Everything from the 4 lessons of this topic on a few pages: the key idea of each lesson, the diagram to remember, the points to learn and the traps examiners set. Read it, cover it, test yourself.
In electrolysis the cathode (negative) reduces and the anode (positive) oxidises. In aqueous solution the ions compete with water: at the cathode the species with the more positive E⊖ is reduced (Cu2+ and Ag+ before H2O; H2 forms instead of Na or K), and at the anode the species with the less positive E⊖ is oxidised, unless a high concentration of Cl− makes chlorine form instead of oxygen. The charge passed is Q = It, and one mole of electrons carries the Faraday constant, F = Le = 96 500 C mol−1. So 2.00 A for 965 s passes 1930 C = 0.0200 mol e−, depositing 0.0100 mol (0.635 g) of copper. Measuring the charge and the mass deposited gives L = F ÷ e.
Time must be in seconds (minutes times 60). Use the half-equation to convert moles of electrons into moles of product: 2 e− per Cu, 3 e− per Al, 1 e− per Ag, 4 e− per O2. In aqueous solution, reactive metals such as Na and K are never deposited: hydrogen forms instead. Dilute chloride solutions give mainly oxygen at the anode, not chlorine.
The standard electrode potential, E⊖, of a half-cell is the potential difference between it and the standard hydrogen electrode (SHE), measured under standard conditions: 298 K, 1.00 mol dm−3 for every ion in solution and 100 kPa (Cambridge also accepts 101 kPa) for any gas. The SHE is H2 gas bubbling over platinum (coated with platinum black) in 1.00 mol dm−3 H+, and is defined as 0.00 V. A metal dips into its own ions; an ion/ion couple such as Fe3+/Fe2+, or a non-metal such as Cl2/Cl−, needs an inert platinum electrode. Half-equations are written as reductions, so the more positive E⊖, the stronger the oxidising agent on the left; the more negative, the stronger the reducing agent on the right.
E⊖ values are always quoted for the reduction. Do not double E⊖ when a half-equation is multiplied up: it is a property of the couple, not of the amount. An ion/ion or gas/ion half-cell needs a platinum electrode and, for ion/ion, equal 1.00 mol dm−3 concentrations of both ions. A more positive E⊖ means the species on the left is a stronger oxidising agent, not that the species on the right is.
When two half-cells are joined, the one with the more positive E⊖ is the positive electrode: reduction happens there. The other is oxidised, and electrons flow through the wire from the negative to the positive electrode. The standard cell potential is E⊖cell = E⊖(positive) − E⊖(negative), e.g. Zn/Cu: +0.34 − (−0.76) = +1.10 V. A reaction is feasible when E⊖cell for it, as written, is positive. E values say nothing about rate (a feasible reaction may have a high activation energy) and apply only under standard conditions. The cell potential links to free energy: ΔG⊖ = −nFE⊖cell, so a positive Ecell gives a negative ΔG.
E⊖cell = E⊖(positive) − E⊖(negative): never add them, and never multiply E⊖ by the coefficients used to balance electrons. Electrons flow in the external wire, from negative to positive; ions, not electrons, carry the current through the salt bridge. In ΔG = −nFE, n is moles of electrons, and the answer in J must be converted to kJ. A feasible reaction may still be too slow to see.
For any half-cell written as a reduction, raising the concentration of the oxidised form pushes the equilibrium to the right, so E becomes more positive; raising the reduced form makes it less positive. The Nernst equation puts a number on this at 298 K: E = E⊖ + (0.059 ÷ z) log ([oxidised] ÷ [reduced]), where z is the number of electrons in the half-equation. Solids are left out, so for a metal E = E⊖ + (0.059 ÷ z) log [Mz+]. For Cu2+ at 0.010 mol dm−3: E = 0.34 + 0.0295 × log 0.010 = 0.34 − 0.059 = +0.281 V. A tenfold dilution changes E by only 0.059 ÷ z volts, which is why E⊖ predictions are reliable unless E⊖cell is small.
Use log to base 10, not ln, with the 0.059 V form. Divide 0.059 by z, the electrons in the half-equation (2 for Cu2+, 1 for Ag+ and Fe3+). Put the oxidised species on the top of the fraction, and raise concentrations to their coefficients (2Cl− gives [Cl−]2). Diluting the oxidised form makes E less positive, never more.