ChemistryCBSE Class 11 · Redox Reactions

Redox Titrations, Electrode Potential and Displacement Reactions Class 11

Updated 17 October 2026 · Fact-checked

In a redox titration the amounts of oxidant and reductant react in the mole ratio of the balanced equation, for example 1 MnO₄⁻ : 5 Fe²⁺ and 1 Cr₂O₇²⁻ : 6 Fe²⁺ in acid. In an electrochemical cell E°cell = E°(reduction at the cathode) − E°(reduction at the anode); a positive E°cell means the reaction is spontaneous, and ΔG° = −nFE°cell. A species with a more negative reduction potential is a stronger reducing agent and displaces a metal with a higher potential from its salt solution.

Oxidation at the Zn anode sends electrons to the Cu cathode; E°cell = 0.34 − (−0.76) = +1.10 V, so the reaction is spontaneous.

Illustration drawn for QuestionBuddy.

Understand Redox Titrations, Electrode Potential and Displacement Reactions

Mole-ratio method for titrations. Write the balanced equation, convert volumes and molarities into moles of the reagent whose amount is known, and multiply by the ratio from the equation. Useful ratios (acidic medium): $M n O_{4}^{-} : F e^{2 +} = 1 : 5$ , $M n O_{4}^{-} : C_{2} O_{4}^{2 -} = 2 : 5$ , $M n O_{4}^{-} : H_{2} O_{2} = 2 : 5$ , $C r_{2} O_{7}^{2 -} : F e^{2 +} = 1 : 6$ , $I_{2} : S_{2} O_{3}^{2 -} = 1 : 2$ .

n-factor. It is the number of electrons exchanged per formula unit. $M n O_{4}^{-}$ has $n = 5$ in strongly acidic medium ( $M n^{2 +}$ ), $3$ in neutral or weakly basic medium ( $M n O_{2}$ ) and $1$ in strongly basic medium ( $M n O_{4}^{2 -}$ ). $C r_{2} O_{7}^{2 -}$ has $n = 6$ . For a titration, $n_{1} M_{1} V_{1} = n_{2} M_{2} V_{2}$ (equivalence of electrons).

Self indicator and iodometry. Permanganate is purple and its first excess gives a faint pink end point (a self indicator); dichromate needs an indicator. In iodometry an oxidant liberates iodine from excess iodide and the iodine is titrated with thiosulphate using starch (blue to colourless).

Electrode potential. The standard reduction potential $E^{⊖}$ measures the tendency to be reduced, relative to the standard hydrogen electrode (0 V). For a cell, $E_{ce l l}^{⊖} = E_{c a t h o d e}^{⊖} - E_{an o d e}^{⊖}$ with both values as reduction potentials. Oxidation occurs at the anode, reduction at the cathode, and electrons flow from anode to cathode through the wire.

Spontaneity. $Δ G^{⊖} = - n F E_{ce l l}^{⊖}$ with $F = 96500$ C mol⁻¹, and $lo g K = \frac{n E ^{⊖}}{0.059}$ at 298 K. A positive $E_{ce l l}^{⊖}$ means a negative $Δ G^{⊖}$ and $K > 1$ .

Electrochemical series and displacement. Arrange the half-reactions in order of $E^{⊖}$ . A species on the lower (more negative) side reduces any ion above it: zinc displaces copper from copper sulphate but copper does not displace zinc. Metals with negative $E^{⊖}$ displace hydrogen from dilute acids; metals with positive $E^{⊖}$ do not.

Redox Titrations, Electrode Potential and Displacement Reactions: formulas and key facts

Electron equivalence in titration: $n_{1} M_{1} V_{1} = n_{2} M_{2} V_{2}$; n is the electrons exchanged per formula unit.
Cell potential: $E_{ce l l}^{⊖} = E_{c a t h o d e}^{⊖} - E_{an o d e}^{⊖}$; Use reduction potentials for both.
Gibbs energy: $Δ G^{⊖} = - n F E_{ce l l}^{⊖}$; F = 96500 C mol⁻¹.
Equilibrium constant: $lo g K = \frac{n E _{ce l l}^{⊖}}{0.059}$; At 298 K.
Permanganate with iron(II): $M n O_{4}^{-} + 5 F e^{2 +} + 8 H^{+} \to M n^{2 +} + 5 F e^{3 +} + 4 H_{2} O$; Ratio 1 : 5.
Dichromate with iron(II): $C r_{2} O_{7}^{2 -} + 6 F e^{2 +} + 14 H^{+} \to 2 C r^{3 +} + 6 F e^{3 +} + 7 H_{2} O$; Ratio 1 : 6.

