Cambridge International AS & A Level Chemistry 9701 · A Level topic 24

Electrochemistry

What this chapter covers24.1–24.2

Every tonne of aluminium, every coil of pure copper wire and every cylinder of chlorine begins with electrolysis: an electric current forcing a redox reaction that would not happen by itself. A torch battery or a phone cell does the opposite, letting a feasible redox reaction push electrons round a circuit. This chapter treats the two directions together, because both rest on the same idea — electrons transferred at an electrode — and both are made quantitative by the same constants.

At AS you described redox in terms of electron transfer and oxidation numbers. Here the transfer is organised so that the electrons travel through a wire. That makes the chemistry measurable: the charge passed tells you how many moles of product form, and the voltage of a cell tells you how strongly the reaction is driven.

Electrolysis: what happens at each electrode24.1.1

Electrolysis is the decomposition of an electrolyte — an ionic compound, molten or in solution — by a direct current. The power supply pumps electrons onto one electrode and draws them off the other.

• The cathode is the negative electrode. Electrons arrive there from the supply and are gained by species in the electrolyte: reduction happens at the cathode.
• The anode is the positive electrode. Species in the electrolyte lose electrons to it, and those electrons return to the supply: oxidation happens at the anode.

Inside the electrolyte the current is carried by ions: cations drift towards the cathode, anions towards the anode. In the external wires it is carried by electrons. No electrons travel through the solution.

Molten electrolytes

A molten binary compound contains only its own two kinds of ion, so there is no choice to make: the cation is reduced at the cathode and the anion oxidised at the anode. Molten sodium chloride gives sodium and chlorine; molten lead(II) bromide gives lead and bromine; aluminium oxide dissolved in molten cryolite gives aluminium and oxygen. An examiner reported hydrogen at the cathode for molten PbBr₂ as a common error: with no water present, there is nothing to form hydrogen from.

Aqueous electrolytes: position in the redox series and concentration24.1.1

In an aqueous solution, water molecules are present at every electrode alongside the ions of the salt, and water can itself be reduced or oxidised:

(in acid, the cathode reaction is simply 2H⁺(aq) + 2e⁻ → H₂(g)). Each electrode therefore has a competition, and the syllabus names the three factors that settle it.

1 · Position in the redox series (electrode potential)

At the cathode, the species that is most easily reduced — the one whose half-equation has the most positive E° — gains the electrons. Ions of metals with E° more positive than that for reducing water (copper, silver) are discharged as the metal. Ions of reactive metals with very negative E° (Na⁺ −2.71 V, K⁺, Ca²⁺, Mg²⁺ −2.38 V, Al³⁺ −1.66 V) are far harder to reduce than water, so hydrogen is formed instead.

At the anode, the species that is most easily oxidised — the reduced form in the half-equation with the least positive E° — loses electrons. Iodide (E° I₂/I⁻ = +0.54 V) and bromide (+1.07 V) are oxidised more easily than water (+1.23 V), so iodine and bromine form. Sulfate and nitrate are very hard to oxidise, so from their solutions oxygen is formed from water.

2 · Concentration

Chloride sits close to water: E°(Cl₂/Cl⁻) = +1.36 V against +1.23 V for oxygen. Here concentration decides. In concentrated chloride solution, chlorine is the main product at the anode; in very dilute solution, oxygen is. Raising the concentration of an ion makes it easier to discharge — an effect you will be able to explain quantitatively once you meet the Nernst equation later in this chapter.

3 · The state of the electrolyte

Molten means no water and no competition; aqueous means water competes at both electrodes. Always check which you have before predicting.

The Faraday constant and F = Le24.1.2

Every electrode reaction is written with electrons, so the amount of product depends only on how many moles of electrons pass. The charge on one electron is e = 1.60 × 10⁻¹⁹ C. The charge on one mole of electrons is the Faraday constant, F:

where L is the Avogadro constant. The relationship is simply a count: a mole of electrons is L electrons, each carrying charge e. It also means that if F and e are measured, L can be found — which is exactly how electrolysis gives a value of the Avogadro constant.

