Standard electrode potentials

Every half-cell has a tendency to gain electrons (to be reduced). We cannot measure that tendency on its own — only a difference — so we measure everything against a reference: the standard hydrogen electrode (SHE), which is defined as E^⦵ = 0.00 V.

The standard electrode potential, E^⦵, of a half-cell is the voltage measured when it is connected to a standard hydrogen electrode under standard conditions:

• All solutions at 1.00 mol dm⁻³;
• Any gases at 100 kPa;
• Temperature 298 K;
• An inert platinum electrode where the half-cell contains no metal (e.g. Fe³⁺/Fe²⁺), and a high-resistance voltmeter so that almost no current flows and the maximum potential difference is measured.

Half-equations are always written as reductions, with the electrons on the left:

Why a salt bridge? It completes the electrical circuit and allows ions to move to balance the charge building up in each beaker, without the two solutions mixing. It is usually saturated KNO₃ — chosen because neither K⁺ nor NO₃⁻ reacts with the electrolytes (KCl would precipitate AgCl in a silver half-cell).

Calculating E^⦵_cell

Put two half-cells together. The one with the more positive E^⦵ is reduced (it runs forwards, as written); the one with the more negative E^⦵ is oxidised (it runs backwards) and is the negative electrode, from which electrons flow through the wire.

Do not multiply E^⦵ values. Electrode potentials are intensive: even if you double a half-equation to balance the electrons, E^⦵ is unchanged.

Your turn — E^⦵_cell

Standard conditions

Predicting whether a redox reaction will go

Combine the two half-equations so that E^⦵_cell comes out positive. That combination is the thermodynamically feasible direction (it corresponds to ΔG being negative — the link is ΔG = −nFE^⦵_cell).

Three limitations you must be able to state:

• Kinetics. E^⦵_cell says nothing about rate. A reaction with a large positive E^⦵_cell may have a very high activation energy and be immeasurably slow.
• Non-standard conditions. Real reactions are rarely at 1 mol dm⁻³ and 298 K. Changing concentration shifts the electrode potential (Le Chatelier applied to the half-equation), so a value close to zero can flip sign.
• Not in solution. E^⦵ values apply to aqueous half-cells; they cannot predict reactions of solids or gases in other phases.

A positive E^⦵_cell — but no reaction

Storage cells and fuel cells

A commercial cell is just two half-cells packaged together. The e.m.f. is the difference of the two electrode potentials; a bigger difference gives a bigger voltage.

• Non-rechargeable (e.g. alkaline zinc/manganese): the reaction goes to completion and cannot be reversed.
• Rechargeable (e.g. lithium-ion): applying an external voltage drives the reaction backwards, regenerating the reactants.
• Fuel cell: the reactants are supplied continuously from outside, so the cell never goes flat as long as fuel flows.

The hydrogen–oxygen fuel cell in alkaline conditions:

Check that overall equation the way an examiner does: add the two half-equations, cancel the 4e⁻ and cancel 4OH⁻ on each side; 4H₂O − 2H₂O leaves 2H₂O. Atoms and charges balance.

Balanced judgement, please. Fuel cells emit only water at the point of use and are more efficient than a combustion engine — but most hydrogen is currently made from methane (releasing CO₂), and hydrogen is hard to store and transport. That nuance is what separates a 3-mark answer from a 1-mark one.

What makes a fuel cell different?

Manganate(VII) titrations

Potassium manganate(VII) is a powerful oxidising agent in acid. Learn this half-equation cold:

Against iron(II), Fe²⁺ → Fe³⁺ + e⁻. To cancel the electrons you need five Fe²⁺ per MnO₄⁻:

• Self-indicating: MnO₄⁻ is deep purple, Mn²⁺ is almost colourless. The end point is the first permanent pale pink from one drop of excess.
• Acidify with dilute sulfuric acid — an excess of it. Never hydrochloric acid (MnO₄⁻ would oxidise Cl⁻ to Cl₂, using up the titrant and giving a falsely high titre) and never nitric acid (itself an oxidising agent).
• Too little acid and you get a brown MnO₂ precipitate instead of Mn²⁺.

Why sulfuric acid?

Your turn — concentration of Fe²⁺

Your turn — percentage of iron in steel

The other big oxidising agent: dichromate(VI)

Acidified dichromate(VI) is orange and is reduced to green Cr³⁺. It is not self-indicating, so it is used with a redox indicator — but you must still be able to balance it.

Against iron(II) the ratio is 1 : 6 (six electrons, six Fe²⁺):

Ratios to memorise: MnO₄⁻ : Fe²⁺ = 1 : 5 · Cr₂O₇²⁻ : Fe²⁺ = 1 : 6 · I₂ : S₂O₃²⁻ = 1 : 2. Get the ratio wrong and every mark after it goes.

Balancing the dichromate equation

Iodine–thiosulfate titrations

This is the standard way of measuring an oxidising agent. The oxidising agent liberates iodine from excess iodide; the iodine is then titrated with sodium thiosulfate of known concentration.

Sulfur is oxidised from +2 in S₂O₃²⁻ to +2.5 (an average) in the tetrathionate ion S₄O₆²⁻ — one of the few places you will meet a fractional oxidation number.

• Indicator: starch, giving a blue-black complex with iodine. Add it only near the end point, when the solution has faded to pale straw — starch adsorbs iodine strongly and if added early the colour is slow to discharge, giving a false, high titre.
• End point: the blue-black colour disappears sharply.

Your turn — concentration of iodine

Your turn — copper in an alloy

When do you add the starch?

Oxidation numbers

Work out the oxidation number of the element in bold, then tap the right bin. Remember: O is −2, H is +1, and the numbers must add up to the overall charge.

Match the piece of apparatus to its job

Tap an item on the left, then its partner on the right.

The big ideas to know

You've now covered Topic 14: Redox II of the Edexcel A-level Chemistry (9CH0) specification. Press Finish to see your score.