can be zero, negative or fractional.

Alkali metals are always +1 and alkaline earth metals +2 in compounds. For a compound, set up one equation and solve.

Fractional ONs are averages. In S₄O₆²⁻ the real structure is O₃S–S–S–SO₃. The two terminal sulphur atoms are +5 and the two middle ones are 0, which averages to +5/2. In C₃O₂ (O=C=C=C=O) the two end carbons are +2 and the middle one is 0, giving +4/3. Fe₃O₄ is FeO·Fe₂O₃ (Fe²⁺ and Fe³⁺, average +8/3), and Pb₃O₄ is 2PbO·PbO₂ (two Pb²⁺ and one Pb⁴⁺).

Stock notation writes the ON of a metal as a Roman numeral after its symbol: Au(I)Cl, Au(III)Cl₃, Sn(II)Cl₂, Sn(IV)Cl₄, Mn(IV)O₂, Fe₂(III)O₃. It tells you which oxidised or reduced form you mean.

What the ON tells you about behaviour. An element in its highest ON can only be reduced, so it can only be an oxidising agent. An element in its lowest ON can only be oxidised, so it can only be a reducing agent. In between, it can act as either. The highest ON of a main group element is its group number (or group number minus 10 for groups 13 to 17): Na +1, Mg +2, Al +3, Si +4, P +5, S +6, Cl +7.

Beyond the textbook. For the exam: the ON of C in CO₂ is +4 and of H in HCl is +1. In reality: these are not real charges. In CO₂ and HCl the electrons are shared unequally, not handed over, so carbon is far from a true C⁴⁺ ion. Real charge transfer happens only in ionic compounds like NaCl. Why it matters: the ON is a bookkeeping rule that gives correct stoichiometry, but you should not use it to predict bond polarity or geometry.

Types of redox reactions

Almost every redox reaction you will meet fits one of four patterns.

Not every combination or decomposition is redox. CaCO₃ → CaO + CO₂ has no change in ON anywhere, and no product is an element. All combustion of dioxygen is redox.

Pause and predict: in 2NaH → 2Na + H₂, which element is oxidised?

Check your answer: hydrogen. In the hydride H is −1, and it ends as H₂ at 0, so it loses electrons. Sodium goes from +1 to 0 and is reduced. Students who assume H is always +1 get this backwards.

Displacement reactions come in two kinds.

The halogen displacement is the basis of the layer test: shake with CCl₄ and the organic layer turns orange brown for Br₂ or violet for I₂.

Beyond the textbook. For the exam: a more active metal displaces a less active metal from its compound, e.g. Al displaces Cr from Cr₂O₃. In reality: the same idea is the thermite reaction, Fe₂O₃ + 2Al → Al₂O₃ + 2Fe, which releases so much heat that the iron comes out molten and is used to weld railway tracks. Why it matters: the reaction runs because aluminium is a much stronger reducing agent than iron.

Disproportionation needs an element in an intermediate ON that can go both up and down, so it must have at least three possible oxidation states.

The chlorine reaction is the basis of household bleach. Two cases cannot disproportionate. Fluorine has no positive ON, since it is the most electronegative element, so with alkali it gives a different reaction (2F₂ + 2OH⁻ → 2F⁻ + OF₂ + H₂O) in which fluorine is only reduced. Among the chlorine oxyanions ClO⁻ (+1), ClO₂⁻ (+3), ClO₃⁻ (+5) and ClO₄⁻ (+7), the last cannot disproportionate because chlorine is already at its highest ON of +7 and can only be reduced.

Balancing redox equations

There are two systematic methods. Both obey the same law: electrons lost equals electrons gained.

Worked example: oxidation number method. Balance Fe₂O₃ + CO → Fe + CO₂.

Worked example: half reaction method in acid. Balance Cr₂O₇²⁻ + SO₃²⁻ → Cr³⁺ + SO₄²⁻ in acid.

Worked example: basic medium. Balance MnO₄⁻ + I⁻ → MnO₂ + I₂ in base. Balance as if acidic first, then neutralise.

The common wrong step: adding OH⁻ to one side only, or leaving H⁺ in the final equation of a basic medium.

Pause and predict: can H⁺ appear in the final balanced equation of a reaction in basic solution?

Check your answer: no. Whenever H⁺ appears during balancing, add the same number of OH⁻ to both sides, combine H⁺ + OH⁻ into H₂O and cancel any water that repeats. The final basic equation shows OH⁻ and H₂O only.

Redox titrations

Redox reactions can be used to measure concentrations, and the main problem is spotting the end point. Three situations cover the exam.

The thiosulphate reaction is I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻, in which S goes from +2 to the average +5/2.

Worked example: a permanganate titration. How many mL of 0.020 M KMnO₄ are needed to oxidise 20 mL of 0.10 M FeSO₄ in acid? MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O.

The common wrong step: treating the reaction as 1 : 1. Always take the ratio from the balanced equation (here each MnO₄⁻ gains 5 electrons and each Fe²⁺ loses 1).

Beyond the textbook. For the exam: KMnO₄ titrations are done in acid, using dilute sulphuric acid. In reality: the acid is never HCl, because MnO₄⁻ (E° = +1.51 V) is strong enough to oxidise Cl⁻ (Cl₂/Cl⁻ is +1.36 V) and some of the permanganate would be wasted on the acid. Nitric acid is avoided because it is itself an oxidant. Why it matters: the choice of acid shows how electrode potentials decide which reactions can compete.

Redox and electricity: the Daniell cell

When Zn is placed in CuSO₄ solution, electrons jump directly from Zn to Cu²⁺ and the energy is lost as heat. Separate the two halves into two beakers, zinc rod in ZnSO₄ and copper rod in CuSO₄, join them with a wire and a salt bridge, and the electrons are forced through the wire. That is the Daniell cell: the same reaction, now producing current.

Each half cell is a redox couple: the oxidised and reduced forms of the same species, written oxidised first (Zn²⁺/Zn, Cu²⁺/Cu). Each has an electrode potential, the tendency of the couple to stay in its reduced or oxidised form. Under standard conditions (unit concentration, 1 bar for gases, 298 K) it is the standard electrode potential E°, measured against the standard hydrogen electrode, whose E° (H⁺/H₂) is fixed at 0.00 V.

Read the table with two rules. A more negative E° means the metal (reduced form) is a stronger reducing agent. A more positive E° means the oxidised form is a stronger oxidising agent. So the oxidised form of a couple with higher E° will oxidise the reduced form of a couple with lower E°. A negative E° also means the metal is a better reducing agent than hydrogen, which is why it can liberate H₂ from acids.

Worked example: will it react? Use the E° values above to decide.

The common wrong step: reading the sign backwards and saying that the metal with a more positive E° is the better reducing agent. It is the better oxidising agent (as an ion).

Exam traps

Reaction sheet

60-second revision

Where this chapter connects

• Before this: Equilibrium covered the balance of proton transfer, and Classification of Elements and Periodicity in Properties explains why elements have the oxidation states and electronegativity values you have just used.
• Next: Organic Chemistry: Some Basic Principles and Techniques moves to carbon compounds, where oxidation and reduction appear as gain of oxygen or loss of hydrogen.
• Forward link: Electrochemistry builds cell potentials and the Nernst equation on the Daniell cell, and The d and f Block Elements uses oxidation states and KMnO₄ and K₂Cr₂O₇ chemistry heavily.