Calculating charge, moles, mass and volume24.1.3

A current of I amperes flowing for t seconds transfers a charge Q coulombs:

The last step comes from the half-equation. Cu²⁺ + 2e⁻ → Cu needs two moles of electrons per mole of copper; 2H₂O → O₂ + 4H⁺ + 4e⁻ releases one mole of O₂ per four moles of electrons. Time must be in seconds.

Determining the Avogadro constant by electrolysis24.1.4

Because F = Le, a measurement of the charge needed to deposit or dissolve a known amount of a metal gives L directly, provided the charge on the electron is known. The standard method uses copper electrodes.

Procedure and the reason for each step

1. Clean and dry the copper anode, then weigh it. Grease or oxide would react or add mass; water would be weighed as copper.
2. Set up the series circuit with the d.c. supply, a variable resistor and an ammeter, and immerse both copper electrodes in the electrolyte.
3. Switch on and start the stopwatch; keep the current constant by adjusting the variable resistor. Q = It is valid only for a steady current.
4. After a measured time, switch off, remove the anode, rinse with distilled water, dry carefully (for example with propanone) and reweigh. Rinsing removes electrolyte that would otherwise add mass; drying removes water.

Processing the results

Errors and improvements

• Anode not completely dry when reweighed: the mass loss appears smaller, so L comes out too large. Dry to constant mass.
• Loose copper falling off the anode rather than dissolving: the loss in mass is too large and L too small. Use a clean, smooth anode and a modest current. Weighing the cathode as well gives a check, since the two changes in mass should be equal.
• Current drifting: Q calculated from the starting current is wrong. Keep the current constant with the variable resistor, or record it regularly and use the mean.
• Short run or small current: the change in mass is small, so the percentage uncertainty of the weighing is large. Pass the current for longer, or use a larger current within the limit that avoids loose deposits.

Past-paper practice · Set 24A · Electrolysis and Faraday calculations

Examination questions on this part of the unit. Try each one on paper before opening the answer.

Question 24A.1[8]

Answer and marking guidance

(a) PbBr₂(l): anode Br₂, cathode Pb; concentrated NaCl(aq): anode Cl₂, cathode H₂; Cu(NO₃)₂(aq): anode O₂, cathode Cu — two correct for one mark, four for two, six for three. (b)(i) F = Le ✔. (ii) moles of Cu = 0.350 ÷ 63.5 = 5.51 × 10⁻³; Q = It = 0.600 × 30 × 60 = 1080 C; electrons = 1080 ÷ 1.60 × 10⁻¹⁹ = 6.75 × 10²¹; Cu²⁺ ions = 6.75 × 10²¹ ÷ 2 = 3.375 × 10²¹; L = 3.375 × 10²¹ ÷ 5.51 × 10⁻³ = 6.12 × 10²³ mol⁻¹ — all five points for four marks. Examiner insight: errors in (a) were H₂ at the cathode for molten PbBr₂, O₂ and Na for concentrated NaCl, and NO₂ and H₂ for Cu(NO₃)₂. In (b)(ii) most found the moles of copper and the charge, but some then used the printed value of F instead of calculating L from e.

Question 24A.2[5]

Answer and marking guidance

(a) anode O₂ AND cathode H₂ ✔. (b) Q = 1.5 × 3600 × 4.5 = 24 300 C ✔; moles of electrons = 24 300 ÷ 96 500 = 0.252 ✔; volume of Cl₂ = 0.252 ÷ 2 × 24 = 3.02 dm³ ✔; mass of Na = 0.252 × 23 = 5.79 g ✔. Examiner insight: common errors in (a) were SO₂ at the anode and Na at the cathode. In (b) the commonest faults were stopping after 24 300 C and 0.2518 mol of electrons, and the wrong ratios (6.04 dm³ of Cl₂, 2.90 g of Na). The clearest answers explained each step.

Question 24A.3[5]

Answer and marking guidance

(a) moles of H₂ = 462 ÷ 24 000 = 0.01925 ✔; molecules = 0.01925 × 6.02 × 10²³ = 1.16 × 10²² ✔. (b) electrons = 2 × 1.16 × 10²² = 2.32 × 10²² ✔ (2H⁺ + 2e⁻ → H₂). (c) Q = 2.32 × 10²² × 1.6 × 10⁻¹⁹ = 3.71 × 10³ C ✔. (d) x = 3.71 × 10³ ÷ (14 × 60) = 4.4 A ✔. Examiner insight: in (a) some stopped at moles and did not use the Avogadro constant; in (b) omitting the ×2 was common.

