Educational Galaxy

Chapter 7: Redox Reactions — EG's own summary, in simple words

Book: Chemistry — Chemistry — Part II (CBSE, Class 11)  |  Chapter 7  |  Medium: English

Oxidation numbers, balancing redox equations, and the electrochemical series.

Diagram (labels on the image): Zinc strip, Wire, Copper strip. Labels were added by Educational Galaxy on an originally generated illustration; see the caption below the second image for an honest note on what the picture shows.

1. 1. Oxidation and reduction, old and new

Classically, oxidation meant gaining oxygen or losing hydrogen, and reduction the reverse. The modern definition is cleaner and covers everything: oxidation is loss of electrons, reduction is gain of electrons. Remember OIL RIG — Oxidation Is Loss, Reduction Is Gain. In every redox reaction, one species is oxidised and another is reduced, and the electrons lost equal the electrons gained.

The oxidising agent is the species that causes oxidation — it is itself reduced, because it gains the electrons. The reducing agent causes reduction and is itself oxidised. Keeping those two the right way round is half the chapter.

2. 2. Oxidation number — the bookkeeping system

Oxidation number is the charge an atom would have if all its bonds were ionic. The rules, in the order to apply them: an element in its free state is 0; fluorine is always −1; oxygen is usually −2 (except in peroxides, −1, and in OF2, +2); hydrogen is usually +1 (except in metal hydrides, −1); group 1 metals are +1 and group 2 are +2; and the sum of oxidation numbers in a neutral compound is 0, in an ion it equals the ion's charge.

Oxidation number is not the same as real charge in a covalent compound — it is a bookkeeping device. Its value is that it lets you spot redox instantly: if any element's oxidation number changes during a reaction, the reaction is redox. It also lets you balance equations without tracking electrons directly.

Caption: An originally generated illustration for this chapter — a general scene for the topic, not a photograph of a real experiment or a copy of a textbook figure. The labelled diagram above is the one to study for the named parts.

3. 3. Balancing redox equations — two reliable methods

Oxidation-number method: write the oxidation numbers, find which elements change, work out the total increase and total decrease, and choose coefficients so the total increase equals the total decrease. Then balance the remaining atoms by inspection, adding H+ (acidic medium) or OH− (basic medium) and water as needed.

Ion-electron (half-reaction) method: split the reaction into an oxidation half and a reduction half. Balance each half for atoms (add H2O for oxygen, H+ for hydrogen), balance charge by adding electrons, then multiply the halves so the electrons cancel and add them together. In basic medium, add OH− to both sides to neutralise H+ into water. Both methods give the same balanced equation; use whichever you find more reliable and check that atoms and charge both balance at the end.

4. 4. Types of redox reactions and the electrochemical series

Combination and decomposition reactions can be redox (burning carbon is oxidation; heating potassium chlorate is internal redox). Displacement reactions are redox by nature: a more reactive metal displaces a less reactive one from its compound, and the displacing metal is oxidised. Disproportionation is special: one element is both oxidised and reduced, as when chlorine reacts with cold dilute alkali to give chloride and hypochlorite.

The electrochemical series ranks electrodes by their standard reduction potentials. A metal higher in the series (more negative potential) displaces metals below it from their salt solutions, and acts as the stronger reducing agent. This is why zinc displaces copper from copper sulphate, but copper cannot displace zinc — and why the series predicts the direction of displacement reactions without memorising each one.

5. 5. Redox in everyday life

Redox chemistry is everywhere: rusting is slow oxidation of iron; bleaching, photography and the working of batteries are all redox processes. In a galvanic cell, oxidation happens at the anode and reduction at the cathode, and electrons flow through the external wire from anode to cathode — the cell converts chemical energy into electrical energy. Corrosion prevention (painting, galvanising with zinc, sacrificial anodes) works by protecting the metal from oxidation or by giving oxidation an easier target.

