Electrochemical cells

Put the two halves of a redox reaction in separate beakers and the electrons have to travel through a wire to get from one to the other, where they can be made to do work.

The previous lesson balanced redox equations and ranked the agents by displacement experiments, and ended with three questions it could not answer: why copper(I) disproportionates, why permanganate is weaker in neutral solution, and how far any of these reactions go. All three need a number attached to each half reaction. This lesson gets that number from a voltmeter.

Separating the halves

Drop zinc into copper sulfate and the reaction $Zn + C \to Z + Cu$ happens on the metal surface, releasing $219$ kJ mol⁻¹ as heat and nothing else. The electrons go straight from zinc atom to copper ion across a distance of a few tenths of a nanometre.

John Daniell's arrangement of 1836 separates them. Zinc sits in zinc sulfate, copper in copper sulfate, and the only electrical path between the two solutions is a salt bridge, a tube of electrolyte that carries ions but not electrons. The electrons must now go the long way, through an external wire, and on that journey they can turn a motor.

The vocabulary is fixed by function, not by sign. Oxidation happens at the anode, reduction at the cathode, always. In this cell the zinc is the anode, dissolving as $Z$ and leaving electrons behind, so it is the negative terminal. The salt bridge exists because without it the zinc solution would build up positive charge and the copper solution negative, and after a few nanocoulombs the resulting electric field would stop the reaction dead. Sulfate migrating one way and potassium the other keeps both solutions neutral.

The conventional shorthand is $Zn(s) | Z(aq) ‖ C(aq) | Cu(s)$ , with single bars for phase boundaries, the double bar for the salt bridge, and the anode written on the left.

A scale needs a zero

A voltmeter measures a difference, so no single electrode has a potential that can be measured on its own. The convention assigns zero to one of them: the standard hydrogen electrode, hydrogen gas at $10^{5}$ Pa bubbling over platinised platinum in a solution of unit hydrogen ion activity, is defined as $0.000$ V at all temperatures.

Every other half reaction is then measured against it and tabulated as a standard reduction potential $E^{ominus}$ , always written as a reduction. Zinc measures $- 0.76$ V, copper $+ 0.34$ V, silver $+ 0.80$ V, and the permanganate couple in acid $+ 1.51$ V. A positive value means the couple takes electrons more readily than hydrogen ions do, so a strong oxidising agent sits at the top of the table and a strong reducing agent at the bottom.

This is the activity series of the previous lesson made numerical, and the extra content is that the numbers can be subtracted. For any pair,

E_{cell}^{ominus} = E_{cathode}^{ominus} - E_{anode}^{ominus}

with both taken from the table as reductions. A positive cell potential means the reaction as written is spontaneous under standard conditions.

One point causes endless confusion and is worth stating flatly: potentials are not multiplied when a half reaction is scaled. Doubling $A + e^{-} \to Ag$ leaves $E^{ominus}$ at $0.80$ V. Potential is energy per unit charge, an intensive quantity, and doubling the reaction doubles both the energy and the charge.

Example. Find the standard cell potential of the Daniell cell, and of a cell combining copper with silver, given $E^{ominus} = - 0.76$ V for $Z / Zn$ , $+ 0.34$ for $C / Cu$ and $+ 0.80$ for $A / Ag$ .

For the Daniell cell, copper is reduced and zinc oxidised, so $E_{cell}^{ominus} = 0.34 - (- 0.76) = 1.10$ V. For the copper and silver cell, silver is the stronger oxidising agent, so silver is reduced and copper oxidised: $E_{cell}^{ominus} = 0.80 - 0.34 = 0.46$ V. The overall reaction is $Cu + 2 A \to C + 2 Ag$ , and the silver potential is not doubled to match the two electrons.

Now you. Using $E^{ominus} = - 0.44$ V for $F / Fe$ and $+ 0.15$ V for $S / S$ , find the cell potential and say which way the reaction runs.

Answer

The tin couple has the higher potential, so it is reduced and iron oxidised: $E_{cell}^{ominus} = 0.15 - (- 0.44) = 0.59$ V, positive, so $Fe + S \to F + S$ is spontaneous under standard conditions.

Potential is free energy per coulomb

The third lesson noted that $Δ G$ is the maximum non-expansion work a process can deliver. An electrochemical cell is a machine for collecting exactly that work, so the two quantities must be connected.

