Redox Reactions Class 11: Oxidation Number, Balancing Methods and Electrode Processes

What Redox Reactions is about

Weightage: This chapter is Unit VII, Redox Reactions, which the syllabus gives 4 of the 70 theory marks in the school annual exam.Source: CBSE Chemistry (043) syllabus 2026-27, Class XI course structure, Unit VII Redox Reactions

Key concepts

Classical idea of oxidation and reduction

NCERT section 7.1 Classical Idea of Redox Reactions – Oxidation and Reduction Reactions

Originally, oxidation meant adding oxygen to a substance: $2Mg+O_2\to 2MgO$. The idea was then widened:

• Oxidation: addition of oxygen or an electronegative element (Cl, F, S), or removal of hydrogen or an electropositive element. Example: $H_{2}S+C{l}_2\to 2HCl+S$ (H₂S loses hydrogen, so it is oxidised).
• Reduction: removal of oxygen or an electronegative element, or addition of hydrogen or an electropositive element. Example: $2HgO\to 2Hg+O_2$ (HgO loses oxygen, so it is reduced).

Oxidation and reduction always occur together, so such reactions are called redox reactions.

• Oxidising agent (oxidant): the substance that causes oxidation and is itself reduced.
• Reducing agent (reductant): the substance that causes reduction and is itself oxidised.

In $CuO+H_2\to Cu+H_{2}O$, CuO is reduced (it is the oxidant) and H₂ is oxidised (it is the reductant).

Worked example

Using the classical definitions, identify the substance oxidised and the substance reduced in (a) $ZnO+C\to Zn+CO$ and (b) $2H_{2}S+S{O}_2\to 3S+2H_{2}O$. Name the oxidising agent in each. (2 marks)

Try it first, then show the solutionSolution

1. (a) ZnO loses oxygen, so it is reduced and is the oxidising agent; carbon gains oxygen, so it is oxidised (the reducing agent). (1)
2. (b) H₂S loses hydrogen, so it is oxidised; SO₂ loses oxygen, so it is reduced and is the oxidising agent. (1)

Answer: (a) C oxidised, ZnO reduced (oxidant ZnO) (b) H₂S oxidised, SO₂ reduced (oxidant SO₂)

Redox reactions as electron transfer; competitive electron transfer

NCERT section 7.2 Redox Reactions in Terms of Electron Transfer Reactions (7.2.1 Competitive Electron Transfer Reactions)

In $2Na+C{l}_2\to 2NaCl$, each sodium atom gives one electron to chlorine. Splitting the reaction into half-reactions:

• $2Na\to 2N{a}^{+}+2e^{-}$ (oxidation half: loss of electrons)
• $C{l}_2+2e^{-}\to 2C{l}^{-}$ (reduction half: gain of electrons)

So oxidation is loss of electrons and reduction is gain of electrons. The oxidant accepts electrons; the reductant donates them.

Competitive electron transfer: a strip of zinc in copper sulphate solution gets coated with reddish copper and the blue colour fades: $Zn(s)+C{u}^{2+}(aq)\to Z{n}^{2+}(aq)+Cu(s)$. Zinc gives electrons to Cu²⁺. The reverse does not happen: copper in zinc sulphate shows no change. Copper, in turn, displaces silver from silver nitrate solution. Hence the tendency to release electrons is Zn > Cu > Ag. Arranging metals by this tendency gives the activity series (and, quantitatively, the electrochemical series).

This competition also exists at equilibrium; for Zn and Cu²⁺ the equilibrium lies so far to the right that it seems complete.

