The electrochemical series and predicting feasibility

The series as a ranking

Gold lies in a river bed unchanged for thousands of years, while a freshly cut surface of sodium tarnishes in seconds. Both facts are in the same table. List standard electrode potentials in order, each half-equation written as a reduction, and the list is the electrochemical series.

The table reads in two directions. On the left of each half-equation is an oxidised form. The further down the table it sits, the more positive its E° and the more readily it gains electrons, so fluorine and manganate(VII) are powerful oxidising agents and Li⁺ and K⁺ oxidise almost nothing. On the right is a reduced form. The further up it sits, the more negative its E° and the more readily it gives electrons away, so lithium and potassium are powerful reducing agents and F⁻ and Cl⁻ are very weak ones.

The two rankings run in opposite directions because they describe the same equilibria. A metal that gives up its electrons easily forms an ion with little pull on them, and an ion with a strong pull on electrons comes from an element that holds on to them. Sodium, near the top, is found in nature only as Na⁺ in compounds; gold, near the bottom, is found as the metal.

Describe the drawing

The electrochemical series as a table against a vertical E° scale, most negative at the top. Thirteen half-equations run from Li⁺/Li at −3.04 V through Mg, Zn, Fe, hydrogen at 0.00, Cu at +0.34, iodine, Fe³⁺/Fe²⁺, Ag, bromine, chlorine and manganate(VII) at +1.51, to F₂/F⁻ at +2.87 V, each joined to its place on a scale drawn to proportion. Oxidised forms are on the left, reduced forms on the right. A broad arrow down the left reads oxidising agents get stronger, and one up the right reads reducing agents get stronger. A curved arrow from Cu²⁺ on the left runs up to Zn on the right, marking a reaction that goes because E°cell is positive.

Name the species exactly when ranking. Chlorine, Cl₂, is the oxidising agent; chloride, Cl⁻, is the reducing agent of the same couple. 'Chlorine is a weak reducing agent' mixes the two forms up.

Electrochemical series

A list of half-equations written as reductions, in order of their standard electrode potentials.

Oxidising agent

A species that accepts electrons from another species and is itself reduced.

Reducing agent

A species that donates electrons to another species and is itself oxidised.

Predicting the direction of a reaction

Of any two half-equations from the series, the one with the more positive E° goes forwards, as written, and gains electrons. The one with the less positive E° goes backwards and gives them up. The arithmetic gives the same answer: put the half-cell being reduced on the right, and E°cell = E°right − E°left comes out positive.

The same test works when a question names the reaction. Take the half-cell in which reduction happens as the right-hand electrode and calculate E°cell. If it is positive, the reaction is feasible under standard conditions. If it is negative, the reaction as written is not feasible, and the reverse reaction is. The scheme credits 'feasible'; 'the reaction happens' claims something E° cannot show.

Describe the drawing

Two half-equations written one above the other: Zn²⁺ plus two electrons in equilibrium with Zn, E° −0.76 V, on top, and Cu²⁺ plus two electrons in equilibrium with Cu, E° +0.34 V, below. A loop of arrows runs anticlockwise: right along the lower equation from Cu²⁺ to Cu, labelled forwards: reduction, up to Zn, left along the upper equation to Zn²⁺, labelled backwards: oxidation, and down again. Cu²⁺ is marked as the oxidising agent and Zn as the reducing agent. Beneath, Zn plus Cu²⁺ gives Zn²⁺ plus Cu with E°cell +1.10 V, and a note that with the more positive equation on top the loop runs clockwise.

The anticlockwise rule is a way of seeing the same thing in a pair of half-equations. Write the two with the more negative one on top. Start at the bottom left and trace an anticlockwise loop: along the lower equation to the right, up, and along the upper equation to the left. The species met in order are the oxidising agent, the product of its reduction, the reducing agent and the product of its oxidation. On a table printed with the most positive value at the top, the loop runs clockwise instead, so check the order before using it.

The overall equation comes from the two half-equations as any redox equation does: reverse the one that goes backwards, multiply until the electrons match, and add. The E° values are never multiplied, because a potential does not depend on how many electrons are written.

The test applies when one species appears in both half-equations. Copper(I) ions are reduced in Cu⁺(aq) + e⁻ ⇌ Cu(s), E° = +0.52 V, and oxidised in the reverse of Cu²⁺(aq) + e⁻ ⇌ Cu⁺(aq), E° = +0.15 V. E°cell = 0.52 − 0.15 = +0.37 V, so copper(I) ions in water are not stable: white copper(I) sulfate dissolves to give a blue solution and a precipitate of copper. One species is both oxidised and reduced, which is disproportionation.

2Cu⁺(aq) → Cu(s) + Cu²⁺(aq)

Feasible

Able to take place; a redox reaction is feasible under standard conditions when E°cell, calculated with the half-cell in which reduction occurs on the right, is positive.

Disproportionation

A redox reaction in which the same element in one species is simultaneously oxidised and reduced.

When a feasible reaction is not seen

A positive E°cell means a reaction is energetically possible. It says nothing about how fast the reaction goes, because E° comes from the energies of reactants and products and not from the activation energy between them. A mixture of hydrogen and oxygen has E°cell = +1.23 V for forming water, and it sits unchanged in a balloon until a spark arrives.

Aluminium is the everyday case. E°(Al³⁺/Al) = −1.66 V, more negative than zinc, so the reduction of H⁺ in dilute acid by aluminium is even more feasible than by zinc, and an aluminium pan might be expected to dissolve in vinegar. It does not, because a layer of aluminium oxide a few nanometres thick covers the metal and keeps the acid off it. Once sodium hydroxide solution has dissolved the layer, the metal underneath reacts vigorously, giving off hydrogen.