How to solve Redox Titrations, Electrode Potential and Displacement Reactions questions

Balance the equation first; every titration and cell question depends on the mole ratio or the number of electrons, n.

1Write the balanced ionic equation for the medium used.
2Convert volumes and molarities of the known solution into moles (M × V in litres).
3Multiply by the ratio from the equation to get moles of the unknown, then divide by its volume.
4For cells identify the cathode (higher E°) and anode, compute E°cell and then ΔG° = −nFE° or log K.
5State the conclusion in words: spontaneous or not, which metal is a stronger reducing agent.

Redox Titrations, Electrode Potential and Displacement Reactions in real life

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.

Checking a vitamin tablet

Suppose a tablet of ascorbic acid, C₆H₈O₆ (176 g mol⁻¹), is dissolved in water and needs 15.0 mL of 0.020 M iodine solution to reach the first permanent blue with starch. The reaction is C₆H₈O₆ + I₂ → C₆H₆O₆ + 2HI, a 1 : 1 ratio.

Find the mass of ascorbic acid in the tablet.

Try it first, then show the solutionSolution

Moles of I₂ = 0.020 × 0.0150 = 3.0 × 10⁻⁴ mol, equal to the moles of ascorbic acid.
Mass = 3.0 × 10⁻⁴ × 176 = 0.0528 g.

Answer: 0.0528 g (52.8 mg) of ascorbic acid.

Memory hook: Use the ratio from the equation, here 1 : 1.

A dry cell on the shelf

Suppose a cell made from two metals M and N has E°(M²⁺/M) = −0.40 V and E°(N²⁺/N) = +0.30 V (assumed values).

Which metal dissolves and what is the cell potential?

Try it first, then show the solutionSolution

N has the higher reduction potential, so N²⁺ is reduced at the cathode; M is the anode and dissolves.
E°cell = 0.30 − (−0.40) = 0.70 V.

Answer: M dissolves; E°cell = 0.70 V.

Memory hook: The more negative metal is the one that is used up.

Tricks to remember Redox Titrations, Electrode Potential and Displacement Reactions

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

MnO4 equals 5, Cr2O7 equals 6
Electrons gained per permanganate (acid) are five and per dichromate are six.
Use it when: Titration ratios.
Cathode minus anode
E°cell is always the higher reduction potential minus the lower one.
Use it when: A positive value confirms the choice.
Negative E is a strong reductant
The most negative E° is the strongest reducing agent; the most positive E° is the strongest oxidising agent.
Use it when: Displacement questions.
An OIL RIG for cells
Anode Oxidation, Cathode Reduction: An Ox, Red Cat.
Use it when: Labelling electrodes.

Redox Titrations, Electrode Potential and Displacement Reactions: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: n-factor of permanganate · 2 marks · Very short answerBoard level

State the n-factor of KMnO₄ in (a) strongly acidic, (b) neutral and (c) strongly basic medium, with the manganese species formed.

Try it first, then show the solutionSolution

(a) Mn(+7) to Mn²⁺ (+2): n = 5. (1)
(b) Mn(+7) to MnO₂ (+4): n = 3, and (c) Mn(+7) to MnO₄²⁻ (+6): n = 1. (1)

Answer: n = 5, 3 and 1 respectively.

Example 2: Strength of an iron(II) solution · 3 marks · Short answerHarder board level

25.0 mL of an iron(II) sulphate solution needs 20.0 mL of 0.020 M KMnO₄ in acidic medium. Find the molarity of the Fe²⁺ solution and the mass of FeSO₄ (152 g mol⁻¹) in one litre.

Try it first, then show the solutionSolution

Moles of MnO₄⁻ = 0.020 × 0.0200 = 4.0 × 10⁻⁴ mol. (1)
From MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O, Fe²⁺ = 5 × 4.0 × 10⁻⁴ = 2.0 × 10⁻³ mol in 25.0 mL, so M = 0.080 mol L⁻¹. (1)
Mass of FeSO₄ per litre = 0.080 × 152 = 12.16 g. (1)

Answer: 0.080 M; 12.2 g of FeSO₄ per litre.