Worked example

Write the oxidation and reduction half-reactions for $2F{e}^{3+}+S{n}^{2+}\to 2F{e}^{2+}+S{n}^{4+}$, and name the oxidant and reductant. (2 marks)

Try it first, then show the solutionSolution

1. Oxidation: $S{n}^{2+}\to S{n}^{4+}+2e^{-}$; reduction: $2F{e}^{3+}+2e^{-}\to 2F{e}^{2+}$. (1)
2. Fe³⁺ accepts electrons, so it is the oxidant; Sn²⁺ donates electrons, so it is the reductant. (1)

Answer: Oxidant Fe³⁺; reductant Sn²⁺

Oxidation number: rules and calculation

NCERT section 7.3 Oxidation Number

The oxidation number (oxidation state) is the charge an atom would have if all its bonds were fully ionic, with shared electrons given to the more electronegative atom. Rules:
1. An element in its free state has oxidation number 0 (H₂, O₂, Cl₂, P₄, S₈, Na, Cu).
2. For a monatomic ion, it equals the charge (Na⁺ = +1, Cl⁻ = −1, Al³⁺ = +3). Alkali metals are +1 and alkaline earth metals +2 in compounds; Al is +3.
3. Oxygen is usually −2; in peroxides (H₂O₂, Na₂O₂) it is −1; in superoxides (KO₂) −½; in OF₂ +2 and in O₂F₂ +1.
4. Hydrogen is +1 with non-metals and −1 in metal hydrides (NaH, CaH₂).
5. Fluorine is always −1; other halogens are −1 in halides, but positive when bonded to O (as in oxoacids).
6. The sum of oxidation numbers is 0 in a neutral compound and equals the charge in a polyatomic ion.

For an element in a compound, its highest oxidation number often equals its group valence electrons (S up to +6, Cl up to +7, Mn up to +7). Fractional values (Fe in Fe₃O₄: +8/3) are averages over atoms in different states.

Stock notation shows the oxidation number of a metal in Roman numerals: FeO is iron(II) oxide, Fe₂O₃ is iron(III) oxide, Cu₂O is copper(I) oxide.

Redox in terms of oxidation number: oxidation is an increase in oxidation number; reduction is a decrease. The oxidant contains the element whose oxidation number falls.

Worked example

Find the oxidation number of (a) Cr in K₂Cr₂O₇, (b) S in Na₂S₄O₆ (average), and (c) Mn in KMnO₄. (3 marks)

Try it first, then show the solutionSolution

1. (a) 2(+1) + 2x + 7(−2) = 0 → 2x = 12 → x = +6. (1)
2. (b) 2(+1) + 4x + 6(−2) = 0 → 4x = 10 → x = +2.5 (average of S atoms in different states). (1)
3. (c) +1 + x + 4(−2) = 0 → x = +7. (1)

Answer: (a) +6 (b) +2.5 (average) (c) +7

Types of redox reactions

NCERT section 7.3.1 Types of Redox Reactions

1. Combination: two substances combine, at least one being an element in its free state. $C+O_2\to C{O}_2$; $3Mg+N_2\to M{g}_3{N}_2$.
2. Decomposition: a compound breaks into components, at least one being an element. $2NaH\to 2Na+H_2$; $2KCl{O}_3\to 2KCl+3O_2$. Not every decomposition is redox: $CaC{O}_3\to CaO+C{O}_2$ involves no change in oxidation number.
3. Displacement: an ion or atom in a compound is replaced by another element. $X+YZ\to XZ+Y$.

• Metal displacement: $Zn+CuS{O}_4\to ZnS{O}_4+Cu$; also used in metallurgy, e.g. $C{r}_2{O}_3+2Al\to A{l}_2{O}_3+2Cr$.
• Non-metal displacement: hydrogen displacement by active metals from water or acids ($2Na+2H_{2}O\to 2NaOH+H_2$), and halogen displacement: F₂ > Cl₂ > Br₂ > I₂ as oxidants, so chlorine displaces bromine and iodine from their salts ($C{l}_2+2KBr\to 2KCl+B{r}_2$). Fluorine is so reactive that it even oxidises water to oxygen.

4. Disproportionation: the same element in one oxidation state is simultaneously oxidised and reduced. The element must have at least three possible oxidation states, and it must be in an intermediate one.

