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H2 Chemistry · JC 1 and 2

Electrochemistry

Electrochemistry links redox reactions to electricity. In H2 Chemistry you use standard electrode potentials to calculate cell potentials and predict whether redox reactions are feasible, connect cell potential to Gibbs free energy, and use Faraday's laws to calculate the amounts of substances produced in electrolysis. The topic draws together redox from earlier years, equilibrium, and the energetics of ΔG, and its reasoning appears in questions on batteries, corrosion and industrial electrolysis.

What the syllabus expects

  • Define standard electrode potential and describe the standard hydrogen electrode
  • Calculate standard cell potentials from E° values and identify the positive and negative electrodes
  • Use E° values to predict whether a redox reaction is feasible and to compare oxidising and reducing agents
  • Use ΔG° = −nFE°cell to relate cell potential to Gibbs free energy
  • Predict qualitatively how changes in concentration affect electrode potentials
  • Predict the products of electrolysis and calculate quantities using Q = It and the Faraday constant

Electrode potentials and the hydrogen electrode

When a metal is placed in a solution of its ions, an equilibrium is set up, such as Cu²⁺(aq) + 2e⁻ ⇌ Cu(s), and a potential difference develops. This cannot be measured on its own, so every half-cell is measured against the standard hydrogen electrode. That electrode has hydrogen gas at 1 bar bubbling over platinum in 1 mol dm⁻³ H⁺ at 298 K, and is given a potential of 0.00 V. The standard electrode potential, E°, of a half-cell is its potential relative to the standard hydrogen electrode under standard conditions: 298 K, 1 bar for gases and 1 mol dm⁻³ for ions. E° values are written for reduction. A more positive E° means the oxidised form is more easily reduced, so it is a stronger oxidising agent. A more negative E° means the reduced form is more easily oxidised, so it is a stronger reducing agent. For example, E°(Zn²⁺/Zn) = −0.76 V and E°(Cu²⁺/Cu) = +0.34 V.

Cell potentials and feasibility

Joining two half-cells with a wire and a salt bridge makes an electrochemical cell. The half-cell with the more positive E° is the positive electrode, where reduction happens, and the more negative one is the negative electrode, where oxidation happens. E°cell = E°(reduction half-cell) − E°(oxidation half-cell). For the zinc and copper cell, E°cell = +0.34 − (−0.76) = +1.10 V. Electrons flow through the external circuit from zinc to copper, while the salt bridge, often saturated potassium nitrate, lets ions move to keep each solution electrically neutral. A reaction is feasible under standard conditions if E°cell is positive. When combining half-equations to write the overall equation, multiply them to balance electrons, but never multiply the E° values, because electrode potential does not depend on the amount of substance.

Linking E°cell to ΔG and concentration

Cell potential and Gibbs free energy are linked by ΔG° = −nFE°cell, where n is the number of moles of electrons transferred in the balanced equation and F is the Faraday constant, 96 500 C mol⁻¹. A positive E°cell therefore gives a negative ΔG°, which is why it signals a feasible reaction. For the zinc and copper cell, n = 2, so ΔG° = −2 × 96 500 × 1.10 = −212 300 J mol⁻¹, about −212 kJ mol⁻¹. A feasible reaction may still not be observed, because the rate may be very slow, and real conditions may not be standard. Changes in concentration shift each half-cell equilibrium. Increasing [Cu²⁺] shifts Cu²⁺ + 2e⁻ ⇌ Cu to the right, making the electrode potential more positive. Decreasing [Zn²⁺] makes the zinc electrode potential more negative. Both changes would make the zinc and copper cell potential larger.

Electrolysis and Faraday calculations

In electrolysis an external power supply drives a non-spontaneous reaction. Oxidation happens at the anode, which is now positive, and reduction at the cathode, which is negative. To predict products, compare E° values: at the cathode, the species with the most positive E° is reduced first, and at the anode, the species most easily oxidised reacts first. Concentration and the electrode material can change the outcome. Concentrated chloride solutions give chlorine at an inert anode even though E°(Cl₂/Cl⁻) = +1.36 V is more positive than E°(O₂/H₂O) = +1.23 V, and a copper anode dissolves rather than releasing oxygen. Quantities follow from charge: Q = It, with I in amperes and t in seconds, and the moles of electrons = Q ÷ F. The Faraday constant is the charge on one mole of electrons, F = Le, where L is the Avogadro constant and e is the charge on one electron. Use the half-equation to convert moles of electrons into moles of product.

