Redox Reactions • Topic 3 of 3

Redox Titrations & Applications

Redox reactions can be split into two physically separate redox couples, each made of an oxidised and a reduced form (e.g. Zn2+/Zn, Cu2+/Cu). When the two halves are connected in a galvanic cell, electrons flow through an external wire.

Electrode processes and electrode potential

At each electrode a redox couple reaches its own tendency to gain or lose electrons, measured as the electrode potential. By convention these are reduction potentials measured against the standard hydrogen electrode (SHE), taken as 0.00 V. Arranging couples in order of their standard reduction potentials gives the electrochemical series. A couple with a more positive potential (e.g. Cu2+/Cu, +0.34 V) is a better oxidising agent; a more negative one (e.g. Zn2+/Zn, −0.76 V) is a better reducing agent. A metal higher (more negative) in the series displaces one below it from solution.

Redox titrations

In a redox titration the equivalence point is where the moles of electrons supplied by the reducing agent equal those accepted by the oxidising agent.

  • Permanganometry (KMnO4): KMnO4 is a strong, self-indicating oxidant in acidic medium — the first permanent pink colour marks the end point. Each MnO4- accepts 5 electrons.
  • Dichromatometry (K2Cr2O7): a primary standard, stable in air; each Cr2O72- accepts 6 electrons. It needs an external indicator (e.g. diphenylamine).
  • Iodometry / iodimetry: uses the I2/I- couple; liberated I2 is titrated with thiosulphate, S2O32-, using starch (blue→colourless) as indicator: I2 + 2S2O32- → 2I- + S4O62-.

Equivalent concept

The n-factor in a redox reaction is the number of electrons gained or lost per formula unit. The equivalent mass = molar mass / n-factor. For KMnO4 in acid n = 5, for K2Cr2O7 n = 6. At the equivalence point, milliequivalents of oxidant = milliequivalents of reductant: N1V1 = N2V2.

Applications

Redox chemistry powers batteries and fuel cells, metal extraction (reduction of ores) and corrosion (oxidation of iron to rust), as well as bleaching, water treatment and respiration. Quantitative titrations let chemists measure the iron content of an ore, the available chlorine in bleach or the strength of hydrogen peroxide.

Daniell cell: Zn is oxidised at the anode, Cu²⁺ is reduced at the cathodee⁻ flow →salt bridgeAnode: Zn → Zn²⁺ + 2e⁻Cathode: Cu²⁺ + 2e⁻ → Cu(−)(+)
Solved Examples
1
Worked Example
Write the n-factor (electrons gained per formula unit) of KMnO4 in acidic medium and its equivalent mass. (Molar mass = 158 g mol-1.)
Solution
  1. In acid, MnO4- + 8H+ + 5e- → Mn2+ + 4H2O, so n = 5.
  2. Equivalent mass = molar mass / n-factor = 158 / 5.
  3. = 31.6 g equiv-1.

Answer: n = 5; equivalent mass = 31.6 g equiv-1.

2
Worked Example
Find the n-factor and equivalent mass of K2Cr2O7 as an oxidant in acidic medium. (Molar mass = 294 g mol-1.)
Solution
  1. Cr2O72- + 14H+ + 6e- → 2Cr3+ + 7H2O, so n = 6.
  2. Equivalent mass = 294 / 6.
  3. = 49 g equiv-1.

Answer: n = 6; equivalent mass = 49 g equiv-1.

3
Worked Example
20 mL of FeSO4 solution required 25 mL of 0.02 M KMnO4 in acidic medium. Find the molarity of FeSO4.
Solution
  1. Reaction: MnO4- + 5Fe2+ + 8H+ → Mn2+ + 5Fe3+ + 4H2O, so 1 mol KMnO4 ≡ 5 mol Fe2+.
  2. Moles of KMnO4 = 0.02 × 25/1000 = 5×10-4 mol.
  3. Moles of Fe2+ = 5 × 5×10-4 = 2.5×10-3 mol.
  4. Molarity = 2.5×10-3 / (20/1000) = 0.125 M.

Answer: The FeSO4 solution is 0.125 M.

4
Worked Example
In an iodometric titration, 22.0 mL of 0.10 M Na2S2O3 reacted with liberated I2. How many moles of I2 were present?
Solution
  1. I2 + 2S2O32- → 2I- + S4O62-, so 2 mol thiosulphate ≡ 1 mol I2.
  2. Moles of S2O32- = 0.10 × 22.0/1000 = 2.2×10-3 mol.
  3. Moles of I2 = (2.2×10-3) / 2 = 1.1×10-3 mol.

Answer: 1.1×10-3 mol of I2.

5
Worked Example
Using the electrochemical series (Zn²⁺/Zn = −0.76 V, Cu²⁺/Cu = +0.34 V), predict whether Zn can displace Cu from CuSO4 and the cell EMF.
Solution
  1. Zn has the more negative reduction potential, so it is the stronger reducing agent and is oxidised (anode).
  2. Cu2+ is reduced at the cathode; hence Zn displaces Cu.
  3. EMF = E°cathode − E°anode = (+0.34) − (−0.76).
  4. = +1.10 V (positive, so the reaction is spontaneous).

Answer: Yes, Zn displaces Cu; cell EMF = +1.10 V.

6
Worked Example
10 mL of H2O2 solution required 20 mL of 0.05 M KMnO4 in acid. Find the molarity of H2O2. (5H2O2 react per 2MnO4-.)
Solution
  1. 2MnO4- + 5H2O2 + 6H+ → 2Mn2+ + 5O2 + 8H2O; ratio MnO4- : H2O2 = 2 : 5.
  2. Moles of KMnO4 = 0.05 × 20/1000 = 1.0×10-3 mol.
  3. Moles of H2O2 = (5/2) × 1.0×10-3 = 2.5×10-3 mol.
  4. Molarity = 2.5×10-3 / (10/1000) = 0.25 M.

Answer: The H2O2 solution is 0.25 M.

Key Points

  • A redox reaction is built from two redox couples; in a galvanic cell electrons flow from the anode (oxidation) to the cathode (reduction).
  • Electrode potentials (vs SHE = 0 V) ranked as the electrochemical series; more positive = stronger oxidant, more negative = stronger reductant.
  • KMnO₄ (n = 5, self-indicating) and K₂Cr₂O₇ (n = 6) are key acidic-medium oxidants in redox titrations.
  • Iodometry titrates liberated I₂ with thiosulphate using starch indicator: I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻.
  • Equivalent mass = molar mass / n-factor; at the end point milliequivalents of oxidant equal those of reductant (N₁V₁ = N₂V₂).
Quick Check — 4 MCQs
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Q1.The standard hydrogen electrode is assigned a potential of:
Explanation: By convention the SHE reduction potential is defined as exactly 0.00 V.
Q2.The n-factor of K2Cr2O7 acting as an oxidant in acidic medium is:
Explanation: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O, so 6 electrons are gained per formula unit.
Q3.The indicator used in iodometric titration is:
Explanation: Starch gives a sharp blue-to-colourless change at the end point with iodine.
Q4.KMnO4 is called self-indicating because:
Explanation: The first permanent pink from excess MnO₄⁻ signals the end point — no separate indicator is needed.
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