Electrochemistry

Electrolysis: predicting what appears at each electrode

Electrolysis is the decomposition of an electrolyte, either molten or in aqueous solution, by passing a direct current through it. The ions carry the charge through the electrolyte and electrons carry it through the wires, so the two halves of the redox reaction happen on separate surfaces: reduction at the cathode and oxidation at the anode. In an electrolytic cell the cathode is the negative electrode, because the power supply pushes electrons onto it, and the anode is the positive electrode. That is the reverse of a simple cell, where the electrode at which reduction happens is the positive terminal because it is driven by a spontaneous reaction instead of by a power pack.

With a molten electrolyte there is nothing to compete with the ions of the compound. Molten lead(II) bromide contains only Pb2+ and Br- ions, so the cathode reaction is Pb2+ + 2e- -> Pb and silvery lead forms on the cathode, while the anode reaction is 2Br- -> Br2 + 2e- and brown bromine vapour leaves the anode. Molten sodium chloride, used in the Downs cell with calcium chloride added to lower the melting point to about 600 degrees Celsius, gives sodium at the cathode and chlorine at the anode. Solubility and water play no part at all.

In aqueous solution water itself can be oxidised or reduced, so two reductions compete at the cathode and two oxidations compete at the anode. The winner is decided by the redox series: at the cathode the half-equation with the more positive (or less negative) standard electrode potential is the one that runs, and at the anode the half-equation whose reduction potential is the least positive is the one that is reversed, because that needs the least energy pushed into it.

A worked comparison makes the rule concrete. In 1.00 mol dm-3 copper(II) sulphate with platinum electrodes, copper is deposited at the cathode because E(Cu2+/Cu) is +0.34 V while E(2H2O/H2 + 2OH-) is -0.83 V, and oxygen is released at the anode from 2H2O -> O2 + 4H+ + 4e-, so the solution around the anode becomes more acidic as the electrolysis continues and the blue colour fades as copper(II) ions are used up. Sodium ions are never reduced from aqueous solution, because E(Na+/Na) is -2.71 V, far below that of water: electrolysis of aqueous sodium chloride at the cathode gives hydrogen, not sodium.

Concentration decides the close cases, and this is the part examiners test most often. The two anode reactions in aqueous sodium chloride are only 0.13 V apart: E(Cl2/Cl-) is +1.36 V and E(O2/H2O) is +1.23 V, both quoted for standard conditions of 1.00 mol dm-3. In dilute solution the chloride concentration is far below 1.00 mol dm-3, so the oxygen couple is the one that is reached and oxygen is given off; in concentrated brine the chloride concentration is high, the chlorine couple becomes easier to reverse, and chlorine is the product, which is exactly what the chlor-alkali industry depends on. The same reasoning explains why a change of electrode material or a change of temperature can switch the product, and why overpotential means a reaction predicted by potential alone may need a little extra voltage to start.

Charge, the Faraday constant and how much substance is liberated

The Faraday constant, F, is the charge carried by one mole of electrons, 96500 C mol-1, and it is the bridge between the electricity you pass and the amount of substance you make. It is not an independent constant: it equals the Avogadro constant multiplied by the charge on a single electron, F = Le, so 6.02 x 10^23 mol-1 x 1.60 x 10^-19 C = 9.63 x 10^4 C mol-1. The small difference from the tabulated 9.65 x 10^4 C mol-1 comes only from rounding L and e, and quoting the relationship as F = Le is worth a mark on its own.

The quantity of charge is found from Q = It, where Q is in coulombs, I is the steady current in amperes and t is the time in seconds. Always convert: a current of 2.00 A passed for 5.00 minutes is 2.00 x 300 = 600 C, not 2.00 x 5. Never leave the time in minutes, and remember that coulombs, not faradays, are what the equation gives you.

Dividing the charge by the Faraday constant gives the moles of electrons: moles of electrons = Q / F = It / F. The half-equation then tells you how many electrons one unit of the substance needs, so moles of substance = It / (nF), where n is the number of electrons in that half-equation. The mass is that number of moles multiplied by the molar mass, M, and the volume of a gas is that number of moles multiplied by the molar gas volume, 24.0 dm3 mol-1 at room temperature and pressure.

