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Flashcards · 9701 · A Level Topic 24

Electrochemistry (A Level) flashcards

13 cards on A Level Topic 24, subtopics 24.1 to 24.2. Pick a subtopic or study them all.

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24.1 Electrolysis

Electrolysis: what happens at the cathode and the anode?
Cathode (negative): reduction, cations gain electrons. Anode (positive): oxidation, anions lose electrons.
Equations linking charge, current, time and moles of electrons
Q = It (Q in C, I in A, t in s). Q = nF, with n = moles of electrons and F = 96 500 C mol⁻¹.
Relationship between the Faraday constant, the Avogadro constant and the electron charge
F = L × e: the charge on one mole of electrons.
2.00 A through CuSO₄(aq) for 3860 s. Mass of copper deposited? (Ar Cu = 63.5)
Q = 2.00 × 3860 = 7720 C; n(e⁻) = 7720 ÷ 96 500 = 0.0800 mol; Cu²⁺ + 2e⁻ → Cu, so n(Cu) = 0.0400 mol; mass = 0.0400 × 63.5 = 2.54 g.
Which product forms at the anode in electrolysis of aqueous NaCl, and why does concentration matter?
Concentrated NaCl: chlorine. Very dilute NaCl: oxygen. E⦵(Cl₂/Cl⁻) +1.36 V is only a little more positive than E⦵(O₂/H₂O) +1.23 V, so high [Cl⁻] makes Cl⁻ discharge.
Selective discharge at the cathode in aqueous solution
The cation with the more positive E⦵ is discharged: metals below hydrogen (e.g. Cu) deposit; for reactive metals (e.g. Na) hydrogen is given off instead.

24.2 Standard electrode potentials, standard cell potentials and the Nernst equation

Standard electrode potential, E⦵
The e.m.f. of a half-cell measured against the standard hydrogen electrode (E⦵ = 0.00 V by definition) under standard conditions: 298 K, 1 mol dm⁻³ solutions, 101 kPa for gases.
What does a more positive or more negative E⦵ tell you?
More positive: the left-hand species is more readily reduced (stronger oxidising agent). More negative: the right-hand species is more readily oxidised (stronger reducing agent).
E⦵cell for Zn | Zn²⁺ || Cu²⁺ | Cu (Cu²⁺/Cu +0.34 V, Zn²⁺/Zn −0.76 V)
E⦵cell = 0.34 − (−0.76) = +1.10 V. Zinc is the negative electrode (oxidised); electrons flow from Zn to Cu in the external circuit.
Can Fe³⁺ oxidise Cl⁻ to Cl₂? (Fe³⁺/Fe²⁺ +0.77 V, Cl₂/Cl⁻ +1.36 V)
E⦵cell = E(reduced) − E(oxidised) = 0.77 − 1.36 = −0.59 V. Negative, so not feasible. Cl₂ oxidising Fe²⁺ is the feasible reaction.
Link between ΔG⦵ and E⦵cell
ΔG⦵ = −nFE⦵cell, with n = moles of electrons transferred. Positive E⦵cell → negative ΔG⦵ → feasible.
The Nernst equation, and its qualitative effect
E = E⦵ + (0.059 ÷ z) log([oxidised species] ÷ [reduced species]). Raising [oxidised species] makes E more positive; raising [reduced species] makes it less positive.
Two limitations of predicting reactions from E⦵ values
They say nothing about rate (a feasible reaction may be very slow), and they apply only at standard conditions: concentration changes alter E.

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