Electrode Potentials & Cells

Half cells & the standard hydrogen electrode

Dip a metal into a solution of its own ions and an equilibrium is set up between the metal and its aqueous ions — this is a half cell, and the metal strip is the electrode.

Zn²⁺(aq) + 2e⁻ ⇌ Zn(s)

The position of that equilibrium sets up a potential difference between the metal and the solution — the electrode potential. A single electrode potential cannot be measured on its own, so every value is measured relative to a reference: the standard hydrogen electrode (SHE), defined as exactly 0.00 V.

The SHE is a piece of inert platinum (coated in platinum black) dipping into 1.00 mol dm⁻³ H⁺(aq), with hydrogen gas at 100 kPa bubbled over it at 298 K. It sets up the equilibrium 2H⁺(aq) + 2e⁻ ⇌ H₂(g), whose potential is defined as exactly 0.00 V — so every other E^⦵ is a reading taken against it.

Half-equations are written by IUPAC convention as reductions, with the electrons on the left, for example Cu²⁺(aq) + 2e⁻ ⇌ Cu(s), E^⦵ = +0.34 V.

Half cells with no metal — using a platinum electrode

Not every half cell contains a metal that can act as the electrode. When both species are ions in the same solution, you dip inert platinum into a solution containing both ions at 1.00 mol dm⁻³. The platinum carries the electrons but takes no part in the reaction.

Fe³⁺(aq) + e⁻ ⇌ Fe²⁺(aq)  E^⦵ = +0.77 V

If the half-equation also needs H⁺ — as many ion/ion couples do — then the solution must be acidified to 1.00 mol dm⁻³ H⁺ as well:

VO₂⁺(aq) + 2H⁺(aq) + e⁻ ⇌ VO²⁺(aq) + H₂O(l)  E^⦵ = +1.00 V

The standard hydrogen electrode is itself this kind of half cell — a gas and an ion, with nothing metallic to conduct — which is exactly why it needs a platinum electrode too.

📌 Recap writing and balancing half-equations and oxidation states from AS redox.

Representing cells & Required Practical 8

Join two half cells with a wire and a salt bridge and you have an electrochemical cell. A standard shorthand — the cell notation — records it:

Zn(s) | Zn²⁺(aq) ‖ Cu²⁺(aq) | Cu(s)

A single line | is a phase boundary (solid–solution); the double line ‖ is the salt bridge. By convention the more negative electrode is written on the left, where oxidation occurs; reduction occurs at the right-hand (more positive) electrode.

Cells with an inert electrode

When a half cell has no metal of its own, the inert electrode — Pt or C — goes on the outside of the notation, and species sharing the same solution are separated by a comma rather than a line, because there is no phase boundary between them:

Fe(s) | Fe²⁺(aq) ‖ VO²⁺(aq), H⁺(aq), V³⁺(aq) | Pt(s)

Build it from the outside in: the electrode, the species it touches, the salt bridge, then the same again in reverse. A fuel cell is written the same way, with carbon or platinum at each end:

C(s) | C₆H₁₂O₆(aq), H⁺(aq) | CO₂(g) ‖ O₂(g) | H⁺(aq), H₂O(l) | Pt(s)

Calculating the EMF of a cell

The cell’s EMF is the difference between the two standard electrode potentials. Subtract the more negative from the more positive so the answer comes out positive:

E^⦵_cell = E^⦵(positive electrode) − E^⦵(negative electrode)

Standard electrode potentials are listed together as an electrochemical series — simply a list of electrode potentials in numerical order — running from the most negative (best reducing agents) to the most positive (best oxidising agents).

When the conditions are not standard

Every E^⦵ assumes 1.00 mol dm⁻³ at 298 K. Change a concentration and the electrode equilibrium shifts, so that electrode potential — and the cell EMF — changes with it. Apply Le Chatelier to the electrode equilibrium:

Mg²⁺(aq) + 2e⁻ ⇌ Mg(s)  E^⦵ = −2.38 V

Dilute the Mg²⁺ solution and the equilibrium shifts to the left, so the magnesium releases electrons even more readily and its electrode potential becomes more negative. Measured against the standard hydrogen electrode, the EMF of the cell therefore increases.

The same reasoning explains what happens as a cell runs: the species being oxidised is used up and its ion concentration rises, while the ion being reduced is consumed, so the EMF gradually falls to zero as the cell goes flat.

Predicting the feasibility of reactions

Standard electrode potentials rank how readily each species is reduced. The more positive the E^⦵, the more readily that species is reduced (so it is the better oxidising agent); the more negative the E^⦵, the more readily it is oxidised (the better reducing agent).

Questions also ask for the weakest agent, which is the same ranking read the other way. The weakest oxidising agent is the left-hand species of the most negative couple; the weakest reducing agent is the right-hand species of the most positive couple. Watch which side of the half-equation you are quoting — oxidising agents come from the left, reducing agents from the right.

To combine two half cells: the more positive one runs as reduction (forward), the more negative one reverses to oxidation. The reaction is feasible when E^⦵_cell is positive.

Commercial & fuel cells

Electrochemical cells power everyday devices. They fall into three types: non-rechargeable (the reaction cannot be reversed), rechargeable (the reaction is reversed by an external current), and fuel cells.