A full calculation: a steady 2.00 A is passed through aqueous copper(II) sulphate with platinum electrodes for 30.0 minutes. The charge is Q = 2.00 x 1800 = 3600 C, so the moles of electrons are 3600 / 96500 = 0.0373 mol. The half-equation Cu2+ + 2e- -> Cu needs two electrons for each copper atom, so the moles of copper are 0.0373 / 2 = 0.0187 mol and, with M(Cu) = 63.5 g mol-1, the mass deposited is 0.0187 x 63.5 = 1.19 g. The same charge at a platinum anode gives 3600 / (4 x 96500) = 0.00933 mol of oxygen, because 2H2O -> O2 + 4H+ + 4e- needs four electrons for each molecule, and 0.00933 x 24.0 = 0.224 dm3 of oxygen at r.t.p.

Two more consequences are worth knowing because they are so often examined. In a series circuit the same charge passes through every cell, so the amounts liberated are in the ratio of their n values, which is how an electrolysis cell can be used as a coulometer to measure charge. In the electrolysis of dilute sulphuric acid the volume of hydrogen at the cathode is exactly twice the volume of oxygen at the anode, because hydrogen needs 2 electrons per molecule and oxygen needs 4.

Determining the Avogadro constant by an electrolytic method

Because F = Le, a measured value of the Faraday constant gives the Avogadro constant as soon as the charge on the electron is known. Electrolysis is an excellent way to measure F, because it counts atoms one electron at a time: the charge passed and the mass of substance deposited are linked by a known, whole number of electrons per atom.

The classic apparatus is a silver coulometer. A clean silver cathode (and a silver anode) is weighed on a four-figure balance, then suspended in silver nitrate solution and connected to a steady direct current. The current is read from an ammeter in series and the time is taken with a stopwatch, so Q = It is known. When the current is switched off the cathode is rinsed with distilled water, dried gently without rubbing, and weighed again, and the gain in mass is the silver deposited. Because the half-equation is Ag+ + e- -> Ag, one electron is transferred for every silver atom, so the moles of electrons equal the moles of silver.

Take typical results: a steady current of 1.00 A is passed for 1000 s, so Q = 1000 C, and the cathode gains 1.118 g. The moles of silver are 1.118 / 107.9 = 0.01036 mol, and since moles of electrons = moles of silver, F = Q / moles of electrons = 1000 / 0.01036 = 9.65 x 10^4 C mol-1. Dividing by the charge on the electron gives L = F / e = 9.65 x 10^4 / 1.60 x 10^-19 = 6.03 x 10^23 mol-1, which agrees with the accepted value of 6.02 x 10^23 mol-1 to the precision the experiment can reach.

A copper coulometer works the same way with copper electrodes in copper(II) sulphate solution: the anode loses mass and the cathode gains it. The half-equation Cu2+ + 2e- -> Cu needs two electrons for each atom, so the moles of electrons are twice the moles of copper and F = Q x M / (2 x m). The method is less accurate than the silver coulometer because copper can be oxidised further or flake off the anode. Whichever metal is used, the precision of the result is limited by the balance, the ammeter and the clock, and by whether the current really stayed constant: a current that drifts invalidates Q = It. Gently drying the electrode matters too, because a silver deposit is soft and is easily rubbed off the cathode with the water.

Standard electrode potentials and the standard hydrogen electrode

A single electrode has no measurable potential on its own: only a difference between two electrodes can be measured. Chemists therefore fix one electrode as the reference and quote everything else against it. A standard electrode (reduction) potential is the potential difference of a cell in which the half-cell is connected to the standard hydrogen electrode under standard conditions, with the half-cell written as a reduction. Standard conditions are 298 K, a pressure of 100 kPa for any gas, and a concentration of 1.00 mol dm-3 for every aqueous species, and the standard hydrogen electrode is defined as exactly 0.00 V. The value therefore says how readily the species is reduced relative to H+.

The standard hydrogen electrode (SHE) is a platinum electrode coated with platinum black, dipped into 1.00 mol dm-3 H+(aq), with hydrogen gas at 100 kPa bubbled over the platinum surface at 298 K. The platinum black gives a very large surface area and catalyses the reversible equilibrium 2H+ + 2e- <-> H2, so the electrode responds to both species present. Platinum is used because it is inert and conducts electrons; hydrogen gas on its own is a poor conductor, so the inert metal is essential rather than decorative. The electrode is assigned E = 0.00 V by definition, which is why every other electrode potential in the table carries a sign.

