Electrochemistry

Every objective in this topic is printed under "A Level subject content" in the 9701 syllabus and carries the tier "A Level". None of it is AS, and it is a separate topic from the AS topic 6 of the same name. It is examined on Paper 4. Paper 1 is the AS multiple-choice paper and the whole 9701 bank on this site comes from it, so no practice here is tagged to this topic.

Syllabus points

24.1 Electrolysis

• Predict the identities of substances liberated during electrolysis from the state of electrolyte (molten or aqueous), position in the redox series (electrode potential) and concentration.
• State and apply the relationship F = Le between the Faraday constant, F, the Avogadro constant, L, and the charge on the electron, e.
• Calculate the quantity of charge passed during electrolysis, using Q = It, and the mass and/or volume of substance liberated during electrolysis.
• Describe the determination of a value of the Avogadro constant by an electrolytic method.

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

• Define the terms standard electrode (reduction) potential and standard cell potential.
• Describe the standard hydrogen electrode.
• Describe methods used to measure the standard electrode potentials of metals or non-metals in contact with their ions in aqueous solution, and of ions of the same element in different oxidation states.
• Calculate a standard cell potential by combining two standard electrode potentials.
• Use standard cell potentials to deduce the polarity of each electrode and hence explain or deduce the direction of electron flow in the external circuit of a simple cell, and to predict the feasibility of a reaction.
• Deduce from E values the relative reactivity of elements, compounds and ions as oxidising agents or as reducing agents.
• Construct redox equations using the relevant half-equations.
• Predict qualitatively how the value of an electrode potential, E, varies with the concentrations of the aqueous ions.
• Use the Nernst equation, E = E° + (0.059/z) log([oxidised species]/[reduced species]), to predict quantitatively how the value of an electrode potential varies with the concentrations of the aqueous ions; examples include Cu²⁺(aq) + 2e⁻ ⇌ Cu(s) and Fe³⁺(aq) + e⁻ ⇌ Fe²⁺(aq).
• Understand and use the equation ΔG° = -nE°_cell F.

Predicting the products of electrolysis

At the cathode, the negative electrode, reduction occurs. At the anode, the positive electrode, oxidation occurs. Two mnemonics that never fail: reduction happens at the cathode, and oxidation is loss.

Molten electrolyte. There is only one cation and one anion, so the products are simply the metal at the cathode and the non-metal at the anode. Molten lead(II) bromide gives lead and bromine.

Aqueous electrolyte. Water is present, so there is a competition at each electrode, and three factors decide it.

• Position in the redox series. The species with the more positive electrode potential is reduced at the cathode; the species with the more negative electrode potential is oxidised at the anode. So a metal below hydrogen in reactivity, such as copper or silver, is deposited, while for a reactive metal such as sodium or potassium the water is reduced instead and hydrogen is given off.
• Concentration. A concentrated halide solution gives the halogen at the anode, even though on electrode potentials alone oxygen would be expected. A dilute solution gives oxygen. Concentrated sodium chloride solution gives chlorine; dilute gives oxygen.
• The nature of the electrode. An inert electrode such as platinum or graphite takes no part. A copper anode in copper(II) sulfate solution dissolves instead of releasing oxygen, which is the basis of electroplating and of copper purification.

Quantitative electrolysis

Three relationships, used in sequence:

Q = It

where Q is charge in coulombs, I current in amperes and t time in seconds.

F = Le

The Faraday constant F is the charge on one mole of electrons, 96500 C mol⁻¹. It is the product of the Avogadro constant L and the charge on one electron e.

Check that. 6.02 × 10²³ × 1.60 × 10⁻¹⁹ = 96320, which rounds to the 96500 C mol⁻¹ quoted with more precise values.

The number of moles of electrons passed is

moles of electrons = Q / F

and the amount of substance liberated follows from the half-equation, which tells you how many electrons each mole of product needs.

Worked example. A current of 1.50 A is passed through copper(II) sulfate solution for 30.0 minutes. What mass of copper is deposited? A_r(Cu) = 63.5.

Time in seconds:

30.0 × 60 = 1800

Charge:

Q = 1.50 × 1800 = 2700

Moles of electrons:

2700 / 96500 = 0.02798

The half-equation is Cu²⁺ + 2e⁻ → Cu, so each mole of copper needs two moles of electrons:

0.02798 / 2 = 0.01399

Mass:

0.01399 × 63.5 = 0.8884

so about 0.888 g of copper.

Worked example, a gas volume. What volume of oxygen, at room conditions where one mole occupies 24.0 dm³, is released at the anode in the same experiment? The half-equation is 2H₂O → O₂ + 4H⁺ + 4e⁻, so four electrons per mole of oxygen.