• $2H_2{O}_2\to 2H_{2}O+O_2$ (O: −1 → −2 and 0)
• $C{l}_2+2O{H}^{-}\to Cl{O}^{-}+C{l}^{-}+H_{2}O$ (Cl: 0 → +1 and −1)
• $P_4+3O{H}^{-}+3H_{2}O\to P{H}_3+3H_{2}P{O}_2^{-}$ (P: 0 → −3 and +1)

Species in the highest oxidation state (ClO₄⁻, Cl +7) or lowest (F⁻) cannot disproportionate. Fluorine, having only −1 and 0 in compounds, does not show disproportionation in this way.

Worked example

Which of Cl₂, F₂, NO₂ and MnO₄⁻ can undergo disproportionation, and why? Write the disproportionation of NO₂ in water, which gives nitric acid and nitrous acid. (3 marks)

Try it first, then show the solutionSolution

1. Cl₂ (Cl = 0) and NO₂ (N = +4) are in intermediate oxidation states, so they can be both oxidised and reduced. (1)
2. F₂ cannot: fluorine has no positive oxidation state, so it can only be reduced. MnO₄⁻ cannot: Mn is +7, its highest state, so it can only be reduced. (1)
3. $2N{O}_2+H_{2}O\to HN{O}_3+HN{O}_2$ (N: +4 → +5 and +3). (1)

Answer: Cl₂ and NO₂ can; F₂ and MnO₄⁻ cannot. 2NO₂ + H₂O → HNO₃ + HNO₂

Balancing redox reactions: oxidation number method

NCERT section 7.3.2 Balancing of Redox Reactions (a) Oxidation Number Method

Steps:
1. Write the skeletal equation with correct formulas of reactants and products.
2. Assign oxidation numbers and identify the atoms whose oxidation numbers change.
3. Find the increase for the oxidised atom and the decrease for the reduced atom (per formula, multiplying by the number of such atoms).
4. Multiply by suitable coefficients so that the total increase equals the total decrease.
5. Balance the charges: in acidic solution add H⁺ to the deficient side; in basic solution add OH⁻.
6. Balance H by adding water; check that O balances.

This method is quick when only two elements change oxidation number.

Worked example

Balance by the oxidation number method, in acidic solution: $C{r}_2{O}_7^{2-}+I^{-}\to C{r}^{3+}+I_2$ (skeletal). (3 marks)

Try it first, then show the solutionSolution

1. Cr: +6 → +3, decrease 3 per Cr, so 6 per Cr₂O₇²⁻. I: −1 → 0, increase 1 per I⁻. (1)
2. Equalise: 1 Cr₂O₇²⁻ needs 6 I⁻: Cr₂O₇²⁻ + 6I⁻ → 2Cr³⁺ + 3I₂. (1)
3. Charge: left −2 − 6 = −8, right +6. Add 14H⁺ on the left (−8 + 14 = +6), then 7H₂O on the right: $C{r}_2{O}_7^{2-}+6I^{-}+14H^{+}\to 2C{r}^{3+}+3I_2+7H_{2}O$. Check O: 7 = 7. (1)

Answer: Cr₂O₇²⁻ + 6I⁻ + 14H⁺ → 2Cr³⁺ + 3I₂ + 7H₂O

Balancing redox reactions: half-reaction (ion-electron) method

NCERT section 7.3.2 Balancing of Redox Reactions (b) Half Reaction Method

Steps:
1. Write the unbalanced ionic equation and split it into an oxidation half and a reduction half.
2. In each half, balance the atoms other than O and H.
3. Balance O by adding H₂O; then balance H by adding H⁺ (acidic solution).
4. In basic solution, add as many OH⁻ to both sides as there are H⁺, combine H⁺ + OH⁻ into H₂O, and cancel extra water.
5. Balance charge by adding electrons to the more positive side.
6. Multiply the halves so that the electrons lost equal the electrons gained, add them, and cancel common species.
7. Check that atoms and charges balance.

This method shows the electrons explicitly and works well for reactions in solution.