Manganate(VII) and ethanedioate ions are a third example. E°cell for the reaction is large and positive, yet at room temperature the purple colour fades only slowly, and the titration is done warm for that reason. Mixtures like these are kinetically stable: the reaction is feasible but its rate is too low to observe.

The second limit is that E° values apply under standard conditions. Concentrations of exactly 1.00 mol dm⁻³ are rare outside a data book, and a reaction run with concentrated reagents, or hot, can go the other way from the prediction when E°cell is small, as dichromate and hot concentrated hydrochloric acid show.

Describe the drawing

Four test tubes in a row, each labelled above with its mixture and E°cell and below with the prediction and what is seen. Chlorine water with iron(II), E°cell +0.59 V, feasible, pale green turns yellow-brown, as predicted. Iodine with iron(II), E°cell −0.23 V, not feasible, no change, as predicted. Aluminium foil in dilute hydrochloric acid, +1.66 V, feasible, but almost nothing happens at first, with a magnified inset showing the oxide layer on the foil. Acidified manganate(VII) with ethanedioate, about +2.0 V, feasible, but the purple fades only slowly. The last two are marked feasible but slow.

Kinetically stable

Describes a mixture that does not react at a measurable rate although the reaction is feasible, because the activation energy is high.

Concentration and the electrode potential

Each electrode potential belongs to an equilibrium, and an equilibrium responds to a change in concentration. In the copper half-cell, Cu²⁺(aq) + 2e⁻ ⇌ Cu(s), raising [Cu²⁺] above 1.00 mol dm⁻³ shifts the position to the right. The electrode takes electrons more readily and its potential becomes more positive than +0.34 V. Lowering [Cu²⁺] shifts the equilibrium to the left and the potential becomes less positive.

Describe the drawing

A graph of the electrode potential of the copper half-cell, from +0.24 to +0.36 V, against copper(II) concentration on a logarithmic axis from 0.001 to above 1 mol dm⁻³. The points lie on a straight line rising from left to right, about 0.03 V for every tenfold rise in concentration. A dashed line at +0.34 V meets the line at 1.00 mol dm⁻³, marked as the standard value. A note says that at higher [Cu²⁺] the equilibrium Cu²⁺ + 2e⁻ ⇌ Cu shifts right and E becomes more positive.

The reasoning extends to a whole cell. In the zinc–copper cell, anything that makes the right-hand electrode more positive, or the left-hand one more negative, raises the EMF. A more concentrated copper(II) sulfate solution raises it. A more concentrated zinc sulfate solution lowers it, because a higher [Zn²⁺] shifts Zn²⁺(aq) + 2e⁻ ⇌ Zn(s) to the right and makes the zinc electrode less negative.

The same shifts make a cell go flat. As the zinc–copper cell delivers a current, [Cu²⁺] falls and [Zn²⁺] rises. Both changes lower the EMF, and the current stops when the two electrode potentials are equal, which is when the cell reaction has reached equilibrium.

For a half-equation that includes H⁺, the hydrogen ion concentration counts too. Raising [H⁺] shifts Cr₂O₇²⁻(aq) + 14H⁺(aq) + 6e⁻ ⇌ 2Cr³⁺(aq) + 7H₂O(l) to the right and makes dichromate(VI) a stronger oxidising agent, which is part of the reason hot concentrated hydrochloric acid gives chlorine with it. Temperature changes E as well, and standard values are quoted at 298 K.

An answer on concentration earns its marks in three steps: name the equilibrium, say which way it shifts, and say whether the electrode potential becomes more or less positive. Only the direction of the change is needed here, reasoned from Le Chatelier's principle.

Other boards: the Nernst equation and free energy

Cambridge asks how far an electrode potential moves as well as which way, and gives the answer by the Nernst equation, which at 298 K reads:

E = E° + (0.059 / z) log₁₀([oxidised form] / [reduced form])z is the number of electrons in the half-equation. Solids and water are left out of the concentration term.

For a metal in a solution of its ions the reduced form is the solid, so only the ion appears. The copper half-cell gives E = 0.34 + 0.0295 log₁₀[Cu²⁺]: at 1.00 mol dm⁻³ the logarithm is zero and E equals E°, and each tenfold change in [Cu²⁺] moves E by 0.0295 V, the slope of the copper graph. The sign agrees with Le Chatelier's principle: more of the oxidised form, more positive E.

An EMF is also a measure of energy. One mole of electrons carries a charge of 96 500 C, the Faraday constant, F. When the equation as written moves n moles of electrons through a potential difference E°cell, the electrical work available is nFE°cell, and Cambridge links it to the standard free-energy change:

ΔG° = −nFE°cellWith F in C mol⁻¹ and E°cell in V, ΔG° comes out in J mol⁻¹.

The minus sign makes it the feasibility test in another form: a positive E°cell gives a negative ΔG°. The two differ in one way. ΔG° belongs to the equation as written, so doubling the equation doubles n and doubles ΔG°. E°cell is energy per unit of charge, the same however many electrons are written, and E° values are never multiplied for that reason.

Edexcel makes the same link through entropy and the equilibrium constant. Since ΔStotal = −ΔG/T and ΔG° = −RT ln K, both ΔStotal and ln K are proportional to E°cell. A large positive E°cell means a large positive total entropy change and a K so large that the reaction goes to completion. An E°cell close to zero means K close to 1, with reactants and products both present at equilibrium, so a change in concentration can push the reaction either way, as dichromate and chloride ions show.

Nernst equation

E = E° + (0.059/z) log₁₀([oxidised form] / [reduced form]) at 298 K, giving the electrode potential of a half-cell at concentrations that are not standard.

Faraday constant

The charge on one mole of electrons, 96 500 C mol⁻¹.