0.02798 / 4 = 0.006995

Volume:

0.006995 × 24.0 = 0.16788

so about 0.168 dm³, or 168 cm³.

The step that decides most of these questions is the number of electrons in the half-equation. Write the half-equation first, every time, and read the ratio off it.

Determining the Avogadro constant by electrolysis

The method is a direct application of F = Le, run backwards.

1. Electrolyse copper(II) sulfate solution using copper electrodes, with an ammeter and a variable resistor in the circuit to keep the current steady.
2. Clean, dry and weigh the cathode before the experiment.
3. Pass a known steady current for a measured time, recording both.
4. Remove the cathode, wash it gently with distilled water, dry it without rubbing off the deposit, and reweigh it.
5. The increase in mass gives the moles of copper deposited, and doubling that gives the moles of electrons, since Cu²⁺ needs two.
6. Q = It gives the total charge, so F = Q divided by the moles of electrons.
7. Finally L = F/e, using the known charge on an electron.

Worked example. A current of 0.500 A for 20.0 minutes deposits 0.197 g of copper.

Q = 0.500 × 1200 = 600

Moles of copper:

0.197 / 63.5 = 0.003102

Moles of electrons, twice that:

0.003102 × 2 = 0.006204

F = 600 / 0.006204 = 96712

L = 96712 / (1.60 × 10⁻¹⁹) = 6.0445 × 10²³

so about 6.04 × 10²³ mol⁻¹, which is within one per cent of the accepted value.

Standard electrode potentials

An electrode potential is measured, not calculated, and the definitions must be exact.

• Standard electrode (reduction) potential, E°: the potential of a half-cell relative to a standard hydrogen electrode, measured under standard conditions, with all solutions at 1.00 mol dm⁻³, any gas at 101 kPa and a temperature of 298 K.
• Standard cell potential, E°_cell: the potential difference between two half-cells under those same standard conditions.

The standard hydrogen electrode

The reference against which everything is measured, and it is defined as exactly 0.00 V.

It consists of hydrogen gas at 101 kPa bubbled over a platinum electrode coated with platinum black, dipping into a solution of H⁺ ions at 1.00 mol dm⁻³, all at 298 K.

Two details carry marks: the platinum is inert, so it conducts without reacting, and it is coated with platinum black to give a large surface area, which catalyses the establishment of the equilibrium H⁺ + e⁻ ⇌ ½H₂.

Measuring an electrode potential

Connect the half-cell to a standard hydrogen electrode with a salt bridge, usually filter paper or a tube of agar soaked in saturated potassium nitrate, which completes the circuit by allowing ions to move without the two solutions mixing. Potassium nitrate is chosen because neither of its ions forms a precipitate with common cell contents.

Connect a high-resistance voltmeter between the electrodes. It must be high resistance so that effectively no current flows, since a flowing current would change the ion concentrations and therefore the potential being measured.

For a metal in contact with its ions, the metal itself is the electrode. For a non-metal, or for two ions of the same element in different oxidation states, such as Fe³⁺ and Fe²⁺, there is no solid conductor, so an inert platinum electrode is dipped into a solution containing both species at 1.00 mol dm⁻³.

Using standard electrode potentials

E° values are quoted for reduction half-equations, written as reversible:

Zn²⁺(aq) + 2e⁻ ⇌ Zn(s), E° = -0.76 V

Cu²⁺(aq) + 2e⁻ ⇌ Cu(s), E° = +0.34 V

Read them like this:

• A more positive E° means the species on the left is more readily reduced, so it is a better oxidising agent.
• A more negative E° means the species on the right is more readily oxidised, so it is a better reducing agent.

So among the values above, Cu²⁺ is the better oxidising agent and Zn is the better reducing agent.

Combining two half-cells

E°_cell = E°(more positive) - E°(more negative)

so E°_cell is always positive when calculated this way.

Worked example, the Daniell cell.

0.34 - (-0.76) = 1.10

so E°_cell is +1.10 V.

The half-cell with the more negative E° runs in reverse, as oxidation, and is the negative electrode. The half-cell with the more positive E° runs as reduction and is the positive electrode. Electrons flow through the external circuit from the negative electrode to the positive one, so from zinc to copper.

Note the important thing about the value: E°_cell does not depend on how many electrons are involved, so half-equations are never multiplied when adding potentials, even when they must be multiplied to balance the overall equation.

Constructing the overall equation

Reverse the half-equation with the more negative E°, then multiply one or both so the electrons cancel, then add.