Question 24A.4[2]

Answer and marking guidance

moles of Sn = 2.95 ÷ 118.7 = 0.0249; the same charge deposits ⅔ as many moles of Al (Al³⁺ needs 3e⁻, Sn²⁺ needs 2e⁻) = 0.0166 mol ✔; mass of Al = 0.0166 × 27 = 0.447 g (or 0.448 g) ✔. Examiner insight: most found the moles of tin but did not convert to moles of aluminium correctly, or multiplied by the M_r of Al₂O₃.

Question 24A.5[2]

Answer and marking guidance

Q = 4.75 × 10²² × 2 × 1.60 × 10⁻¹⁹ = 15 200 C, or Q = 2 × 96 500 × (4.75 × 10²² ÷ 6.02 × 10²³) = 15 228 C ✔; I = Q ÷ t = 15 200 ÷ (15 × 60) = 16.9 A ✔. Examiner insight: usually done well; a significant number lost a mark by basing the calculation on Mg⁺ instead of Mg²⁺.

Half-cells and electrode potentials24.2.1(a)

Dip a strip of zinc into a solution of zinc ions. At the surface, some zinc atoms lose electrons and enter the solution as Zn²⁺, leaving their electrons behind on the metal; some Zn²⁺ ions collect electrons from the metal and are deposited as atoms. Very quickly the two rates become equal and an equilibrium is set up:

At equilibrium the zinc carries a slight excess of electrons and the nearby solution a slight excess of positive ions, so there is a potential difference between metal and solution. For copper in copper(II) ions the equilibrium lies further to the right, fewer electrons are left on the metal, and the potential difference is different. A metal in contact with its ions — or, more generally, the two forms of any redox couple with an electrode — is a half-cell, and its tendency to gain or lose electrons is expressed as its electrode potential.

The potential difference of a single half-cell cannot be measured, because any measuring wire dipped into the solution becomes a second half-cell. What can be measured is the difference between two half-cells connected together. Electrode potentials are therefore quoted relative to an agreed reference, the standard hydrogen electrode, whose potential is defined as zero.

E^⦵ values are always written for the half-equation as a reduction, with electrons on the left. A large positive E^⦵ means the oxidised form (on the left) is readily reduced; a large negative E^⦵ means the reduced form (on the right) readily gives up electrons.

The standard hydrogen electrode24.2.2

The reference half-cell is the standard hydrogen electrode (SHE), based on the couple 2H⁺(aq) + 2e⁻ ⇌ H₂(g). Hydrogen does not conduct, so the equilibrium is set up on a platinum electrode that is coated with finely divided platinum ("platinum black"). The large surface area and catalytic surface let the hydrogen gas and the hydrogen ions reach equilibrium quickly; platinum itself is inert and takes no part.

• hydrogen gas at 101 kPa (1 atm);
• H⁺(aq) at 1.00 mol dm⁻³ — for example 1.00 mol dm⁻³ HCl;
• 298 K;
• a platinum electrode coated with platinum black.

The SHE is awkward in practice — it needs a supply of pure hydrogen at controlled pressure, and platinum is easily "poisoned" — so laboratories often use a secondary reference electrode whose potential against the SHE is already known. For measurement purposes, though, every E^⦵ is defined by comparison with the SHE.

Measuring standard electrode potentials24.2.3

To measure E^⦵ of any half-cell, connect it to a standard hydrogen electrode:

• the two electrodes are joined through a high-resistance voltmeter, so that almost no current flows. With no current, the half-cells stay at equilibrium and their concentrations do not change, so the reading is the true potential difference;
• the two solutions are joined by a salt bridge — a strip of filter paper soaked in, or a tube filled with, a solution of an unreactive salt such as KNO₃. It completes the circuit by letting ions move between the solutions, without allowing them to mix. KCl must not be used with silver-ion half-cells, because it would precipitate AgCl.