Worked example

Balance by the half-reaction method in basic solution: permanganate oxidises sulphite to sulphate and is reduced to manganese dioxide. (3 marks)

Try it first, then show the solutionSolution

1. Reduction (Mn +7 → +4): MnO₄⁻ has 4 O and MnO₂ has 2, so add 2H₂O on the right, 4H⁺ on the left, and 3e⁻ to balance charge: $Mn{O}_4^{-}+4H^{+}+3e^{-}\to Mn{O}_2+2H_{2}O$. (1)
2. Oxidation: $S{O}_3^{2-}+H_{2}O\to S{O}_4^{2-}+2H^{+}+2e^{-}$. Multiply reduction by 2 and oxidation by 3 (6 electrons each) and add: 2MnO₄⁻ + 3SO₃²⁻ + 2H⁺ → 2MnO₂ + 3SO₄²⁻ + H₂O. (1)
3. Basic solution: add 2OH⁻ to both sides; 2H⁺ + 2OH⁻ = 2H₂O, cancel one water: $2Mn{O}_4^{-}+3S{O}_3^{2-}+H_{2}O\to 2Mn{O}_2+3S{O}_4^{2-}+2O{H}^{-}$. Charge: −8 = −8. (1)

Answer: 2MnO₄⁻ + 3SO₃²⁻ + H₂O → 2MnO₂ + 3SO₄²⁻ + 2OH⁻

Redox titrations and limitations of oxidation number

NCERT section 7.3.3 Redox Reactions as the Basis for Titrations; 7.3.4 Limitations of Concept of Oxidation Number

In a redox titration, an oxidant of known concentration reacts with a reductant (or vice versa), and the end point is detected by a colour change.
1. KMnO₄ (self-indicator): MnO₄⁻ is intensely purple, Mn²⁺ almost colourless. In acidic solution, the first drop of excess permanganate after all the reductant (Fe²⁺ or C₂O₄²⁻) has reacted gives a permanent pale pink colour. $Mn{O}_4^{-}+5F{e}^{2+}+8H^{+}\to M{n}^{2+}+5F{e}^{3+}+4H_{2}O$.
2. K₂Cr₂O₇: the orange-to-green change is not sharp enough, so an external redox indicator such as diphenylamine is used, which gives an intense blue colour at the end point.
3. Iodometric titration: iodine with starch gives an intense blue. An oxidant such as Cu²⁺ liberates I₂ from iodide, and the I₂ is titrated with sodium thiosulphate: $I_2+2S_2{O}_3^{2-}\to 2I^{-}+S_4{O}_6^{2-}$. The blue colour vanishes at the end point.

Limitations of the oxidation number concept: oxidation numbers are formal charges assigned by rules; they do not always describe the real distribution of electrons (for example, in molecules with covalent bonds, or in compounds such as Fe₃O₄ where the average is fractional). Modern theory describes oxidation as a decrease in electron density around an atom, and reduction as an increase.

Worked example

20.0 mL of 0.020 M KMnO₄ is needed to oxidise all the Fe²⁺ in 25.0 mL of a ferrous solution in acid. Find the molarity of Fe²⁺. What colour marks the end point? (3 marks)

Try it first, then show the solutionSolution

1. Moles of MnO₄⁻ = 0.020 × 0.0200 = 4.0 × 10⁻⁴ mol. (1)
2. From MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O, Fe²⁺ = 5 × 4.0 × 10⁻⁴ = 2.0 × 10⁻³ mol; molarity = 2.0 × 10⁻³ ÷ 0.0250 = 0.080 M. (1)
3. End point: the first permanent pale pink colour from a drop of excess KMnO₄. (1)

Answer: 0.080 M Fe²⁺; permanent pale pink end point

Redox reactions and electrode processes

NCERT section 7.4 Redox Reactions and Electrode Processes

When zinc is dipped in copper sulphate, the electrons pass directly. If we separate the two halves, the electrons must flow through a wire, giving an electric current. This is a Daniell cell:

• A zinc rod in ZnSO₄ solution (oxidation: $Zn\to Z{n}^{2+}+2e^{-}$) and a copper rod in CuSO₄ solution (reduction: $C{u}^{2+}+2e^{-}\to Cu$).
• The solutions are connected by a salt bridge (KCl or NH₄NO₃ in agar-agar gel), which completes the circuit and keeps both solutions electrically neutral.
• Electrons flow from zinc to copper through the external wire.

