HookWhy every phone battery says 3.7 V
Check any phone's spec sheet and the battery says 3.7 V. Every brand, every year — because the number is not a design choice. An AA alkaline cell says 1.5 V; a car battery says 12 V only because it is six 2.0 V lead–acid cells wired in series. Each of those voltages is fixed by chemistry: it is the gap between two entries in the electrochemical series — the electrode potentials of the two half-reactions running inside. Lithium sits at \(E^{\ominus} = -3.04\ \text{V}\), the most negative electrode potential of any practical metal, so cells built on lithium open the widest available gap and pack the most volts into a single cell — which, together with lithium's featherweight 6.9 g mol⁻¹, is why it owns portable power.
3.1.11 builds the ruler behind those numbers. You will saw redox reactions into half-cells, measure each half-cell's pull on electrons against an agreed zero — the standard hydrogen electrode — and then subtract two numbers to predict any cell's EMF and any redox reaction's feasibility. The theory gets cashed in twice: once for the batteries and fuel cells in your pocket and garage, and once for Required Practical 8, where you build the cells and take the readings yourself.
ModelHalf-cells — a redox reaction, sawn in half
Drop zinc into copper(II) sulfate solution and copper plates out while the beaker warms: electrons pass from zinc to copper ions on contact, and the energy leaves as heat. An electrochemical cell is that same reaction sawn in half. Put zinc in zinc sulfate in one beaker, copper in copper sulfate in another, join the metals by a wire and the solutions by a salt bridge, and the electrons now travel through the wire — where they can run a motor instead of warming a beaker.
Each beaker is a half-cell. At a metal electrode the equilibrium \(\text{M}^{n+}(\text{aq}) + n\text{e}^- \rightleftharpoons \text{M}(\text{s})\) sets up a charge separation whose size depends on the metal's willingness to shed electrons. Rank that willingness and you have the electrochemical series, with real numbers attached: \(\text{Li}^+/\text{Li}\) at \(-3.04\ \text{V}\), \(\text{Zn}^{2+}/\text{Zn}\) at \(-0.76\ \text{V}\), hydrogen at exactly 0, \(\text{Cu}^{2+}/\text{Cu}\) at \(+0.34\ \text{V}\), \(\text{Ag}^+/\text{Ag}\) at \(+0.80\ \text{V}\). The more negative the potential, the stronger the reduced species is as a reducing agent; the more positive, the stronger the oxidised species is as an oxidising agent.
The salt bridge — typically a strip of filter paper soaked in saturated potassium nitrate — completes the circuit by letting ions drift between the beakers without the solutions mixing. Electrons never cross it.
ModelThe standard hydrogen electrode — sea level for voltage
A voltmeter can only ever read a difference: no instrument measures one half-cell's potential on its own, for the same reason no altitude means anything without sea level. Chemistry's sea level is the standard hydrogen electrode (SHE): hydrogen gas at 100 kPa bubbling over a platinum electrode in a solution with \([\text{H}^+] = 1.00\ \text{mol dm}^{-3}\) at 298 K, hosting the equilibrium \(2\text{H}^+(\text{aq}) + 2\text{e}^- \rightleftharpoons \text{H}_2(\text{g})\). Its potential is defined as exactly 0 V. Platinum earns its place three times over: it is inert, it conducts, and roughened into platinum black it offers a large surface on which the gas equilibrium establishes quickly.
The standard electrode potential \(E^{\ominus}\) of any half-cell is the EMF of a cell in which that half-cell faces the SHE, with every solution at \(1.00\ \text{mol dm}^{-3}\), every gas at 100 kPa, and the whole assembly at 298 K. The standardisation is not pedantry. Each half-cell is an equilibrium, so changing a concentration shifts it and drags the potential along — dilute the copper solution and \(E(\text{Cu}^{2+}/\text{Cu})\) slides below its standard \(+0.34\ \text{V}\). Only under agreed conditions are two tabulated numbers comparable, which is the whole point of tabulating them.