(a) A metal in contact with its ions

The metal is the electrode and dips into a 1.00 mol dm⁻³ solution of its ions: for example Cu(s) in 1.00 mol dm⁻³ Cu²⁺(aq). Connected to the SHE, it reads +0.34 V with the copper as the positive electrode, so E^⦵(Cu²⁺/Cu) = +0.34 V. Zinc gives a reading of 0.76 V with the zinc as the negative electrode, so E^⦵(Zn²⁺/Zn) = −0.76 V.

(a) A non-metal in contact with its ions

A non-metal cannot be the electrode. Chlorine gas at 101 kPa is bubbled over a platinum electrode dipping into 1.00 mol dm⁻³ Cl⁻(aq) — the arrangement mirrors the hydrogen electrode.

(b) Ions of the same element in two oxidation states

For Fe³⁺/Fe²⁺, neither species is a metal, so both are dissolved in the same solution, each at 1.00 mol dm⁻³, and an inert platinum electrode dips into it. For a couple that involves H⁺ — MnO₄⁻/Mn²⁺, for example — H⁺ must also be present at 1.00 mol dm⁻³, because it is part of the half-equation.

Standard cell potentials24.2.1(b), 24.2.4

When two half-cells are connected, the one with the more positive E^⦵ has the greater tendency to gain electrons, so it becomes the positive electrode and reduction occurs there. The other is the negative electrode and its half-equation runs backwards, as an oxidation. The cell potential is the difference between the two:

For a cell that is actually delivering current, E^⦵_cell calculated this way is always positive.

Polarity and the direction of electron flow24.2.5(a)

The polarity of each electrode follows directly from the E^⦵ values:

• the half-cell with the more negative (less positive) E^⦵ is the negative electrode — oxidation releases electrons there;
• the half-cell with the more positive E^⦵ is the positive electrode — electrons are consumed there by reduction;
• electrons flow through the external circuit (the wire) from the negative electrode to the positive electrode. Ions, not electrons, carry the charge through the salt bridge.

Predicting whether a reaction is feasible24.2.5(b)

A redox reaction is feasible under standard conditions if the E^⦵_cell for the reaction as written is positive. In practice: identify the species being reduced and the species being oxidised, then

The limits of the prediction

• Rate. A positive E^⦵_cell says a reaction can occur. If the activation energy is high, it may be too slow to observe — exactly the same limitation as for ΔG.
• Conditions. E^⦵ values apply at 1.00 mol dm⁻³, 1 atm and 298 K. At other concentrations the electrode potentials shift (the Nernst equation, below), and a reaction predicted as unfeasible can become feasible — or the reverse — when E^⦵_cell is small.

Oxidising and reducing agents: the electrochemical series24.2.6

Arranging half-equations in order of E^⦵ gives the electrochemical series, which ranks the strength of every species in it as an oxidising or reducing agent.

• The more positive the E^⦵, the more readily the species on the left is reduced — the stronger the oxidising agent. F₂, MnO₄⁻/H⁺ and Cl₂ are strong oxidising agents.
• The more negative the E^⦵, the more readily the species on the right is oxidised — the stronger the reducing agent. Na, Mg and Al are strong reducing agents.

An oxidising agent can oxidise the reduced form of any couple below it in the series. Chlorine (+1.36 V) oxidises bromide (+1.07 V) and iodide (+0.54 V); bromine oxidises iodide but not chloride; iodine oxidises neither — which is the displacement order of the halogens you met at AS, now explained quantitatively.

Constructing redox equations from half-equations24.2.7

The equation for a redox reaction is built from its two half-equations in three steps:

1. write the half-equation for the species reduced as it appears in the table, and the half-equation for the species oxidised reversed;
2. multiply one or both so that the numbers of electrons are equal;
3. add them, cancel the electrons, and cancel any species (H⁺, H₂O, OH⁻) that appear on both sides.

Then check both atoms and charge. The cell builder below does exactly these three steps for any pair of half-cells, and the ΔG it reports uses the number of electrons from step 2.

Past-paper practice · Set 24B · Standard electrode potentials and cells

Examination questions on this part of the unit. Try each one on paper before opening the answer.