Because every potential is quoted for a reduction, a negative value such as E(Zn2+/Zn) = -0.76 V does not mean zinc cannot be oxidised. It means that the equilibrium Zn2+ + 2e- <-> Zn lies further to the left than the hydrogen equilibrium, so zinc has a greater tendency to release electrons than hydrogen does. The standard cell potential of a cell built from two half-cells is found by subtracting: E(cell standard) = E(standard of the electrode that is reduced) - E(standard of the electrode that is oxidised), with both written as reductions. Combining E(Zn2+/Zn) = -0.76 V and E(Cu2+/Cu) = +0.34 V gives E(cell standard) = +0.34 - (-0.76) = +1.10 V. The two values are never added.

Standard electrode potentials are always quoted for the same reaction direction, which is what makes the table usable, and they are all measured with the SHE on the reference side. Practically, a secondary reference such as the silver/silver chloride electrode is often used in the laboratory because the SHE is awkward to set up, and its potential is then converted into the hydrogen scale.

Using electrode potentials: polarity, electron flow, oxidising power and feasibility

In a simple cell the electrode with the more positive standard electrode potential is the positive terminal, and reduction happens there. Electrons are released at the more negative electrode and travel through the external circuit to the more positive one, so in a zinc-copper cell with a salt bridge the electrons flow from the zinc to the copper through the wire, while conventional current is in the opposite direction. Zinc is oxidised, Zn -> Zn2+ + 2e-, and the zinc electrode loses mass; Cu2+ + 2e- -> Cu runs at the copper, which gains mass and is the positive terminal. Inside the salt bridge, ions drift to keep the charge balanced: negative ions move towards the zinc compartment, where positive zinc ions are being produced, and positive ions move towards the copper compartment, where copper(II) ions are being removed.

Electrode potentials measure oxidising and reducing strength as well. The more positive the reduction potential, the stronger the oxidising agent: fluorine, E(F2/F-) = +2.87 V, and acidified manganate(VII), E(MnO4-/Mn2+) = +1.51 V, are powerful oxidising agents, while sodium ions and aluminium ions, with potentials near the bottom of the table, are extremely weak oxidising agents. Read the same table from the other side and the species on the right of a half-equation with a very negative potential is a strong reducing agent: lithium, potassium, calcium, sodium, magnesium, aluminium and zinc will all displace less reactive metals from their compounds, because each of them can push electrons onto the ions of a metal that sits above it in the series.

This gives a simple test of feasibility under standard conditions: the reaction is feasible when E(cell standard) = E(standard of the couple being reduced) - E(standard of the couple being oxidised) is positive. Zinc will displace copper, because +0.34 - (-0.76) = +1.10 V, but copper will not displace zinc, because the same subtraction the other way round gives -1.10 V. Chlorine oxidises iodide to iodine, +1.36 - (+0.54) = +0.82 V, and bromine oxidises iron(II) to iron(III), +1.09 - (+0.77) = +0.32 V.

Redox equations are built from half-equations. Write both half-equations as reductions, reverse the one that must be oxidised, multiply each so that the electrons lost equal the electrons gained, then add and cancel the electrons. Iron(II) with acidified dichromate(VI) is the standard example: the chromium half-equation Cr2O7^2- + 14H+ + 6e- -> 2Cr3+ + 7H2O supplies six electrons, so six Fe2+ -> Fe3+ + e- steps are reversed and combined to give Cr2O7^2- + 14H+ + 6Fe2+ -> 2Cr3+ + 7H2O + 6Fe3+, which balances in atoms and in charge. A positive cell potential tells you only that the change is energetically possible: it says nothing about how fast the reaction goes, and a large activation energy, a non-standard set of concentrations or an overpotential at the electrode surface can all stop a feasible reaction from being seen in the test tube.

Measuring standard electrode potentials

Every measurement of a standard electrode potential uses the same shape of apparatus: the half-cell of interest, the standard hydrogen electrode, a salt bridge joining the two solutions, and a high-resistance voltmeter or potentiometer in the external circuit. The salt bridge is usually a tube or a strip of filter paper soaked in saturated potassium nitrate solution; it completes the circuit and lets ions move so that charge does not build up, but it prevents the two solutions mixing. A high-resistance voltmeter is essential because it draws almost no current: if current flowed, electrolysis would change the concentrations at the electrode surfaces, the reading would drift downwards and the cell would no longer be standard.

(a) A metal in contact with its ions. A clean rod of the metal is dipped into a 1.00 mol dm-3 solution of its own ions, for example zinc in 1.00 mol dm-3 zinc sulphate, and this half-cell is joined to the SHE. The voltmeter then reads 0.76 V with the SHE as the positive terminal, so the zinc electrode is the negative one and E(Zn2+/Zn) = -0.76 V. A copper electrode in 1.00 mol dm-3 copper(II) sulphate gives a reading of 0.34 V with the copper as the positive terminal, so E(Cu2+/Cu) = +0.34 V. The metal must be abraded with emery paper and rinsed first, because an oxide layer changes the surface and shifts the reading.