Each metal in contact with its ions is a redox couple (Zn²⁺/Zn, Cu²⁺/Cu). The potential developed at the electrode–solution interface is the electrode potential. Measured under standard conditions (1 M ions, 1 bar gases, 298 K), it is the standard electrode potential, E⦵.

It cannot be measured alone, so it is measured against the standard hydrogen electrode (SHE), given E⦵ = 0.00 V: a platinum electrode coated with platinum black, dipped in 1 M H⁺, with H₂ bubbled at 1 bar.

By convention E⦵ values are reduction potentials:

• A large positive E⦵ means the oxidised form is a strong oxidising agent (F₂/F⁻ +2.87 V).
• A large negative E⦵ means the reduced form is a strong reducing agent (Li⁺/Li −3.05 V).
• Zn²⁺/Zn −0.76 V; Cu²⁺/Cu +0.34 V.

The list of E⦵ values is the electrochemical series. The cell's standard emf is $E_{cell}^{\ominus }=E_{cathode}^{\ominus }-E_{anode}^{\ominus }$; for the Daniell cell, 0.34 − (−0.76) = 1.10 V.

Worked example

Using E⦵(Mg²⁺/Mg) = −2.36 V and E⦵(Cu²⁺/Cu) = +0.34 V, (a) predict whether magnesium will displace copper from CuSO₄ solution, (b) write the cell reaction and (c) find the standard emf of the cell. (3 marks)

Try it first, then show the solutionSolution

1. (a) Mg has the much lower (more negative) reduction potential, so it is the better reducing agent and gives electrons to Cu²⁺: yes, magnesium displaces copper. (1)
2. (b) $Mg+C{u}^{2+}\to M{g}^{2+}+Cu$ (Mg is the anode, Cu the cathode). (1)
3. (c) E⦵cell = E⦵cathode − E⦵anode = 0.34 − (−2.36) = 2.70 V. (1)

Answer: Yes; Mg + Cu²⁺ → Mg²⁺ + Cu; E⦵cell = 2.70 V

Chemistry in real life: Redox Reactions

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Oxidation Number and Balancing Redox Equations
• Redox Titrations, Electrode Potential and Displacement Reactions

Redox Reactions formula sheet

Sum of oxidation numbers

$\sum (\text{oxidation numbers})=\text{charge on species}$

Zero for a neutral compound.

Oxidation and reduction

$\text{oxidation}=\text{loss of }{e}^{-}=\text{increase in O.N.};\text{reduction}=\text{gain of }{e}^{-}=\text{decrease in O.N.}$

Oxidant is reduced; reductant is oxidised.

Balancing by oxidation number

$\text{total increase in O.N.}=\text{total decrease in O.N.}$

Then balance charge with H⁺ (acid) or OH⁻ (base), and H with H₂O.

Permanganate–iron(II)

$Mn{O}_4^{-}+5F{e}^{2+}+8H^{+}\to M{n}^{2+}+5F{e}^{3+}+4H_{2}O$

1 mol MnO₄⁻ reacts with 5 mol Fe²⁺.

Dichromate–iodide

$C{r}_2{O}_7^{2-}+6I^{-}+14H^{+}\to 2C{r}^{3+}+3I_2+7H_{2}O$

1 mol dichromate accepts 6 electrons.

Iodine–thiosulphate

$I_2+2S_2{O}_3^{2-}\to 2I^{-}+S_4{O}_6^{2-}$

Starch indicator: blue to colourless.

Standard emf of a cell

$E_{cell}^{\ominus }=E_{cathode}^{\ominus }-E_{anode}^{\ominus }$

Both written as reduction potentials; positive E⦵cell means the reaction is feasible.