Example 3: Dichromate standard solution · 3 marks · Short answerHarder board level

1.47 g of K₂Cr₂O₇ (294 g mol⁻¹) is dissolved in water to make 250 mL. 25.0 mL of it is used to oxidise Fe²⁺ in acidic medium. Find the amount of Fe²⁺ oxidised, and the molarity of an iron(II) solution of which 30.0 mL is just oxidised by that 25.0 mL.

Try it first, then show the solutionSolution

Molarity of dichromate = (1.47/294)/0.250 = 0.0050/0.250 = 0.020 M. (1)
Moles of Cr₂O₇²⁻ in 25.0 mL = 0.020 × 0.0250 = 5.0 × 10⁻⁴ mol; Fe²⁺ oxidised = 6 × 5.0 × 10⁻⁴ = 3.0 × 10⁻³ mol. (1)
Molarity of the iron(II) solution = 3.0 × 10⁻³/0.0300 = 0.10 M. (1)

Answer: 3.0 × 10⁻³ mol of Fe²⁺; 0.10 M.

Example 4: A zinc-copper cell · 4 marks · Long answerHarder board level

Given the standard reduction potentials E°(Zn²⁺/Zn) = −0.76 V and E°(Cu²⁺/Cu) = +0.34 V, for the cell Zn | Zn²⁺ || Cu²⁺ | Cu find (a) E°cell, (b) ΔG° and (c) log K and K at 298 K (F = 96500 C mol⁻¹).

Try it first, then show the solutionSolution

(a) Cathode Cu, anode Zn: E°cell = 0.34 − (−0.76) = 1.10 V. (1)
(b) n = 2, so ΔG° = −2 × 96500 × 1.10 = −2.12 × 10⁵ J = −212 kJ. (1)
(c) log K = 2 × 1.10/0.059 = 37.3. (1)
K ≈ 2 × 10³⁷, so the reaction goes essentially to completion. (1)

Answer: (a) 1.10 V; (b) −212 kJ; (c) log K = 37.3, K ≈ 2 × 10³⁷.

Example 5: Analysis of a bleach solution · 4 marks · Long answerHOTS / competency

10.0 mL of a hypochlorite solution is added to excess acidified KI. ClO⁻ + 2I⁻ + 2H⁺ → Cl⁻ + I₂ + H₂O. The iodine liberated needs 24.0 mL of 0.050 M Na₂S₂O₃ (I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻). Find the molarity of the hypochlorite, its mass concentration as NaOCl (74.5 g mol⁻¹) and the change in oxidation number of chlorine. Name the indicator and the colour change.

Try it first, then show the solutionSolution

Moles of S₂O₃²⁻ = 0.050 × 0.0240 = 1.2 × 10⁻³ mol, so I₂ = 6.0 × 10⁻⁴ mol. (1)
One ClO⁻ gives one I₂, so ClO⁻ = 6.0 × 10⁻⁴ mol in 10.0 mL, and M = 0.060 mol L⁻¹. (1)
Mass concentration = 0.060 × 74.5 = 4.47 g L⁻¹; chlorine goes from +1 to −1. (1)
Starch indicator, added near the end point: blue (starch-iodine) to colourless. (1)

Answer: 0.060 M; 4.47 g L⁻¹; Cl from +1 to −1; starch, blue to colourless.

Where students lose marks in Redox Titrations, Electrode Potential and Displacement Reactions

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

Mistake: Using the 1 : 1 ratio between MnO₄⁻ and Fe²⁺.
Not using the balanced equation.
Do this: Write the equation; the ratio is 1 : 5.
Usually costs: The whole numerical.
Mistake: Subtracting the potentials the wrong way round, giving a negative E°cell for a working cell.
Not identifying the cathode first.
Do this: E°cell = E°cathode − E°anode, with cathode the one of higher reduction potential.
Usually costs: Signs and ΔG°.
Mistake: Forgetting that ΔG° has to be converted to kJ when F is in C mol⁻¹.
The answer is in J.
Do this: Divide by 1000 when reporting kJ.
Usually costs: The unit mark.
Mistake: Multiplying E° by the stoichiometric coefficient (or number of electrons) when a half-reaction is multiplied.
Treating E° as extensive.
Do this: E° is intensive; do not multiply it by the coefficient.
Usually costs: The cell potential.