Question 24B.1[8]

Answer and marking guidance

(i) The voltage produced by a half-cell compared with a standard hydrogen electrode ✔; at 1 mol dm⁻³, 298 K and 1 atm ✔. (ii) Mg wire and Pt wire ✔; voltmeter, salt bridge and complete circuit ✔; solutions of Mg²⁺ and of MnO₄⁻, Mn²⁺ and H⁺ ✔. (iii) Mg negative, Pt positive AND the electron arrow pointing towards the MnO₄⁻/Mn²⁺ half-cell ✔. (iv) E^⦵_cell = 1.52 − (−2.38) = +3.90 V ✔. (v) 5Mg + 2MnO₄⁻ + 16H⁺ → 5Mg²⁺ + 2Mn²⁺ + 8H₂O ✔. (vi) No change (in the direction of the cell reaction) AND dilution makes the Mg²⁺/Mg potential even more negative ✔. Examiner insight: in (i) the commonest errors were "cell" instead of "electrode" or "half-cell", and units such as 101 Pa or 101 atm. In (ii) H⁺ was often omitted. Part (vi) was very difficult: dilution makes E(Mg²⁺/Mg) more negative but E_cell more positive, and answers had to make clear which voltage they meant.

Question 24B.2[7]

Answer and marking guidance

(i) Sn²⁺ and Sn⁴⁺ each at 1 mol dm⁻³ AND 298 K ✔. (ii) Any of: Pt (or C) electrodes in both half-cells; Sn²⁺/Sn⁴⁺ solution and H⁺ solution; a workable H₂ delivery system; H₂ labelled; voltmeter with complete circuit and salt bridge touching both solutions; salt bridge labelled — three marks for the full set. (iii) No reaction ✔. Cl⁻ cannot be reduced, so the only possible change is Cl⁻ reducing Sn²⁺ to Sn: E^⦵_cell = E^⦵(Sn²⁺/Sn) − E^⦵(Cl₂/Cl⁻) = −0.14 − 1.36 = −1.5 V, which is negative ✔. (The mark scheme also accepted the argument that E^⦵(Sn⁴⁺/Sn²⁺) is less positive than E^⦵(Cl₂/Cl⁻).) (iv) Sn²⁺ → Sn⁴⁺ and VO²⁺ → V³⁺ ✔; Sn²⁺ + 2VO²⁺ + 4H⁺ → Sn⁴⁺ + 2V³⁺ + 2H₂O ✔. Examiner insight: diagram errors were a Sn electrode instead of Pt, H⁺ in the tin half-cell, and Sn⁴⁺ and Sn²⁺ in different half-cells. Part (iii) proved difficult: some combined the right E^⦵ values the wrong way round and obtained a positive E_cell. Part (iv) was found hard to construct.

Question 24B.3[8]

Answer and marking guidance

(i) E becomes more positive (less negative) than E^⦵ ✔; because the lower [H₂PO₂⁻] shifts the equilibrium to the right-hand side ✔ (the Nernst equation was an accepted alternative). (ii) E_cell = −0.74 − (−1.57) = +0.83 V ✔. (iii) Any three of Pt electrode, Cr electrode, H⁺(aq), Cr³⁺(aq), H₂(g), voltmeter, labelled salt bridge, conditions of 1 atm and 1 mol dm⁻³, a complete circuit — one mark; six — two; all nine — three. (iv) Pt electrode positive AND electrons flowing (anticlockwise) towards the SHE ✔. (v) H₂PO₂⁻ + 3OH⁻ + Ni²⁺ → HPO₃²⁻ + 2H₂O + Ni ✔. Examiner insight: (i) needed Le Chatelier's principle (or the Nernst equation) applied to the half-equation. In (ii) the common errors were −2.31 and −0.83. In (iii) the errors were omitting the Pt label, labelling the Cr electrode as Pt, and naming the electrolyte "Cr³⁺/Cr". Part (iv) was found difficult, and in (v) equations were often unbalanced for atoms and charge.