For the Daniell cell:

Zn → Zn²⁺ + 2e⁻ and Cu²⁺ + 2e⁻ → Cu

give Zn + Cu²⁺ → Zn²⁺ + Cu.

Where the electron numbers differ, for example combining Fe³⁺ + e⁻ ⇌ Fe²⁺ with Cl₂ + 2e⁻ ⇌ 2Cl⁻, multiply the iron half-equation by two before adding.

Predicting feasibility

A reaction is feasible if the calculated E°_cell for it is positive.

The practical test: write the two half-equations, identify the one that must run in reverse for your proposed reaction, and check that it is the one with the more negative E°. If so, the reaction goes.

Worked example. Will acidified potassium manganate(VII) oxidise Fe²⁺ to Fe³⁺?

MnO₄⁻ + 8H⁺ + 5e⁻ ⇌ Mn²⁺ + 4H₂O, E° = +1.52 V

Fe³⁺ + e⁻ ⇌ Fe²⁺, E° = +0.77 V

For manganate(VII) to oxidise iron(II), the iron half-equation must run in reverse, and it is the less positive of the two, so:

1.52 - 0.77 = 0.75

E°_cell is +0.75 V, positive, so the reaction is feasible. This is the basis of the standard manganate(VII) titration.

Two limitations to state whenever you use E° values in a longer answer. They apply only under standard conditions, and a change in concentration can reverse a marginal prediction. And, as in topic 23, a feasible reaction may still be immeasurably slow, because E° values say nothing about activation energy.

The Nernst equation

Non-standard concentrations shift the electrode potential, and the direction is predictable from Le Chatelier's principle applied to the half-equation.

For M^n+ + ne⁻ ⇌ M, increasing the concentration of the oxidised species pushes the equilibrium to the right, favouring reduction, so E becomes more positive. Decreasing it makes E more negative.

Quantitatively:

E = E° + (0.059/z) log([oxidised species] / [reduced species])

where z is the number of electrons in the half-equation. A solid does not appear in the expression, so for a metal electrode only the ion concentration is used.

Worked example. Find the electrode potential of a copper half-cell in which [Cu²⁺] is 0.100 mol dm⁻³. E° = +0.34 V and z = 2.

log 0.100 = -1, so:

E = 0.34 + (0.059 / 2) × (-1) = 0.3105

so E is about +0.31 V, slightly less positive than the standard value, exactly as Le Chatelier predicts for a lower concentration of the oxidised species.

Worked example, an ion pair. For Fe³⁺ + e⁻ ⇌ Fe²⁺, with [Fe³⁺] = 0.0100 and [Fe²⁺] = 1.00 mol dm⁻³. E° = +0.77 V and z = 1.

The ratio is 0.0100/1.00 = 0.0100, whose log is -2:

E = 0.77 + 0.059 × (-2) = 0.652

so E is about +0.65 V. Note that with z = 1 the effect of a given concentration ratio is twice as large as it would be with z = 2, which is why the factor is there.

Free energy and cell potential

ΔG° = -nE°_cell F

where n is the number of electrons transferred in the balanced overall equation and F is 96500 C mol⁻¹. The answer comes out in joules, so divide by 1000 for kJ mol⁻¹.

The minus sign links the two criteria for feasibility met in this and the previous topic: a positive E°_cell gives a negative ΔG°, and both mean the same thing. That equivalence is the point of the equation, and stating it earns the mark.

Unlike E°_cell, ΔG° does depend on n, because it is an energy for the reaction as written.

Worked example, the Daniell cell. E°_cell = +1.10 V and n = 2.

ΔG° = -2 × 1.10 × 96500 = -212300

so ΔG° is -212 kJ mol⁻¹, comfortably negative, and the reaction is feasible.

Common mistakes

• Saying oxidation happens at the cathode. Reduction is at the cathode, in electrolysis and in a cell.
• Predicting oxygen at the anode from a concentrated halide solution. Concentration overrides the electrode potential there.
• Leaving the time in minutes in Q = It.
• Forgetting to divide the moles of electrons by the number in the half-equation, so a copper mass comes out twice too large.
• Multiplying an E° value when a half-equation is multiplied to balance electrons. E° is intensive and never scales.
• Adding two E° values instead of subtracting.
• Saying the standard hydrogen electrode has a measured potential of zero. It is defined as zero.
• Leaving out the platinum black, the 101 kPa or the 1.00 mol dm⁻³ from a description of the hydrogen electrode.
• Using an ordinary voltmeter rather than a high-resistance one, so that current flows and the concentrations change.
• Saying a positive E°_cell means the reaction is fast.
• Including a solid in the Nernst expression.
• Forgetting that ΔG° from -nE°F comes out in joules.