Exam tips

Write the balanced titration equation with the ratio highlighted.
Give the molarity with unit mol L⁻¹ or M, and the mass concentration in g L⁻¹.
Name the indicator and the colour change at the end point.
In a cell question label cathode and anode before calculating.
Do not multiply E° by the number of electrons.

Practice questions on Redox Titrations, Electrode Potential and Displacement Reactions

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

Q1 · 1 mark · MCQBoard level
Which of the following cannot act as a reducing agent?
(a) H₂SO₄ (b) SO₂ (c) H₂S (d) Na₂S₂O₃
Hint
Sulphur is at its highest oxidation state in one of them.
Answer
(a): sulphur is already +6 and cannot be oxidised further.
Q2 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Zinc displaces copper from copper sulphate solution, but copper does not displace zinc from zinc sulphate solution.
Reason (R): Zinc has a more negative standard reduction potential than copper, so it is the stronger reducing agent.
(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
Compare the two E° values.
Answer
(a).
Q3 · 2 marks · Very short answerBoard level
Write the reduction half-reaction of dichromate in acidic medium and state the n-factor of K₂Cr₂O₇ and the colour change of chromium.
Hint
14H⁺ are needed.
Answer
Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O; n = 6; orange dichromate turns green (Cr³⁺).
Q4 · 3 marks · Short answerHarder board level
20.0 mL of a hydrogen peroxide solution needs 16.0 mL of 0.020 M KMnO₄ in acid: 2MnO₄⁻ + 5H₂O₂ + 6H⁺ → 2Mn²⁺ + 5O₂ + 8H₂O. Find the molarity of the H₂O₂ and its concentration in g L⁻¹ (H₂O₂ = 34 g mol⁻¹).
Hint
Ratio MnO₄⁻ : H₂O₂ = 2 : 5.
Answer
MnO₄⁻ = 3.2 × 10⁻⁴ mol, so H₂O₂ = 3.2 × 10⁻⁴ × 5/2 = 8.0 × 10⁻⁴ mol in 20.0 mL, M = 0.040 M; concentration = 0.040 × 34 = 1.36 g L⁻¹.
Q5 · 4 marks · Long answerHOTS / competency
Given E°(Mg²⁺/Mg) = −2.36 V (assumed), E°(Ag⁺/Ag) = +0.80 V and E°(H⁺/H₂) = 0.00 V. (a) Write the cell reaction and find E°cell for the Mg–Ag cell and its ΔG° (F = 96500 C mol⁻¹). (b) Will silver dissolve in dilute hydrochloric acid? (c) Name the anode and the direction of electron flow.
Hint
Balance the electrons: Mg gives 2, Ag⁺ takes 1.
Answer
(a) Mg + 2Ag⁺ → Mg²⁺ + 2Ag; E°cell = 0.80 − (−2.36) = 3.16 V; ΔG° = −2 × 96500 × 3.16 = −6.10 × 10⁵ J = −610 kJ. (b) No: E° for Ag oxidising and H⁺ reducing = 0.00 − 0.80 = −0.80 V, negative. (c) Magnesium is the anode; electrons flow from Mg to Ag through the external wire.

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.Ratio MnO₄⁻ : Fe²⁺ in acid?Show answer
1 : 5.
2.Ratio Cr₂O₇²⁻ : Fe²⁺?Show answer
1 : 6.
3.E°cell formula?Show answer
E°cathode − E°anode.
4.ΔG° in terms of E°cell?Show answer
−nFE°cell.
5.Indicator in iodometric titrations?Show answer
Starch.
6.Why is KMnO₄ a self indicator?Show answer
Its own purple colour gives a faint pink at the end point.

Redox Titrations, Electrode Potential and Displacement Reactions: frequently asked questions

Why is hydrochloric acid not used to acidify a permanganate titration?

Chloride ions can be oxidised by permanganate, using up extra oxidant and giving wrong results; dilute sulphuric acid is used instead.

What does a positive E°cell mean?

The cell reaction is spontaneous as written, with ΔG° negative and K greater than 1.

Why is the hydrogen electrode taken as zero?

It is the agreed reference point; all other potentials are measured against it.

Does a large positive E° mean a fast reaction?

No. It shows only feasibility, not the rate.

Why can a metal above hydrogen displace hydrogen from acids?

Its negative E° means it is a stronger reducing agent than hydrogen gas.