A non-rechargeable (primary) cell

A non-rechargeable cell runs one way only: once a reactant is used up the cell is spent, so it is discarded or recycled rather than recharged. The everyday zinc–manganese(IV) oxide battery is a good example. Given the two electrode reactions, you can deduce how it works:

2MnO₂(s) + 2NH₄⁺(aq) + 2e⁻ ⇌ Mn₂O₃(s) + 2NH₃(aq) + H₂O(l)  E^⦵ = +0.52 V

Zn²⁺(aq) + 2e⁻ ⇌ Zn(s)  E^⦵ = −0.76 V

MnO₂ is in the more positive couple, so it is reduced — MnO₂ is the oxidising agent — while zinc is oxidised at the negative electrode:

negative electrode:  Zn → Zn²⁺ + 2e⁻

positive electrode:  2MnO₂ + 2NH₄⁺ + 2e⁻ → Mn₂O₃ + 2NH₃ + H₂O

overall:  Zn + 2MnO₂ + 2NH₄⁺ → Zn²⁺ + Mn₂O₃ + 2NH₃ + H₂O   E^⦵_cell = (+0.52) − (−0.76) = +1.28 V

Nothing drives those changes backwards, so once the MnO₂ is used up the cell is finished.

The lithium (rechargeable) cell

The simplified electrode reactions are:

negative electrode:  Li → Li⁺ + e⁻

positive electrode:  Li⁺ + CoO₂ + e⁻ → Li⁺[CoO₂]⁻

Lithium cells must use a non-aqueous electrolyte. Lithium sits far below water in the series (Li⁺/Li is −3.04 V against −0.83 V for 2H₂O + 2e⁻ ⇌ H₂ + 2OH⁻), so in an aqueous electrolyte the lithium would simply reduce the water to hydrogen instead of driving the cell.

On recharging, an external current drives both reactions in reverse. That is the general rule: recharging runs the discharge reaction backwards, so if you are given the discharge equation you simply reverse it. A rechargeable silver–zinc cell discharges by Zn + Ag₂O → ZnO + 2Ag, so while it is charging:

recharging:  2Ag + ZnO → Zn + Ag₂O

A cell can only be recharged if its electrode reactions are reversible — that is why a lithium cell recharges but a zinc–MnO₂ cell does not.

These portable cells bring clear benefits — a compact, on-demand supply of electricity for phones, laptops and medical devices — but also risks to weigh. Non-rechargeable cells are cheap and reliable, yet are thrown away after a single use and add to waste; rechargeable lithium cells are reused hundreds of times but contain reactive, sometimes toxic materials, so they must be recycled or disposed of carefully rather than sent to landfill.

The alkaline hydrogen–oxygen fuel cell

A fuel cell is supplied continuously with fuel (hydrogen) and oxygen, so it produces a current without needing to be recharged.

negative electrode:  H₂ + 2OH⁻ → 2H₂O + 2e⁻

positive electrode:  ½O₂ + H₂O + 2e⁻ → 2OH⁻

overall:  H₂ + ½O₂ → H₂O

The only product is water, so fuel cells are efficient and produce no pollutants at the point of use. But hydrogen is difficult to store (a flammable gas, often stored under high pressure), and it is usually manufactured from hydrocarbons or by electrolysis — processes that themselves consume energy and may release CO₂.

The acidic and alkaline hydrogen–oxygen cells have the same EMF (+1.23 V), because the overall reaction is the same in both — 2H₂ + O₂ → 2H₂O. Only the mobile ion, and the side the water leaves, are different.

Deducing the half-equations for an unfamiliar fuel cell

Fuel cells in exams are often ones you have never met — methanol, glucose. You are told in words what reacts at each electrode, and you build the half-equation by balancing in this order: the main atoms, then oxygen using H₂O, then hydrogen using H⁺, then the charge using electrons.

In a methanol–oxygen fuel cell, methanol reacts with water at the negative electrode to give carbon dioxide and hydrogen ions, while oxygen reacts with hydrogen ions at the positive electrode:

negative electrode:  CH₃OH + H₂O → CO₂ + 6H⁺ + 6e⁻

positive electrode:  O₂ + 4H⁺ + 4e⁻ → 2H₂O

overall:  CH₃OH + 1½O₂ → CO₂ + 2H₂O

Given the cell EMF and one electrode potential, the other follows by rearranging E^⦵_cell = E^⦵(positive) − E^⦵(negative). Here the EMF is +1.20 V and the CO₂/CH₃OH electrode is +0.03 V, so E^⦵(O₂/H₂O) = 1.20 + 0.03 = +1.23 V.

Methanol has one practical advantage over hydrogen in a car: it is a liquid, so it is much easier to store and transport, and a given volume releases more energy than the same volume of hydrogen gas.

Run the cell below with each electrolyte. The alkaline cell is the one AQA examines at A-level; the acidic version is the GCSE cell — switching the electrolyte switches the mobile ion and the side the water leaves.

The cell-reasoning checklist

Electrochemistry questions reward a few consistent habits.

🔗 Electrode potentials return in the redox chemistry of transition metals (3.2.5) later in the A2 course, where E^⦵ values predict which oxidation-state changes are feasible.