How concentration changes an electrode potential24.2.8

An electrode potential belongs to an equilibrium, so Le Chatelier's principle predicts how it responds to a change in concentration. For any half-equation written as a reduction,

• increasing the concentration of the oxidised form (or decreasing the reduced form) shifts the equilibrium to the right. The half-cell takes up electrons more readily and its E becomes more positive;
• decreasing the concentration of the oxidised form (or increasing the reduced form) shifts it to the left, and E becomes more negative (less positive).

Diluting a Cu²⁺/Cu half-cell therefore lowers its E; diluting the Zn²⁺/Zn half-cell of a zinc–copper cell makes E(Zn) more negative and so increases E_cell. When a question asks about "the voltage", say which one — the half-cell E or the overall E_cell — because they can move in opposite directions.

Concentration can reverse a prediction

Copper(II) ions and iodide ions react to give iodine and a white precipitate of copper(I) iodide, even though E^⦵(Cu²⁺/Cu⁺) = +0.15 V is less positive than E^⦵(I₂/I⁻) = +0.54 V, which predicts no reaction under standard conditions. The explanation is concentration: CuI is precipitated, so [Cu⁺] falls far below 1 mol dm⁻³. Removing the reduced form pulls Cu²⁺ + e⁻ ⇌ Cu⁺ to the right, E(Cu²⁺/Cu⁺) becomes more positive than +0.54 V, and the reaction becomes feasible.

The Nernst equation24.2.9

The Nernst equation makes the qualitative argument quantitative:

The equation shows that E rises by 0.059/z volts for every ten-fold increase in the ratio [oxidised]/[reduced]. When the ratio is 1, log 1 = 0 and E = E^⦵. For a metal/ion couple such as Cu²⁺/Cu the metal is a solid, so only [Cu²⁺] appears.

Cell potential and Gibbs free energy24.2.10

A positive E^⦵_cell and a negative ΔG^⦵ are two ways of saying that a reaction is feasible, and they are linked exactly:

The electrical work a cell can do is charge × voltage: nF coulombs pushed through E volts. The minus sign makes a positive E^⦵_cell correspond to a negative ΔG^⦵. Units work out because 1 V × 1 C = 1 J.

Past-paper practice · Set 24C · The Nernst equation and ΔG = −nE°cellF

Examination questions on this part of the unit. Try each one on paper before opening the answer.

Question 24C.1[8]

Answer and marking guidance

(a) The potential difference (voltage) between the two half-cells (electrodes) of a cell under standard conditions ✔. (b) Any three of salt bridge, voltmeter, complete circuit (salt bridge touching both solutions), Cl₂, Cl⁻, a good gas delivery system, Pt, Pt, Fe²⁺ and Fe³⁺ — one mark; six — two; all nine — three. (c) ΔG^⦵ = −nFE^⦵_cell = −2 × 96 500 × 0.59 ✔ = −113 870 J mol⁻¹ = −114 kJ mol⁻¹ ✔. (d)(i) E = 0.77 + (0.059/1) log (1/0.15) = +0.82 V ✔ (z = 1). (ii) E_cell = 1.36 − 0.82 = +0.54 V ✔. Examiner insight: in (a) many referred to the hydrogen electrode and so defined the standard electrode potential. Diagram errors in (b) were missing Pt labels, labelling the Fe³⁺/Fe²⁺ electrode as Fe, and "Cl₂/Cl⁻" as an electrolyte. In (c) the common errors were −56.9 (n = 1), +114 and −113 870. In (d)(i) the commonest error was 0.72, from inverting the ratio.

Question 24C.2[4]

Data-booklet values needed: E^⦵(Cu²⁺/Cu⁺) = +0.15 V; E^⦵(I₂/I⁻) = +0.54 V.

Answer and marking guidance

(c) Cu²⁺/Cu⁺ E^⦵ = +0.15 V AND I₂/I⁻ E^⦵ = +0.54 V ✔; no, because E^⦵_cell is negative (−0.39 V) — or because I₂/I⁻ is the more positive, so I₂ is the stronger oxidant ✔. (d) E(Cu²⁺/Cu⁺) becomes more positive as the equilibrium shifts to the right when Cu⁺ is removed ✔; the new E is more positive than 0.54 V, so I⁻ can now reduce Cu²⁺ ✔. Examiner insight: part (c) was answered well by many.