Moving $n$ moles of electrons through a potential difference $E$ transfers charge $n F$ , where $F = 96485$ C mol⁻¹ is the Faraday constant, the charge on a mole of electrons. The electrical work done is charge times potential, and when the cell is drawn on reversibly, meaning infinitely slowly against an almost equal opposing voltage, that work equals $Δ G$ :

Δ_{r} G = - n F E and at standard conditions Δ_{r} G^{ominus} = - n F E^{ominus}

The minus sign puts a positive potential with a negative free energy change, so a cell that reads positive is a cell whose reaction goes. Combining this with $Δ_{r} G^{ominus} = - R T \ln K$ from the fourth lesson gives a startling shortcut:

\ln K = \frac{n F E^{ominus}}{R T}

A voltmeter reads an equilibrium constant. This is the most sensitive method available for constants far from one, because the logarithm compresses them: at $298$ K, $0.0592$ V of cell potential per electron is a factor of ten in $K$ .

Example. For the Daniell cell, $E^{ominus} = 1.10$ V with $n = 2$ . Find $Δ_{r} G^{ominus}$ and $K$ .

$Δ_{r} G^{ominus} = - 2 \times 96485 \times 1.10 = - 2.12 \times 10^{5}$ J mol⁻¹, or $- 212$ kJ mol⁻¹. Then $\ln K = 212267 / (8.314 \times 298.15) = 85.6$ , so $K = 1.5 \times 10^{37}$ . The reaction goes essentially to completion, which is what a strip of zinc in copper sulfate looks like.

Now you. For the copper and silver cell, $E^{ominus} = 0.46$ V with $n = 2$ . Find $Δ_{r} G^{ominus}$ and $K$ .

Answer

$Δ_{r} G^{ominus} = - 2 \times 96485 \times 0.46 = - 8.88 \times 10^{4}$ J mol⁻¹, or $- 88.8$ kJ mol⁻¹. Then $\ln K = 88766 / 2478.8 = 35.8$ , so $K = 3.6 \times 10^{15}$ . Less extreme than the Daniell cell, and still complete for any practical purpose.

The Nernst equation

Standard conditions mean unit activities, which no working battery has. Take $Δ_{r} G = Δ_{r} G^{ominus} + R T \ln Q$ from the fourth lesson and divide throughout by $- n F$ :

E = E^{ominus} - \frac{R T}{n F} \ln Q

This is the Nernst equation, from Walther Nernst in 1889. At $298$ K, converting to base ten logarithms gives the form used in practice, since $R T \ln 10 / F = 0.0592$ V:

E = E^{ominus} - \frac{0.0592}{n} \log_{10} Q

Everything about a cell under real conditions is in that line. As a battery discharges, products accumulate, $Q$ rises and $E$ falls, reaching zero exactly when $Q = K$ : a flat battery is a cell at equilibrium, and the reason it is flat is that there is no free energy left to extract, not that anything has been used up in the ordinary sense.

Example. A Daniell cell has $[Z] = 1.0$ mol dm⁻³ and $[C] = 0.010$ mol dm⁻³. Find its potential.

The reaction is $Zn + C \to Z + Cu$ , and the solids have activity $1$ , so $Q = [Z] / [C] = 100$ . With $n = 2$ , $E = 1.10 - (0.0592 / 2) \log_{10} (100) = 1.10 - 0.059 = 1.04$ V. A hundredfold change in concentration has cost less than six per cent of the voltage, which is the logarithm at work and the reason cells hold their voltage well until they are nearly exhausted.

Now you. A concentration cell has copper electrodes in $0.0010$ and $1.0$ mol dm⁻³ copper sulfate, joined by a salt bridge. Its $E^{ominus}$ is zero, since both halves are the same couple. What is its potential?

Answer

The cell runs to equalise the two concentrations, so copper dissolves in the dilute half and deposits in the concentrated one, and $Q = 0.0010 / 1.0 = 10^{- 3}$ . Then $E = 0 - (0.0592 / 2) \log_{10} (10^{- 3}) = 0.089$ V. A cell driven entirely by a concentration difference, which is also how a pH meter works.

The Nernst equation also settles the pH question from the previous lesson. Permanganate is reduced as $Mn + 8 + 5 e^{-} \to M + 4 O$ , so $Q$ contains $[]^{- 8}$ and the potential depends on pH as $E = 1.51 - (0.0592 \times 8 / 5) pH$ . At pH $7$ that is $0.85$ V, a loss of two thirds of a volt, which is why permanganate stops at manganese dioxide in neutral solution instead of going to $M$ .