(b) A non-metal in contact with its ions. No metal rod is available, so an inert platinum electrode provides the conducting surface. For the chlorine/chloride couple, chlorine gas at 100 kPa is bubbled over platinum dipping into 1.00 mol dm-3 potassium chloride solution, and the reversible equilibrium Cl2 + 2e- <-> 2Cl- is set up on the platinum surface; the cell is completed with the SHE in the usual way. The same arrangement is used for the bromine/bromide and iodine/iodide couples, and the hydrogen electrode in the SHE is itself an example of this type.

(c) Ions of the same element in two different oxidation states. Both ions are dissolved in the same beaker, each at 1.00 mol dm-3, and an inert platinum electrode is dipped into the mixture, so no metal of the element is involved at all. Iron(II) and iron(III) sulfates give the Fe3+/Fe2+ couple, with E(standard) = +0.77 V for Fe3+ + e- <-> Fe2+. Where the half-equation also contains hydrogen ions, the acid concentration is part of the standard state: for manganate(VII) the solution must be 1.00 mol dm-3 in H+ as well, which is why E(MnO4-/Mn2+) = +1.51 V is quoted for acidified solution and falls in neutral or alkaline conditions. In each case the reading is taken at 298 K, and the sign of the voltmeter reading gives the sign of the standard electrode potential of the half-cell as a reduction.

Concentration, the Nernst equation and the link to Gibbs free energy

An electrode potential depends on concentration, and the trend is easy to reason out without any algebra. Making the oxidised form more concentrated makes reduction easier, so E becomes more positive; making the reduced form more concentrated makes reduction harder, so E becomes less positive. Diluting a copper(II) solution from 1.00 to 0.0100 mol dm-3 lowers E(Cu2+/Cu) from +0.34 V to about +0.28 V, and adding iron(II) sulphate to a mixture of iron(II) and iron(III) lowers the ratio [Fe3+]/[Fe2+] and so lowers E(Fe3+/Fe2+) below +0.77 V. This is why the voltage of a cell falls as it is used: at the positive electrode the oxidised species is being consumed, so its concentration drops, E of that electrode falls, E(cell) shrinks, and the cell is flat when the two electrode potentials have become equal.

The quantitative version is the Nernst equation, E = E(standard) + (0.059 / z) log([oxidised species] / [reduced species]), where z is the number of electrons in the half-equation and the 0.059 V comes from 2.303RT/F at 298 K. Only species whose concentration actually varies appear: a pure solid such as copper metal or silver, and water in dilute aqueous solution, are left out of the ratio. So for Cu2+ + 2e- <-> Cu the expression is E = E(standard) + (0.059 / 2) log[Cu2+], and for Fe3+ + e- <-> Fe2+ it is E = E(standard) + 0.059 log([Fe3+] / [Fe2+]).

Two worked values. For copper at 0.0100 mol dm-3: E = 0.34 + (0.059 / 2) x log(0.0100) = 0.34 + 0.0295 x (-2) = 0.34 - 0.059 = +0.28 V. For a mixture that is 0.100 mol dm-3 in Fe3+ and 1.00 mol dm-3 in Fe2+: E = 0.77 + 0.059 x log(0.100) = 0.77 - 0.059 = +0.71 V. The same equation explains the pH dependence of the hydrogen electrode, since E = 0.00 + 0.059 log[H+], so at pH 7 the hydrogen electrode potential is about -0.41 V.

Finally, the cell potential is the electrical measure of the driving force of the reaction, and it converts directly into Gibbs free energy: delta G(standard) = -nE(cell standard)F, where n is the number of electrons transferred in the equation as written and F is 96500 C mol-1. For Zn(s) + Cu2+(aq) -> Zn2+(aq) + Cu(s), n = 2 and E(cell standard) = +1.10 V, so delta G(standard) = -2 x 1.10 x 96500 = -212000 J mol-1 = -212 kJ mol-1, and the negative sign is exactly the statement that a cell with a positive E(cell) does electrical work and can drive the reaction forward. The link also runs to equilibrium, because delta G(standard) = -RT ln K: the larger the positive cell potential, the further the equilibrium lies to the right, and E(cell) = 0 means delta G = 0 and K = 1. Notice that n doubles if you double the equation, so delta G doubles while E(cell) itself stays unchanged.