Redox Reactions: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the school exam asks Redox Reactions

The question patterns to expect, and the method that works for each.

Tricks to remember Redox Reactions

Short memory aids. Say each one aloud once, then use it on a question the same day.

• OIL RIG

  Oxidation Is Loss, Reduction Is Gain (of electrons).

  Use it when: Defining oxidation and reduction.

• LEO the lion says GER

  Lose Electrons Oxidation; Gain Electrons Reduction.

  Use it when: Half-reactions.

• Agents do the opposite

  The oxidising agent gets reduced; the reducing agent gets oxidised, like a donor who gives away.

  Use it when: Naming oxidant and reductant.

• Middle states split

  Only an element in a middle oxidation state can disproportionate: from the middle it can go up and down.

  Use it when: Disproportionation questions.

• Acid: H⁺ and water; Base: then add OH⁻

  Balance in acid first; for base, add OH⁻ to both sides to cancel H⁺.

  Use it when: Half-reaction method in basic solution.

Redox Reactions: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section E style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 4-mark question split into 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Redox Reactions

The slips examiners see most often, and what to write instead.

• ✗Mistake: Taking O as −2 in peroxides

  The default rule is applied everywhere.

  Wrong working

  H₂O₂: O = −2, so H = +2

  Right working

  In H₂O₂, H = +1, so O = −1

  ✓Do this: Check for O–O bonds: peroxides (H₂O₂, BaO₂, Na₂O₂) have O = −1; KO₂ has O = −½.

  Usually costs: 1 mark.

• ✗Mistake: Naming the oxidised species as the oxidising agent

  The words are confused.

  ✓Do this: If the oxidation number goes up, the species is oxidised and is the reducing agent.

  Usually costs: 1 mark.

• ✗Mistake: Balancing atoms but not charge

  Students stop after atoms balance.

  Wrong working

  MnO₄⁻ + Fe²⁺ + H⁺ → Mn²⁺ + Fe³⁺ + H₂O (charges unbalanced)

  Right working

  MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O (charge +17 on both sides)

  ✓Do this: Always add up charges on both sides as the final check.

  Usually costs: 1-2 marks.

• ✗Mistake: Using a 1 : 1 ratio in titrations

  The balanced equation is skipped.

  Wrong working

  Moles Fe²⁺ = moles MnO₄⁻

  Right working

  Moles Fe²⁺ = 5 × moles MnO₄⁻

  ✓Do this: Write the balanced equation before calculating.

  Usually costs: The answer mark.

• ✗Mistake: Calling every decomposition a redox reaction

  Decomposition is in the list of redox types.

  ✓Do this: Check oxidation numbers: CaCO₃ → CaO + CO₂ has no change, so it is not redox.

  Usually costs: 1 mark.

• ✗Mistake: Subtracting E⦵ the wrong way round

  Students subtract cathode from anode.

  ✓Do this: E⦵cell = E⦵cathode − E⦵anode, both as reduction potentials; the cathode has the higher value.

  Usually costs: 1 mark.

NCERT exercise guide: Redox Reactions

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Practice questions on Redox Reactions

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQExam level

   The oxidation number of nitrogen in NH₄NO₃ is:
   (a) −3 only (b) +5 only (c) −3 and +5 (d) +1 (average)

   Hint

   Two different N atoms.

   Answer

   (c) −3 in NH₄⁺ and +5 in NO₃⁻ (the average is +1, but each N has its own value).

2. Q2 · 1 mark · MCQExam level

   Which reaction is an example of disproportionation?
   (a) $2KCl{O}_3\to 2KCl+3O_2$ (b) $2H_2{O}_2\to 2H_{2}O+O_2$ (c) $Zn+CuS{O}_4\to ZnS{O}_4+Cu$ (d) $C+O_2\to C{O}_2$

   Hint

   One element goes both up and down.

   Answer

   (b) O: −1 → −2 and 0.

3. Q3 · 2 marks · Very short answerExam level

   Write the Stock notation names for: (a) MnO₂ (b) FeSO₄ (c) SnCl₄ (d) Hg₂Cl₂. (2 marks)

   Hint

   Find the metal's oxidation number.