MechanismEMF — subtract two numbers, predict any redox reaction
Pair any two half-cells and the more negative one becomes the negative electrode: its equilibrium sits furthest towards releasing electrons, so it is the site of oxidation, and electrons flow from it through the external circuit to the more positive electrode, where reduction runs. The cell EMF is the gap: \[ E_{\text{cell}} = E^{\ominus}_{\text{positive}} - E^{\ominus}_{\text{negative}} \] In the conventional cell representation the oxidised half goes on the left, the salt bridge is a double bar, and the reduction on the right — \(\text{Zn(s)} \mid \text{Zn}^{2+}(\text{aq}) \parallel \text{Cu}^{2+}(\text{aq}) \mid \text{Cu(s)}\) — so the same rule reads \(E_{\text{cell}} = E_{\text{right}} - E_{\text{left}}\). Where no metal is present to conduct — gas electrodes, or ion/ion couples like \(\text{Fe}^{3+}/\text{Fe}^{2+}\) — a platinum electrode is written in.
The same subtraction predicts chemistry far from batteries: a proposed redox reaction is feasible if the two half-equations give it a positive EMF — equivalently, the intended oxidising agent must belong to the more positive couple. Two caveats guard the top marks: feasibility is thermodynamic, so a feasible reaction may still be immeasurably slow (kinetic stability — a high activation energy); and \(E^{\ominus}\) values hold only at standard conditions, so a marginal prediction can flip when concentrations change.
The Daniell cell: \(E_{\text{cell}} = +0.34 - (-0.76) = +1.10\ \text{V}\), zinc the negative electrode — so zinc dissolves and copper deposits, exactly the displacement seen in the single beaker. Now the feasibility trick on acids: could copper fizz in dilute hydrochloric acid? The proposed pairing is \(2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2\) (0.00 V) oxidising copper (\(+0.34\ \text{V}\)): EMF \(= 0.00 - 0.34 = -0.34\ \text{V}\). Negative — not feasible — and indeed copper sits untouched in dilute acid, which is why it survives as water pipe. Zinc's version gives \(0.00 - (-0.76) = +0.76\ \text{V}\): feasible, and zinc duly fizzes.
CaseCommercial cells — lithium-ion, and the cell you refuel
AQA sorts cells into three families. Non-rechargeable cells run an irreversible reaction until a reagent is spent. Rechargeable cells use reactions that an applied external voltage can drive backwards, restoring the reactants. In the lithium-ion cell in your phone, the negative electrode releases lithium ions from graphite, \(\text{Li} \rightarrow \text{Li}^+ + \text{e}^-\), while at the positive electrode they lodge in a cobalt oxide lattice: \(\text{Li}^+ + \text{CoO}_2 + \text{e}^- \rightarrow \text{Li[CoO}_2\text{]}\). Overall: \(\text{Li} + \text{CoO}_2 \rightarrow \text{Li[CoO}_2\text{]}\); charging reverses both electrodes. Lithium's ferociously negative potential is what makes the roughly 3.7 V per cell possible — and also why the electrolyte is a flammable organic solvent rather than water, which lithium would attack on sight.
A fuel cell is different in kind: reactants are fed in continuously, so it converts rather than stores. In the alkaline hydrogen–oxygen fuel cell, hydrogen is oxidised at the negative electrode, \(\text{H}_2 + 2\text{OH}^- \rightarrow 2\text{H}_2\text{O} + 2\text{e}^-\) (\(E^{\ominus} = -0.83\ \text{V}\)), and oxygen is reduced at the positive, \(\text{O}_2 + 2\text{H}_2\text{O} + 4\text{e}^- \rightarrow 4\text{OH}^-\) (\(E^{\ominus} = +0.40\ \text{V}\)). The overall reaction is simply \(2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}\): water the only product, no recharging ever needed. The honest evaluation belongs alongside: hydrogen must be manufactured — today mostly by steam-reforming methane, which emits carbon dioxide — then compressed, stored and distributed. 'Zero-emission' is a claim about the tailpipe, not the supply chain.