Oxidation-number rules — quick reference

Situation	Oxidation number
Element in free state (e.g. Fe, O2)	0
Fluorine in compounds	−1 always
Oxygen	−2 (peroxides −1, OF2 +2)
Hydrogen	+1 (metal hydrides −1)
Group 1 metals / Group 2 metals	+1 / +2
Sum in a neutral compound / in an ion	0 / the ion's charge

Going further 1 — Redox titrations, and where oxidation numbers stop (NCERT §7.3.3–§7.3.4)

Redox reactions analyse solutions because the electron transfer can be counted. In a permanganate titration, MnO4− oxidises Fe2+ or oxalate in acid, and the permanganate itself is the indicator: the solution stays colourless while reductant remains, and the first lasting pink tinge appears when MnO4− reaches about 10^−6 mol/L, marking the end point with almost no overshoot. Iodine titrations work the other way: iodine liberated from iodide by an oxidant turns starch an intense blue, so the disappearance of the blue colour marks the end point.

The book then warns that oxidation number is only bookkeeping. In recent thinking, oxidation is visualised as a decrease in electron density around an atom and reduction as an increase — a picture that survives even where the rules give awkward answers, such as fractional oxidation numbers or atoms bonded to themselves. Oxidation numbers remain the fastest tool for spotting redox and balancing equations, but they are not real charges, and they must not be mistaken for them.

Exam note: in a permanganate titration the pink colour at the end point is excess MnO4− itself — no separate indicator is added.

Going further 2 — The hydrogen electrode and cell notation (NCERT §7.4)

Electrode potentials need a zero, and the book sets it at the standard hydrogen electrode: the potential of the hydrogen electrode is defined as 0.00 volts, and every other electrode potential is measured against it. A metal with a negative standard potential, like zinc at −0.76 V, loses electrons more readily than hydrogen and acts as a reducing agent; a metal with a positive potential, like copper at +0.34 V, gains electrons more readily.

Cells are written in shorthand: the anode on the left, the cathode on the right, a single line for a phase boundary and a double line for the salt bridge — Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s) is the Daniell cell. The cell voltage is cathode minus anode: 0.34 − (−0.76) = 1.10 V. Electrons flow through the wire from anode to cathode; ions carry the current inside the cell. Any cell with a positive voltage runs spontaneously as written.

Exam note: the anode is always where oxidation happens, in galvanic and electrolytic cells alike — AN OX is the memory aid, not 'positive electrode'.

Practice set — try it yourself

1. In the reaction Zn + CuSO4 → ZnSO4 + Cu, identify the species oxidised, the species reduced, the oxidising agent and the reducing agent.
2. Find the oxidation number of chromium in K2Cr2O7 and of sulphur in H2SO4. Show your working using the rules.
3. Balance by the ion-electron method in acidic medium: MnO4− + Fe2+ → Mn2+ + Fe3+.
4. Chlorine reacting with cold dilute sodium hydroxide gives chloride and hypochlorite. Explain why this is called disproportionation.
5. Why does zinc displace copper from copper sulphate solution, but copper cannot displace zinc from zinc sulphate? Use the electrochemical series.

Questions with Answers — full answers

1. Define oxidation and reduction in terms of electrons. Answer: Oxidation is the loss of electrons by a species; reduction is the gain of electrons. In a redox reaction the electrons lost by the reducing agent are gained by the oxidising agent, so the two processes always occur together.
2. What is the oxidation number of manganese in KMnO4? Answer: Let it be x. Potassium is +1 and each oxygen is −2, and the compound is neutral: (+1) + x + 4(−2) = 0, so x − 7 = 0 and x = +7. Manganese is in the +7 oxidation state.
3. Why is the reaction between hydrogen and oxygen a redox reaction? Answer: Hydrogen's oxidation number rises from 0 to +1 (oxidation, loss of electrons) while oxygen's falls from 0 to −2 (reduction, gain of electrons). Both changes happen in the same reaction, so it is redox.
4. In a galvanic cell, which electrode is the anode, and which way do electrons flow? Answer: The anode is where oxidation occurs. Electrons are released there and flow through the external wire to the cathode, where reduction occurs. Conventional current is in the opposite direction.
5. Give one everyday example of redox and explain the electron transfer in it. Answer: Rusting: iron is oxidised, losing electrons to form Fe2+ and eventually Fe2O3, while oxygen is reduced, gaining those electrons. Painting or galvanising prevents rusting by keeping oxygen and moisture away from the iron, or by letting zinc be oxidised instead.