   Answer

   (a) manganese(IV) oxide (b) iron(II) sulphate (c) tin(IV) chloride (d) mercury(I) chloride (½ each).

4. Q4 · 2 marks · Very short answerHarder exam level

   Justify that $2Na+H_2\to 2NaH$ is a redox reaction. Which is the oxidant? (2 marks)

   Hint

   H is −1 in a metal hydride.

   Answer

   Na: 0 → +1 (oxidised); H: 0 → −1 (reduced), so both occur (1). H₂ accepts electrons, so it is the oxidant (1).

5. Q5 · 3 marks · Short answerHarder exam level

   Balance in acidic medium by the oxidation number method: $Mn{O}_4^{-}+H_{2}S\to M{n}^{2+}+S$ (skeletal). (3 marks)

   Hint

   Mn falls by 5; S rises by 2.

   Answer

   Mn +7 → +2 (−5); S −2 → 0 (+2), so 2MnO₄⁻ and 5H₂S (1). 2MnO₄⁻ + 5H₂S → 2Mn²⁺ + 5S; charge −2 vs +4, so add 6H⁺ on the left and 8H₂O on the right (1): 2MnO₄⁻ + 5H₂S + 6H⁺ → 2Mn²⁺ + 5S + 8H₂O. Check H: 10 + 6 = 16; O: 8 = 8; charge +4 = +4 (1).

6. Q6 · 3 marks · Short answerHarder exam level

   Suppose 0.135 g of oxalic acid (H₂C₂O₄·2H₂O, 126 g mol⁻¹) needs 21.4 mL of KMnO₄ solution in acid. Calculate the molarity of the KMnO₄. (3 marks)

   Hint

   2MnO₄⁻ react with 5 oxalate.

   Answer

   Moles oxalic acid = 0.135 ÷ 126 = 1.071 × 10⁻³ (1). MnO₄⁻ = 2/5 × 1.071 × 10⁻³ = 4.29 × 10⁻⁴ mol (1). M = 4.29 × 10⁻⁴ ÷ 0.0214 = 0.0200 M (1).

7. Q7 · 2 marks · Very short answerHarder exam level

   Arrange the metals Zn, Mg, Cu, Ag in decreasing order of reducing power using E⦵ values (Zn −0.76, Mg −2.36, Cu +0.34, Ag +0.80 V). Which of these can liberate hydrogen from dilute acid? (2 marks)

   Hint

   More negative = stronger reductant.

   Answer

   Mg > Zn > Cu > Ag (1). Mg and Zn (negative E⦵) liberate H₂ from dilute acids; Cu and Ag do not (1).

8. Q8 · 3 marks · Short answerHarder exam level

   Explain why H₂O₂ can act as an oxidising agent and a reducing agent, with one reaction for each. (3 marks)

   Hint

   O is −1, between −2 and 0.

   Answer

   In H₂O₂, O is −1, an intermediate state (1). Oxidant (O → −2): $H_2{O}_2+2F{e}^{2+}+2H^{+}\to 2F{e}^{3+}+2H_{2}O$ (1). Reductant (O → 0): with acidified KMnO₄, $2Mn{O}_4^{-}+5H_2{O}_2+6H^{+}\to 2M{n}^{2+}+5O_2+8H_{2}O$ (1).

9. Q9 · 3 marks · Short answerHarder exam level

   Find the oxidation number of the underlined element: (a) Cr in CrO₂Cl₂ (b) S in Na₂S₂O₃ (average) (c) P in H₃PO₃ (two H bonded to O, one to P). (3 marks)

   Hint

   Watch the structure.

   Answer

   (a) x + 2(−2) + 2(−1) = 0 → x = +6 (1). (b) 2(+1) + 2x + 3(−2) = 0 → x = +2 (average) (1). (c) 3(+1) + x + 3(−2) = 0 → x = +3 (1).