Fuel-cell EMF straight from the data: \(E_{\text{cell}} = +0.40 - (-0.83) = +1.23\ \text{V}\). That one subtraction explains the engineering: a hydrogen car stacks scores of these 1.23 V cells in series to reach a working voltage, exactly as a 12 V car battery stacks six 2.0 V lead–acid cells.
CaseRequired practical 8 — the voltmeter that must not draw current
The task: build cells from metal/metal-ion half-cells, measure their EMFs, and check the results against the electrochemical series. The method that earns the marks: make each solution \(1.00\ \text{mol dm}^{-3}\) and work at 298 K; clean each metal strip with emery paper, because an oxide skin adds a stray potential; use a fresh salt bridge of filter paper soaked in saturated potassium nitrate — chosen because nitrates are soluble and unreactive towards the ions in use, where a chloride bridge would precipitate silver chloride in a silver half-cell; and read the cell with a high-resistance digital voltmeter.
That last item is the physics of the whole practical. EMF is the cell's potential difference when no current flows. Let current flow and the reading sags: the electrode equilibria are disturbed, and the concentrations start drifting from standard the moment the chemistry runs. High resistance keeps the current negligible, so the meter shows the true EMF — and its sign display tells you which electrode is positive. One reading, two pieces of data.
Variables, stated the way the mark scheme wants them: independent — which half-cell is paired with the reference; dependent — the EMF; control — concentrations at \(1.00\ \text{mol dm}^{-3}\), temperature at 298 K, and the same reference half-cell throughout. Error sources worth naming: solutions not exactly standard, temperature drift, residual oxide films, and the small junction potentials where the salt bridge meets each solution.
A student measures the zinc–copper cell at \(1.10\ \text{V}\) with copper positive. Given \(E^{\ominus}(\text{Cu}^{2+}/\text{Cu}) = +0.34\ \text{V}\), the unknown follows by rearrangement: \(E^{\ominus}(\text{Zn}^{2+}/\text{Zn}) = +0.34 - 1.10 = -0.76\ \text{V}\). Prediction test: pair the same zinc half-cell with silver (\(+0.80\ \text{V}\)) and the cell should read \(0.80 - (-0.76) = 1.56\ \text{V}\), silver positive. If the meter insists on 1.49 V, suspect the silver solution's concentration or a tired salt bridge before doubting the series — non-standard conditions are almost always the culprit.
VocabularyKey terms the mark scheme pays for
TrapsMisconceptions that cost marks
ExamWhat examiners want
Electrode potentials sit on Paper 1, and the mark scheme is obsessed with signs. Copy \(E^{\ominus}\) values with their signs attached, subtract in the stated order — positive minus negative, or right minus left of your cell diagram — and show the substitution line: \(+0.34 - (-0.76) = +1.10\ \text{V}\) is itself a mark. A correct magnitude with a dropped sign scores nothing.
Feasibility answers follow a fixed skeleton the examiner can tick: quote both \(E^{\ominus}\) values, identify what is oxidised (the more negative couple), calculate the EMF, conclude 'positive, therefore feasible' — then, where the question invites it, add the AO3 caveat that the reaction may be slow or the conditions non-standard. Cell-representation marks are cheap but exact: oxidation on the left, double bar for the salt bridge, state symbols, and platinum written in wherever no metal conducts.
For commercial cells, learn the lithium-ion and both alkaline fuel-cell half-equations until you can write them cold — they are gift marks — and rehearse one balanced evaluation sentence for each: energy density against a flammable electrolyte for lithium-ion; continuous running against hydrogen sourcing and storage for the fuel cell. RP8 extended responses want the standard-conditions list, saturated potassium nitrate named for the bridge, and the high-resistance voltmeter explained, not just mentioned: negligible current drawn, equilibria undisturbed, so the reading is the true EMF.