Redox Titrations and Electrolysis Calculations

Redox Titrations and Electrolysis Calculations

Redox Titrations with Potassium Manganate(VII)

Potassium manganate(VII) (KMnO₄) is a powerful oxidising agent used in redox titrations. It is its own indicator — the intense purple colour disappears when it reacts, so no separate indicator is needed.

Half-equation (in acidic solution):

MnO₄⁻(aq) + 8H⁺(aq) + 5e⁻ → Mn²⁺(aq) + 4H₂O(l)

Purple MnO₄⁻ is reduced to the very pale pink (virtually colourless) Mn²⁺.

Common titrations with MnO₄⁻:

With iron(II) ions:

• Fe²⁺ → Fe³⁺ + e⁻ (×5)
• Overall: MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺
• The mole ratio is 1:5 (MnO₄⁻ : Fe²⁺)

With ethanedioic acid (oxalic acid):

• C₂O₄²⁻ → 2CO₂ + 2e⁻ (×5)
• Overall: 2MnO₄⁻ + 16H⁺ + 5C₂O₄²⁻ → 2Mn²⁺ + 8H₂O + 10CO₂
• The mole ratio is 2:5 (MnO₄⁻ : C₂O₄²⁻)

Practical procedure:

1. The iron(II) solution (or other reductant) is placed in the conical flask, acidified with dilute sulfuric acid

2. KMnO₄ is added from the burette (it is always in the burette because it stains)

3. Swirl after each addition — the purple colour should disappear

4. The end point is when the first permanent pink/purple colour persists for at least 30 seconds (one drop of excess MnO₄⁻)

Why sulfuric acid? It provides the H⁺ ions needed for the reaction. Hydrochloric acid cannot be used because Cl⁻ would be oxidised by MnO₄⁻. Nitric acid cannot be used because it is itself an oxidising agent.

Worked Example: MnO₄⁻ / Fe²⁺ Titration

25.0 cm³ of a solution of iron(II) sulfate was acidified and titrated with 0.0200 mol dm⁻³ KMnO₄. The mean titre was 23.6 cm³. Calculate the concentration of Fe²⁺.

Step 1: Moles of MnO₄⁻ = c × V = 0.0200 × (23.6/1000) = 4.72 × 10⁻⁴ mol

Step 2: Mole ratio MnO₄⁻ : Fe²⁺ = 1 : 5

Moles of Fe²⁺ = 5 × 4.72 × 10⁻⁴ = 2.36 × 10⁻³ mol

Step 3: Concentration of Fe²⁺ = n / V = 2.36 × 10⁻³ / (25.0/1000) = 0.0944 mol dm⁻³

Iodine-Thiosulfate Titrations

Another common redox titration involves iodine and sodium thiosulfate:

I₂(aq) + 2S₂O₃²⁻(aq) → 2I⁻(aq) + S₄O₆²⁻(aq)

Procedure:

1. The iodine-containing solution is placed in the conical flask

2. Sodium thiosulfate is added from the burette

3. As thiosulfate is added, the brown iodine colour fades to yellow

4. When the solution is pale yellow, add starch indicator (turns blue-black with I₂)

5. Continue adding thiosulfate dropwise until the blue-black colour disappears — this is the end point

Why add starch late? If added too early, the starch-iodine complex is difficult to break down, giving an imprecise end point.

Application: Often used indirectly — an oxidising agent is first reacted with excess I⁻ to liberate I₂, then the I₂ is titrated with thiosulfate. This allows the determination of oxidising agents like Cu²⁺, IO₃⁻, or Cl₂.

Electrolysis Calculations

Electrolysis is the decomposition of an ionic compound using an electric current. The key equation linking charge, current, and time is:

Q = I × t

Where:

• Q = charge in coulombs (C)
• I = current in amperes (A)
• t = time in seconds (s)

Faraday's Laws

One mole of electrons carries a charge of 96,485 C (one Faraday, F).

The number of moles of electrons is:

n(e⁻) = Q / F = (I × t) / 96,485

The moles of substance deposited or liberated depends on the number of electrons in the half-equation:

• Cu²⁺ + 2e⁻ → Cu: 2 mol e⁻ deposits 1 mol Cu
• Al³⁺ + 3e⁻ → Al: 3 mol e⁻ deposits 1 mol Al
• 2Cl⁻ → Cl₂ + 2e⁻: 2 mol e⁻ liberates 1 mol Cl₂

Worked Example: Electrolysis of CuSO₄(aq)

A current of 2.50 A was passed through aqueous copper(II) sulfate for 45.0 minutes. Calculate the mass of copper deposited at the cathode.

Step 1: Convert time to seconds: 45.0 × 60 = 2700 s

Step 2: Calculate charge: Q = I × t = 2.50 × 2700 = 6750 C

Step 3: Moles of electrons: n(e⁻) = 6750 / 96,485 = 0.06995 mol

Step 4: From Cu²⁺ + 2e⁻ → Cu, moles of Cu = 0.06995 / 2 = 0.03498 mol

Step 5: Mass of Cu = 0.03498 × 63.5 = 2.22 g

Worked Example: Volume of Gas

A current of 1.50 A was passed through molten NaCl for 30.0 minutes. Calculate the volume of Cl₂ produced at the anode at RTP (where 1 mol gas = 24.0 dm³).

Step 1: Q = 1.50 × (30.0 × 60) = 2700 C

Step 2: n(e⁻) = 2700 / 96,485 = 0.02798 mol

Step 3: From 2Cl⁻ → Cl₂ + 2e⁻, moles of Cl₂ = 0.02798 / 2 = 0.01399 mol

Step 4: Volume = 0.01399 × 24.0 = 0.336 dm³ (336 cm³)

Electroplating

Electrolysis is used in electroplating — depositing a thin layer of metal onto an object:

• The object to be plated is the cathode
• The plating metal is the anode (dissolves to replenish ions in solution)
• The electrolyte contains ions of the plating metal

Exam Tips

• In manganate(VII) titrations, always state that sulfuric acid is used (not HCl or HNO₃) and explain why
• Show the half-equations and overall equation before doing calculations
• In electrolysis calculations, always convert time to seconds before using Q = It
• Check the number of electrons in the half-equation carefully — this determines the mole ratio
• Units: F = 96,485 C mol⁻¹ (memorise this or note it is given in the data booklet)