Redox ReactionsNEET Chemistry · Class 11 · NCERT Chapter 7

Classical Concept of Oxidation and Reduction

In classical chemistry, oxidation meant combination with oxygen and reduction meant removal of oxygen. Over time, this definition was extended.

Classical definitions:

• Oxidation: addition of oxygen, or removal of hydrogen.
  Example: 2Mg + O₂ → 2MgO (Mg is oxidised by gaining O)
• Reduction: removal of oxygen, or addition of hydrogen.
  Example: CuO + H₂ → Cu + H₂O (CuO is reduced — loses O)

In the reaction above, H₂ is the reducing agent (it gets oxidised), and CuO is the oxidising agent (it gets reduced). Oxidation and reduction always occur together — you cannot have one without the other. This is why they are called redox reactions.

Electronic Concept of Oxidation and Reduction

The modern electronic definition is much broader and covers all redox reactions:

• Oxidation: loss of electrons (electron deficiency).
• Reduction: gain of electrons (electron surplus).

Memory trick: OIL RIG — Oxidation Is Loss, Reduction Is Gain.

Examples:

In the combination reaction Na + Cl₂ → NaCl, sodium is oxidised and chlorine is reduced. Chlorine acts as the oxidising agent (accepts electrons from Na). Sodium acts as the reducing agent (donates electrons to Cl).

Oxidation States

Select a compound to see the oxidation state of each element with step-by-step reasoning using the standard rules.

The oxidation state (or oxidation number) of an atom is a number that represents the charge the atom would have if the compound were fully ionic (electrons were fully assigned to the more electronegative atom).

Rules for Assigning Oxidation States

1. The oxidation state of any free element (in its standard form) is 0. Examples: Na, Fe, O₂, H₂, S₈ all have oxidation state 0.
2. The oxidation state of a monoatomic ion equals its charge. Examples: Na⁺ is +1, Fe³⁺ is +3, Cl⁻ is −1.
3. Fluorine is always −1 in compounds (it is the most electronegative element).
4. Oxygen is usually −2 in compounds. Exceptions: peroxides (H₂O₂, Na₂O₂) where O is −1; superoxides (KO₂) where O is −½; OF₂ where O is +2 (because F is more electronegative).
5. Hydrogen is +1 when bonded to non-metals (e.g., HCl, H₂O) and −1 when bonded to metals (metal hydrides like NaH, CaH₂).
6. The sum of oxidation states of all atoms in a neutral molecule = 0. The sum in an ion = charge of the ion.

Let us find the oxidation state of Cr in K₂Cr₂O₇:

So Cr is +6 in K₂Cr₂O₇.

Useful Oxidation States to Remember

Oxidising and Reducing Agents

In a redox reaction:

• The oxidising agent (oxidant) accepts electrons and gets reduced. Its oxidation state decreases.
• The reducing agent (reductant) donates electrons and gets oxidised. Its oxidation state increases.

Example: In the reaction 2KMnO₄ + 10FeSO₄ + 8H₂SO₄ → K₂SO₄ + 2MnSO₄ + 5Fe₂(SO₄)₃ + 8H₂O:

• Mn goes from +7 (in KMnO₄) to +2 (in MnSO₄): reduction of 5 per Mn → KMnO₄ is the oxidising agent.
• Fe goes from +2 (in FeSO₄) to +3 (in Fe₂(SO₄)₃): oxidation of 1 per Fe → FeSO₄ is the reducing agent.

Common oxidising agents and their reduced forms:

Types of Redox Reactions

1. Combination Reactions

Two reactants combine to form a single product. At least one element must change oxidation state.

Mg: 0 → +2 (oxidised); O: 0 → −2 (reduced).

2. Decomposition Reactions

A single compound breaks down. At least one element must change oxidation state.

O in H₂O₂: −1 → −2 (in H₂O, reduced) and −1 → 0 (in O₂, oxidised). This is a disproportionation.

3. Displacement Reactions

One element displaces another from its compound. The more reactive metal displaces the less reactive one.

Zn: 0 → +2 (oxidised); Cu: +2 → 0 (reduced). Zinc is above copper in the activity series, so it displaces copper.

4. Disproportionation Reactions

The same element is simultaneously oxidised and reduced. The oxidising agent and the reducing agent are the same substance.