Real cells

A lead-acid cell delivers $2.05$ V from $Pb + Pb + 2 S \to 2 PbS + 2 O$ , and six in series give the familiar $12.3$ V. It is a rare case where the state of charge can be read directly, since the reaction consumes sulfuric acid and the electrolyte density falls from about $1.28$ to $1.10$ g cm⁻³ as it discharges. Its virtue is the ability to deliver hundreds of amperes briefly; its vice is that lead is heavy, at around $40$ W h kg⁻¹.

A lithium-ion cell delivers $3.6$ V and $250$ W h kg⁻¹ or more, and works quite differently: nothing dissolves, and lithium ions shuttle between two host lattices, graphite and a metal oxide, in which they sit between layers. Because the electrodes are not consumed, the cell survives thousands of cycles.

A hydrogen fuel cell is the cleanest illustration of the theory. Its reaction is $+ \frac{1}{2} \to O(l)$ , with $Δ_{r} G^{ominus} = - 237.1$ kJ mol⁻¹ from the third lesson and $n = 2$ , so $E^{ominus} = 237100 / (2 \times 96485) = 1.23$ V, which is what the cell measures. And since a fuel cell is limited by $Δ G$ rather than by the Carnot efficiency of a heat engine, its ceiling is $Δ G / Δ H = 237.1 / 285.8 = 83$ per cent, far above what any combustion engine can reach.

Corrosion and electrolysis

Rusting is a cell that nobody wanted. Iron oxidises at $- 0.44$ V where the metal is stressed or the oxygen supply is poor, the electrons travel through the metal itself, and oxygen is reduced elsewhere on the surface at $+ 0.40$ V in neutral water. The iron(II) formed is then oxidised further to hydrated iron(III) oxide, which flakes off rather than protecting what is beneath, unlike aluminium's oxide.

Knowing it is a cell tells you how to stop it. Attach a metal with a more negative potential, such as zinc at $- 0.76$ V or magnesium at $- 2.37$ V, and that metal becomes the anode and corrodes instead. This is sacrificial protection: the zinc blocks bolted to a ship's hull, and the galvanised coating on a steel bucket, which protects the steel even where the coating is scratched through, because protection is electrical rather than physical.

Run a cell backwards by applying a voltage greater than its own and the reaction is forced uphill. That is electrolysis, and the stoichiometry is exact, since electrons are counted in coulombs: the amount of substance transformed is $Q / n F$ , where $Q = I t$ . Michael Faraday established this in 1834, before anyone knew what an electron was, and it is one of the strongest early pieces of evidence that charge is carried in units.

The industrial scale of that arithmetic is worth seeing. Aluminium is produced as $A + 3 e^{-} \to Al$ , so one kilogram, which is $1000 / 26.982 = 37.06$ mol, needs $3 \times 37.06 \times 96485 = 1.07 \times 10^{7}$ C. A Hall-Héroult cell runs at about $4.5$ V, so the energy is $4.5 \times 1.07 \times 10^{7} = 4.8 \times 10^{7}$ J, or $13.4$ kW h per kilogram. Real smelters use $13$ to $15$ , the excess being resistive heating, and this is why aluminium plants are built next to hydroelectric dams and why recycling aluminium, which needs only melting, saves about $95$ per cent of the energy.

Where this leaves you

The course set out to answer three questions about a reaction: whether it goes, how far, and how fast. All three now have machinery behind them. Free energy decides direction, and $Δ_{r} G^{ominus} = - R T \ln K$ converts a table of enthalpies and entropies into a predicted yield. A rate law, measured rather than assumed, gives the time, and its temperature dependence gives a barrier height that belongs to a step rather than to an equation. Proton transfer, dissolution and electron transfer are the same equilibrium arithmetic applied three times.

What has been set aside throughout is worth naming. Activity coefficients have been taken as one, which is good to a few per cent in dilute solution and poor in seawater or in a battery. Rate constants have been treated as numbers to measure rather than to calculate, and calculating them from the potential energy surface is the subject of chemical dynamics. And nothing here says how to make a particular molecule rather than a mixture, which is synthesis, a discipline built on top of everything in this course rather than contained in it.