Vidaara.orgClass 11 · Chemistry
CodeVID-C11-08-T3-01
Assignment — Redox Titrations & Applications
Name: ____________________
Roll No.: __________
Date: ____________
General Instructions
- All questions are compulsory.
- Section A carries 1 mark each, Section B 2 marks, Section C 3 marks and Section D 5 marks.
- Show all working for Sections B, C and D. Only final answers are given at the end — for full solutions, raise your doubts with your teacher.
Section A — Multiple Choice Questions
5 × 1 = 5 marks
1.
Oxidation in a galvanic cell occurs at the:
- A.cathode
- B.anode
- C.salt bridge
- D.wire
2.
Equivalent mass of KMnO4 (M = 158) in acidic medium is:
- A.158
- B.79
- C.52.7
- D.31.6
3.
In iodometry, I2 is titrated against:
- A.KMnO₄
- B.Na₂S₂O₃
- C.K₂Cr₂O₇
- D.NaOH
4.
A metal with a more negative reduction potential is a:
- A.stronger oxidising agent
- B.stronger reducing agent
- C.weaker reducing agent
- D.neutral species
5.
Rusting of iron is essentially a process of:
- A.reduction
- B.oxidation
- C.neutralisation
- D.sublimation
Section B — Short Answer (2 marks)
3 × 2 = 6 marks
6.
Why is KMnO4 called a self-indicating titrant?
7.
Write the reaction of iodine with thiosulphate and name the indicator.
8.
Define equivalent mass in terms of n-factor for a redox reagent.
Section C — Short Answer (3 marks)
2 × 3 = 6 marks
9.
25 mL of FeSO4 required 20 mL of 0.02 M KMnO4 in acid. Calculate the molarity of FeSO4.
10.
Using Zn²⁺/Zn = -0.76 V and Cu²⁺/Cu = +0.34 V, calculate the EMF of the Daniell cell and state the spontaneous direction.
Section D — Long Answer (5 marks)
1 × 5 = 5 marks
11.
Explain the principle of a redox titration using KMnO4 against oxalic acid, including the balanced ionic equation and how the end point is detected.
Answer Key
Section A — Multiple Choice Questions
- (B) anode
- (D) 31.6
- (B) Na₂S₂O₃
- (B) stronger reducing agent
- (B) oxidation
Section B — Short Answer (2 marks)
- Its intense purple colour is discharged while it reacts; the first permanent pink tinge from a slight excess of MnO₄⁻ marks the end point, so no separate indicator is needed.
- I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻; the indicator is starch (blue to colourless).
- Equivalent mass = molar mass / n-factor, where the n-factor is the number of electrons gained or lost per formula unit.
Section C — Short Answer (3 marks)
- Moles KMnO₄ = 0.02 × 20/1000 = 4×10⁻⁴; moles Fe²⁺ = 5 × 4×10⁻⁴ = 2×10⁻³; molarity = 2×10⁻³ / 0.025 = 0.08 M.
- EMF = E°(cathode) - E°(anode) = +0.34 - (-0.76) = +1.10 V; positive EMF means the cell reaction Zn + Cu²⁺ → Zn²⁺ + Cu is spontaneous.
Section D — Long Answer (5 marks)
- KMnO₄ (oxidant, n = 5) oxidises oxalate to CO₂ in hot acidic medium: 2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O. At the equivalence point milliequivalents of MnO₄⁻ equal those of oxalate. KMnO₄ is self-indicating: the solution stays colourless while oxalate remains; the first permanent faint pink from a slight excess of MnO₄⁻ signals the end point. The titration is done warm (~60 °C) to speed the slow start, and acid is provided by dilute H₂SO₄ (not HCl, which itself reacts with MnO₄⁻).
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