Cl: 0 → −1 (in NaCl, reduced) and 0 → +1 (in NaOCl, oxidised).

5. Comproportionation Reactions

Two different oxidation states of the same element combine to give an intermediate oxidation state. The reverse of disproportionation.

Balancing Redox Equations: Oxidation Number Method

Steps:

1. Assign oxidation states to all atoms and identify which element is oxidised and which is reduced.
2. Calculate the change in oxidation state (increase for oxidation, decrease for reduction).
3. Multiply the formulas of the oxidised and reduced species so that the total increase in oxidation state equals the total decrease (electrons lost = electrons gained).
4. Balance the remaining atoms by inspection (first O, then H using H₂O).
5. Check the overall charge balance.

Example: Balance Fe²⁺ + MnO₄⁻ → Fe³⁺ + Mn²⁺ (in acidic medium).

Fe²⁺ → Fe³⁺: change = +1 (oxidised). Mn⁷⁺ → Mn²⁺: change = −5 (reduced). To balance: 5 Fe²⁺ for every 1 MnO₄⁻.

Balancing Redox Equations: Ion-Electron (Half-Reaction) Method

Walk through the half-reaction method for balancing redox equations. Reveal each step in sequence.

This method works well in acidic or basic media. Steps:

1. Split the reaction into two half-reactions: one for oxidation, one for reduction.
2. Balance each half-reaction:
   • Balance all atoms except H and O.
   • In acidic medium: add H₂O to balance O, then add H⁺ to balance H.
   • In basic medium: add OH⁻ to balance O and H.
   • Add electrons to balance the charge.
3. Multiply each half-reaction by an integer so that electrons lost = electrons gained.
4. Add the two half-reactions and cancel species that appear on both sides.

Example (acidic medium): Cr₂O₇²⁻ + Fe²⁺ → Cr³⁺ + Fe³⁺.

Reduction half: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O

Oxidation half: Fe²⁺ → Fe³⁺ + e⁻ (multiply by 6)

Electrochemical Cells

An electrochemical cell converts chemical energy into electrical energy using a spontaneous redox reaction. It is also called a galvanic cell or voltaic cell.

Structure of a Galvanic Cell (Daniell Cell)

The Daniell cell uses the reaction: Zn + Cu²⁺ → Zn²⁺ + Cu.

• Anode (negative electrode): Zinc electrode dipped in ZnSO₄ solution. Oxidation occurs here: Zn → Zn²⁺ + 2e⁻.
• Cathode (positive electrode): Copper electrode dipped in CuSO₄ solution. Reduction occurs here: Cu²⁺ + 2e⁻ → Cu.
• Salt bridge: Connects the two half-cells. Contains an inert electrolyte (like KCl or KNO₃) that completes the circuit by allowing ions to flow, maintaining electrical neutrality in both solutions.
• External circuit: Electrons flow from anode (Zn, where they are produced) through the external wire to cathode (Cu, where they are consumed).

Cell notation: Zn | Zn²⁺ (1 M) || Cu²⁺ (1 M) | Cu. The double line (||) represents the salt bridge. Anode is written on the left.

Electrode Potential and Electrochemical Series

The electrode potential (E) is the tendency of an electrode to gain or lose electrons. It is measured relative to the standard hydrogen electrode (SHE), which has E° = 0 V by convention.

The standard electrode potential E° is measured at 298 K, 1 atm, and 1 M concentration of all solutions.

The standard cell potential (electromotive force, EMF) is:

For a spontaneous cell reaction, E°_cell must be positive (ΔG = −nFE°_cell, so positive E means negative ΔG).

Electrochemical Series (Activity Series)

The electrochemical series lists standard reduction potentials from most negative (top, best reducing agents) to most positive (bottom, best oxidising agents).

Key uses of the electrochemical series:

• A metal with a more negative E° can displace a metal with a more positive E° from its salt solution. Zn (E° = −0.76 V) can displace Cu (E° = +0.34 V), but Cu cannot displace Zn.
• Metals above hydrogen in the series (E° < 0) can displace H₂ from dilute acids.
• For the Daniell cell: E°_cell = E°_Cu − E°_Zn = 0.34 − (−0.76) = 1.10 V.

Worked NEET Problems

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