Extension Questions with Answers — going further

1. Find the oxidation numbers of nitrogen in ammonium nitrate, NH4NO3. Answer: The compound has two nitrogen atoms in different ions. In NH4+, nitrogen is −3 (x + 4(+1) = +1). In NO3−, nitrogen is +5 (x + 3(−2) = −1). One compound, two oxidation states — which is why the average (+1) is meaningless here.
2. Balance: Cu + HNO3 → Cu(NO3)2 + NO2 + H2O (concentrated acid). Answer: Copper goes 0 to +2 (loses 2 electrons); nitrogen in nitric acid goes +5 to +4 in NO2 (gains 1 electron), so 2 HNO3 molecules act as oxidant. The balanced equation is Cu + 4HNO3 → Cu(NO3)2 + 2NO2 + 2H2O. Check: atoms and charge both balance.
3. Is 2H2O2 → 2H2O + O2 a disproportionation? Explain. Answer: Yes. Oxygen in H2O2 is −1. In H2O it is −2 (reduced) and in O2 it is 0 (oxidised). The same element is both oxidised and reduced, which is the definition of disproportionation.
4. Why is fluorine the strongest oxidising agent? Answer: Fluorine has the highest standard reduction potential, about +2.87 V: it gains electrons more readily than any other element. That is why it oxidises even water, liberating oxygen.
5. Calculate the EMF of the Daniell cell from E°(Zn2+/Zn) = −0.76 V and E°(Cu2+/Cu) = +0.34 V. Answer: EMF = E°cathode − E°anode = 0.34 − (−0.76) = 1.10 V. The positive value confirms the cell runs spontaneously as written, zinc oxidising and copper ions reducing.

Board-pattern Questions with Answers

1. In 2FeCl3 + SnCl2 → 2FeCl2 + SnCl4, identify the oxidising and reducing agents. Answer: Fe3+ is reduced to Fe2+, so FeCl3 is the oxidising agent. Sn2+ is oxidised to Sn4+, so SnCl2 is the reducing agent. The agent is always the opposite of what it does.
2. Balance by the oxidation-number method in acid: MnO4− + C2O4^2− → Mn2+ + CO2. Answer: Mn goes +7 to +2 (gain of 5 electrons); each carbon goes +3 to +4 (loss of 1 electron per carbon, 2 per oxalate). Multiplying to equalise electrons (2 permanganate, 5 oxalate) and balancing H+ and water gives 2MnO4− + 5C2O4^2− + 16H+ → 2Mn2+ + 10CO2 + 8H2O.
3. Given E° values Zn2+/Zn = −0.76 V, Cu2+/Cu = +0.34 V, Ag+/Ag = +0.80 V, arrange the metals in order of reducing power. Answer: Zn > Cu > Ag. The more negative the reduction potential, the stronger the metal as a reducing agent — it loses electrons most readily.
4. What is the oxidation number of oxygen in potassium superoxide, KO2? Answer: Potassium is +1 and the compound is neutral, so 2 oxygen atoms share −1: each oxygen is −1/2. Fractional oxidation numbers are allowed — another sign that oxidation number is bookkeeping, not charge.
5. In a Daniell cell, which electrode is the anode, and what happens to the mass of the zinc rod? Answer: Zinc is the anode, where oxidation happens: Zn → Zn2+ + 2e−. Zinc atoms leave the rod as ions, so the rod loses mass while the cell runs.