Worked example

A current of 0.500 A is passed through aqueous silver nitrate for 32 minutes 10 seconds using inert electrodes. Calculate the mass of silver deposited at the cathode and the volume of oxygen produced at the anode at room conditions. (F = 96 500 C mol⁻¹, Ar of Ag = 108, molar volume of gas = 24.0 dm³ mol⁻¹ at r.t.p.)

  1. Time in seconds: t = 32 × 60 + 10 = 1930 s.
  2. Charge: Q = It = 0.500 × 1930 = 965 C.
  3. Moles of electrons = Q ÷ F = 965 ÷ 96 500 = 0.0100 mol.
  4. Cathode: Ag⁺ + e⁻ gives Ag, so 0.0100 mol of Ag forms, and mass = 0.0100 × 108 = 1.08 g.
  5. Anode: 2H₂O gives O₂ + 4H⁺ + 4e⁻, so moles of O₂ = 0.0100 ÷ 4 = 0.00250 mol.
  6. Volume of O₂ = 0.00250 × 24.0 = 0.0600 dm³ = 60.0 cm³.

Answer: 1.08 g of silver is deposited and 60.0 cm³ of oxygen is produced at r.t.p.

Common mistakes

  • Multiplying E° values by the coefficients used to balance electrons. E° does not depend on the amount of substance, so it stays the same.
  • Subtracting the wrong way round. E°cell = E°(reduction) − E°(oxidation), and a negative result means the reaction is not feasible as written.
  • Leaving time in minutes in Q = It. Convert to seconds first.
  • Forgetting the number of electrons per ion. One mole of Cu²⁺ needs two moles of electrons, so moles of copper = moles of electrons ÷ 2.
  • Saying the salt bridge carries electrons. It carries ions, while electrons flow only through the external wire.
  • Assuming the anode is always positive. In an electrochemical cell the anode (oxidation) is the negative electrode, while in electrolysis it is positive.

Exam tips

  • Write both half-equations with their E° values before calculating anything, then mark which one is reversed for oxidation. This keeps the sign of E°cell straight.
  • In Faraday questions, set out the chain charge, moles of electrons, moles of product, then mass or volume, with the half-equation written beside the mole ratio step.
  • When asked to explain feasibility, say both that E°cell is positive and that this makes ΔG° negative, and add a note on rate or non-standard conditions if the question hints that the reaction is not observed.

Practice questions

1. Easy: Calculate E°cell for a cell made from Mg²⁺/Mg (E° = −2.38 V) and Ag⁺/Ag (E° = +0.80 V) half-cells, and identify the positive electrode.

Show answer

E°cell = +0.80 − (−2.38) = +3.18 V. The silver electrode, with the more positive E°, is the positive electrode where Ag⁺ is reduced.

2. Medium: Given E°(Fe³⁺/Fe²⁺) = +0.77 V, E°(I₂/I⁻) = +0.54 V and E°(Br₂/Br⁻) = +1.07 V, predict whether Fe³⁺ can oxidise I⁻ and Br⁻ under standard conditions. Calculate ΔG° for the feasible reaction.

Show answer

For 2Fe³⁺ + 2I⁻ gives 2Fe²⁺ + I₂, E°cell = 0.77 − 0.54 = +0.23 V, so it is feasible. For Br⁻, E°cell = 0.77 − 1.07 = −0.30 V, so it is not feasible. For the iodide reaction n = 2, so ΔG° = −2 × 96 500 × 0.23 = −44 390 J mol⁻¹, about −44.4 kJ mol⁻¹.

3. Harder: How long would it take to deposit 0.635 g of copper from aqueous copper(II) sulfate using a current of 2.00 A? (F = 96 500 C mol⁻¹, Ar of Cu = 63.5)

Show answer

965 s (16 minutes 5 seconds). Moles of Cu = 0.635 ÷ 63.5 = 0.0100 mol. From Cu²⁺ + 2e⁻ gives Cu, moles of electrons = 0.0200 mol. Q = 0.0200 × 96 500 = 1930 C. t = Q ÷ I = 1930 ÷ 2.00 = 965 s.

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