10. Q10 · 3 marks · Short answerHOTS / competency

    Suppose you are given a solution containing Fe²⁺ and asked to use either KMnO₄ or K₂Cr₂O₇ to find its concentration. (a) Write both balanced ionic equations. (b) Which titration needs an external indicator and why? (c) If 1 mol of each oxidant is used, how many moles of Fe²⁺ does each oxidise? (3 marks)

    Hint

    Count electrons accepted.

    Answer

    (a) MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O; Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O (1). (b) Dichromate: orange to green is not sharp, so diphenylamine is used (1). (c) 5 mol and 6 mol of Fe²⁺ respectively (1).

11. Q11 · 5 marks · Long answerHOTS / competency

    (a) Balance by the half-reaction method in acidic medium: $Cu+N{O}_3^{-}\to C{u}^{2+}+NO$ (skeletal). (b) Identify the oxidant and reductant. (c) Why does copper dissolve in nitric acid but not in dilute HCl? (d) Calculate the volume of NO at STP (22.7 L mol⁻¹) produced when 6.35 g of Cu dissolves. (Cu = 63.5) (5 marks)

    Hint

    N: +5 → +2 takes 3e⁻.

    Answer

    (a) Oxidation: Cu → Cu²⁺ + 2e⁻; reduction: NO₃⁻ + 4H⁺ + 3e⁻ → NO + 2H₂O. ×3 and ×2: 3Cu + 2NO₃⁻ + 8H⁺ → 3Cu²⁺ + 2NO + 4H₂O (2). (b) Oxidant NO₃⁻; reductant Cu (1). (c) Cu has positive E⦵ and cannot reduce H⁺; the nitrate ion is a much stronger oxidant (1). (d) Cu = 0.100 mol; NO = 2/3 × 0.100 = 0.0667 mol × 22.7 = 1.51 L (1).

12. Q12 · 1 mark · Assertion-reasonHOTS / competency

    Assertion (A): Fluorine shows only the oxidation state −1 in its compounds.
    Reason (R): Fluorine is the most electronegative element.
    (a) Both A and R are true and R is the correct explanation of A.
    (b) Both A and R are true but R is not the correct explanation of A.
    (c) A is true but R is false.
    (d) A is false but R is true.

    Hint

    Who wins the shared electrons?

    Answer

    (a)

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.Define oxidation in terms of electrons.Show answer

   Loss of electrons.

2. 2.What is the oxidation number of oxygen in OF₂?Show answer

   +2.

3. 3.What is the oxidation number of hydrogen in CaH₂?Show answer

   −1.

4. 4.Give the Stock name of Cu₂O.Show answer

   Copper(I) oxide.

5. 5.What is a disproportionation reaction?Show answer

   A reaction in which the same element in one oxidation state is simultaneously oxidised and reduced.

6. 6.Why can't ClO₄⁻ disproportionate?Show answer

   Cl is in its highest oxidation state (+7) and cannot be oxidised further.

7. 7.Which halogen is the strongest oxidising agent?Show answer

   Fluorine.

8. 8.Why is KMnO₄ called a self-indicator?Show answer

   Its own intense purple colour shows the end point as a pale pink.

9. 9.What is the E⦵ of the standard hydrogen electrode?Show answer

   0.00 V by convention.

10. 10.What is the role of a salt bridge?Show answer

    It completes the circuit and keeps both half-cell solutions electrically neutral.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you identify oxidant and reductant in a given reaction using oxidation numbers?
• ☐Can you find the oxidation number of S in H₂SO₅ and Na₂S₂O₃?
• ☐Can you write Stock names for FeCl₃, MnO₂ and SnCl₂?
• ☐Can you give one example of each type of redox reaction?
• ☐Can you predict whether a species can disproportionate?
• ☐Can you balance MnO₄⁻ + C₂O₄²⁻ in acid by both methods?
• ☐Can you calculate a concentration from a KMnO₄ titration?
• ☐Can you find E⦵cell and predict displacement from E⦵ values?

Redox Reactions: frequently asked questions