Misconceptions and the examiner's view24.1–24.2

How the topic is assessed

Self-test

Twelve questions across the whole unit, each with the reasoning behind the answer.

Definitions to learn

Data used in this chapter

Standard electrode potentials in the text, the figure of the electrochemical series and the models are reference values. Where an examination question printed a value it is the one used (for example MnO₄⁻/Mn²⁺ +1.52 V, Mg²⁺/Mg −2.38 V, Cr³⁺/Cr −0.74 V, Fe³⁺/Fe²⁺ +0.77 V, Cu²⁺/Cu⁺ +0.15 V, I₂/I⁻ +0.54 V); other values are standard reference values. They are not the official data booklet, and different sources differ in the second decimal place (Br₂/Br⁻ appears as +1.07 and +1.09 V, for example). Always use the value a question gives. Constants: F = 9.65 × 10⁴ C mol⁻¹, L = 6.022 × 10²³ mol⁻¹, e = 1.60 × 10⁻¹⁹ C, V_m = 24.0 dm³ mol⁻¹ at room conditions — the values printed with the examination papers.

The electrolysis model uses clear-cut cases only. Where two electrode processes are close in potential (zinc, iron or nickel ions at the cathode, for example), the outcome in practice also depends on factors outside this syllabus, so those cases are not included.

Summary

24.1 Electrolysis

• Cathode (−): reduction. Anode (+): oxidation. Ions carry charge in the electrolyte, electrons in the wires.
• Molten: the salt's own ions are discharged. Aqueous: water competes at both electrodes.
• Cathode: metal if its E^⦵ is more positive than for water (Cu, Ag); otherwise H₂. Anode: I⁻ and Br⁻ give halogens; concentrated Cl⁻ gives Cl₂, dilute gives O₂; sulfate and nitrate give O₂; a copper anode dissolves.
• F = Le; Q = It; moles of electrons = Q/F; moles of product = moles of electrons ÷ electrons per particle.
• L from a copper anode: L = (It ÷ e ÷ 2) ÷ (Δm ÷ A_r).

24.2 Electrode and cell potentials

• E^⦵ is measured against the SHE under standard conditions with a high-resistance voltmeter and a salt bridge.
• Metal/ion: the metal is the electrode. Non-metal/ion or two ions of one element: platinum electrode, both species present, H⁺ included if in the half-equation.
• E^⦵_cell = E^⦵(positive) − E^⦵(negative); the more positive half-cell is the positive electrode; electrons flow from negative to positive in the wire.
• A reaction is feasible if E^⦵_cell (reduced − oxidised) is positive — subject to rate and concentration.
• More positive E^⦵: stronger oxidising agent on the left. More negative: stronger reducing agent on the right.
• Raising [oxidised] makes E more positive; the Nernst equation quantifies it: E = E^⦵ + (0.059/z) log([ox]/[red]).
• ΔG^⦵ = −nE^⦵_cellF: positive E^⦵_cell ↔ negative ΔG^⦵.

Examination checklist

• Have I checked whether the electrolyte is molten or aqueous, and considered water at both electrodes?
• Did I write the electrode half-equation before calculating an amount of product?
• Is time in seconds, and have I carried the calculation all the way to the quantity asked for?
• In an Avogadro calculation, did I use e (not F) and divide the electrons by two for Cu²⁺?
• Does my E^⦵ definition say "half-cell", compare with the SHE, and give 1 mol dm⁻³, 1 atm (101 kPa) and 298 K?
• Does my cell diagram show Pt where needed, both ions of a couple in one beaker, H⁺ only where it belongs, a salt bridge, a voltmeter and the conditions?
• For feasibility, have I quoted both E^⦵ values and subtracted (reduced − oxidised)?
• Is my redox equation balanced for atoms and charge, with E^⦵ values unscaled?
• For a concentration change, have I said which way the equilibrium shifts, whether E becomes more or less positive, and which voltage I mean?
• In ΔG = −nE^⦵F, is n the electrons in the balanced equation, is the sign right, and is the unit converted to kJ?

Knowledge organiser