Topic 14 Flashcards
Define standard electrode potential (E°).
The voltage measured under standard conditions (298 K, 1 atm, 1 mol dm⁻³) when a half-cell is connected to a standard hydrogen electrode (SHE).
How is cell potential (E_cell) calculated?
E_cell = E_right° - E_left° (more positive E° is the cathode).
When is a redox reaction feasible?
If E_cell > 0 (positive), the reaction is spontaneous. Use ΔG = -nFE_cell to confirm.
Describe the iodine-thiosulfate titration.
- Oxidizing agent: I₂ (iodine).
- Reducing agent: S₂O₃²⁻ (thiosulfate ions).
- Endpoint: Starch indicator turns blue-black → colourless.
What is disproportionation? Give an example.
A species is both oxidized and reduced in the same reaction.
Example: 2Cu⁺ → Cu²⁺ + Cu.
Balance MnO₄⁻ → Mn²⁺ in acidic conditions.
5Fe²⁺ + MnO₄⁻ + 8H⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O.
What does the electrochemical series show?
The tendency of species to act as oxidizing agents. More positive E° → stronger oxidizing agent.
Write equations for a hydrogen-oxygen fuel cell.
- Anode: 2H₂ → 4H⁺ + 4e⁻.
- Cathode: O₂ + 4H⁺ + 4e⁻ → 2H₂O.
How does balancing redox equations differ in alkaline conditions?
Add OH⁻ instead of H⁺ to neutralize H⁺.
Example: O₂ + 2H₂O + 4e⁻ → 4OH⁻.
Give two common oxidizing agents and their reduction half-equations.
- MnO₄⁻/H⁺: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O.
- Cr₂O₇²⁻/H⁺: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O.
How is concentration determined in redox titrations?
Use n = cV and mole ratios from the balanced equation.
Describe the standard hydrogen electrode.
Platinum electrode in 1 mol dm⁻³ H⁺, H₂ gas at 1 atm, 298 K. E° = 0 V.
How is Al extracted using redox?
Electrolysis of Al₂O₃ (dissolved in cryolite):
- Al³⁺ + 3e⁻ → Al (reduction at cathode).
Why is F₂ a stronger oxidizing agent than Cl₂?
Higher E° value (F₂: +2.87 V vs Cl₂: +1.36 V).
Explain rusting of iron as a redox process.
- Oxidation (anode): Fe → Fe²⁺ + 2e⁻.
- Reduction (cathode): O₂ + 2H₂O + 4e⁻ → 4OH⁻.
What is the purpose of acidifying KMnO₄ in redox titrations?
To provide H⁺ ions for the reduction of MnO₄⁻ to Mn²⁺.
Calculate moles of Fe²⁺ oxidized by 0.02 mol MnO₄⁻.
From 5Fe²⁺ : 1MnO₄⁻ ratio → 0.02 × 5 = 0.1 mol.
Why might a reaction with E_cell > 0 not occur?
Kinetic factors (high activation energy) or non-standard conditions.
Write a redox equation for catalytic converters.
2CO + 2NO → 2CO₂ + N₂ (Pt/Rh catalyst).
Define oxidation and reduction.
- Oxidation: Loss of electrons.
- Reduction: Gain of electrons (OIL RIG).
