Work from Entropy as the Driving Force.

Processes which occur with only an entropy change as the driving force (that is DH = 0) are few.

Examples are:

expansion of an ideal gas;

mixing of liquid or gaseous components;

diffusion through a barrier.

And making such a process produce work is often difficult.

Except in the case of a voltaic cell: it is possible to create easily a voltaic cell that depends entirely on an entropy change to produce work.

The cell depends on the entropy change when two solutions of different concentrations are mixed and so is termed a concentration cell.

A Concentration Cell.

Shorthand notation: Cu | Cu 2+ (0.1 M) || Cu 2+ (1 M) | Cu


Cu2+(cathode) + Cu(anode) Cu2+(anode) + Cu(cathode)

From thermodynamics (Nernst equation):

Ecell = Ecell - {0.0257 / n} ln Q

Ecell = 0 - {0.0257 / 2} ln { Cu2+(anode) / Cu2+(cathode)}

Ecell = 0 - {0.01285} ln {0.1 / 1} = 0.03 v

Potentiometric Titrations.

Because the voltage of a cell varies with the concentrations of the species present, it is possible to follow a titration using a cell which has the species being titrated as a component of the cell.

A well known example a potentiometric titration is use of the pH meter to find the [H3O+].

Other electrodes could be used to follow other reactions:

One such electrode is the Ag(s)/AgCl(s) electrode for chloride ion titration. The second electrode is often a calomel electrode:

KCl(satd) | Hg2Cl2(s), Hg(l). This electrode has a fixed potential which does not vary during the titration.

Consequently we have:

EAg = EAg - 0.0257 ln [Cl-]

for the one electrode.

By adding AgNO3 solution to precipitate out the chloride ion, the cell potential will vary as the [Cl-] drops.

Potentiometric titration of chloride ion using a silver electrode.

The following titration curve was obtained for the titration of 2.433 mmol of chloride ion with 0.1000 M silver nitrate:

The end point of the reaction is indicated in the usual way in the centre of the S portion of the curve.

Practical Cells.

I. Primary Cells: a cell that can be discharged once.

(the chemical reaction is not reversible as arranged in this cell)

Examples: any non-rechargeable "battery"

Zn | NH4Cl(aq) | MnO2(s) | C

maximum voltage = 1.55 v.

anode: Zn Zn2+ + 2 e-

cathode 2 MnO2 + H2O + 2 e- Mn2O3 + 2 OH-

also: NH4+ + OH- NH3 + OH-

and: Zn2+ + NH3 + Cl- [Zn(NH3)2]Cl2(s)

The alkaline non-rechargeable battery is very similar, but using and alkaline electrolyte (KOH) instead of an acidic one (NH4+).

Zn | KOH(aq) | MnO2 | C

maximum voltage = 1.55 v.

anode: Zn Zn2+ + 2 e-

cathode 2 MnO2 + H2O + 2 e- Mn2O3 + 2 OH-

also: Zn2+ + 2 OH- Zn(OH)2(s)

The Mercury Cell (primary)

Zn/Hg | KOH (saturated with ZnO) | HgO(s) / C | Hg(l)

voltage: 1.34 v.

anode: Zn(s) + 2 OH- ZnO(s) + H2O + 2 e-

cathode: HgO(s) + H2O + 2 e- Hg(l) + 2 OH-

A button cell battery useful because of its constant voltage. This comes about due to all net reactants and products being pure solids and liquids and thus always having an activity of 1.

The Silver-Zinc Cell (primary)

Zn(s), ZnO(s) | (KOH(sat) | Ag2O(s), Ag(s)

voltage = 1.8 v.

anode: Zn(s) + 2 OH- ZnO(s) + H2O + 2 e-

cathode: Ag2O(s) + H2O + 2e- 2 Ag(s) + 2 OH-

Another useful button battery which has a constant voltage during its operation. Again, no solutions are involved in the redox reaction, so activities remain constant at 1.

Solid Electrolyte Batteries.

Li(s) | Li+(solid electrolyte) | TiS2

voltage = 3 v ( to 3.6 v)

anode: Li Li+

cathode TiS2 + e- TiS2-

II. Secondary Cell: a cell that can be discharged then charged a

number of times.

(the chemical reaction is reversible as arranged in this cell)

Examples: NiCad, lead-acid (automobile) batteries.

The Nickel-Cadmium Storage Cell.

Cd(s), CdO(s) | KOH(20%) | NiO(OH)(s), Ni(s)

voltage = 1.35 v.

anode: Cd(s) + 2 OH- Cd(OH)2(s) + 2 e-

cathode: NiO(OH)(s) + H2O + e- Ni(OH)2(s) + OH-

Here the anode and cathode net reactions all involve solids, thereby not allowing any chemical to diffuse away from the electrode where it is produced. This allows the reverse reactions to occur on charging. "Memory" is a complex phenomenon in which charging a battery at a level which is not completely run down causes the battery to discharge only to this level.

The Edison Storage Cell (NiFe)

Ni(s), Fe(s) | KOH | NiO2(s), Ni(s)

voltage = 1.4 v

anode: Fe(s) + 2 OH- Fe(OH)2(s) + 2 e-

cathode: 2 NiO2(s) + H2O + 2 e- Ni2O3(s) + 2 OH-

The Lead-Acid Storage Battery.

Pb(s), PbSO4(s) | H2SO4(35%) | PbSO4(s), PbO2(s)

voltage = 2 v.

anode: Pb(s) + SO4 2- PbSO4(s) + 2 e-

cathode: PbO2(s) + SO4 2- + 4 H+ + 2 e- PbSO4(s) + 2 H2O

Again, both reactants and products of the process are solids and remain in place at their respective electrodes, thereby allowing the process to be reversed on charging.

Automobile batteries consist of six of these cells to give the 12 v used in the electrical sysem.

These batteries do not suffer from "memory" so that they can be constantly held at full charge by a generator whilst the engine is running. A very large power output is required to turn over a car engine to start it, and these batteries can supply it.

For an electric car, the batteries must also be able to provide a power output, and the lead acid battery is used. Unfortunately, this battery is extremely heavy and not environmentally friendly. Research to find other cells which could provide the high power required but which have a lower mass and perhaps which are more environmentally friendly is in progress.

III. Fuel Cells: A fuel cell is a cell that is constantly provided with reactants and has the reaction products removed.

(continuous operation is made possible by supplying reactants and removing products as needed)

Example: the hydrogen-oxygen fuel cell.

Basically the problem here is to take an atom transfer oxidation and carry it out using electron transfer processes. This is possible, but often the conditions needed for a successful process are highly corrosive (either acidic or basic)

The Hydrogen-Oxygen Fuel Cell.

maximum voltage: 1.2 v

operating voltage: 0.81 v

anode: H2(g) + 2 OH- Ž 2 H2O + 2 e-

cathode: O2(g) + 2 H2O + 4 e- Ž 4 OH-

Conditions: temperature: 200C

gas pressures: 2000 to 4000 kPa

KOH solution: 75 to 80%

The Methane Fuel Cell.

Possible redox reactions are:

anode: CH4(g) + 2 H2O(l) Ž CO2(g) + 8 H+ + 8 e-

cathode: O2(g) + 4 H+ + 4 e- Ž 2 H2O(l)

As shown this cell requires an acid electrolyte. The temperature of operation is high (about 200C) and this results in a very corrosive mixture. A cell made up of these half reactions has been operated under research conditions but it has not yet been commercialized.


Corrosion is the spontaneous oxidation of metals.


1. Cover the metal so that its surface is not in contact with oxygen or moisture:

2. Use a sacrificial anode.

This consists of electrically connecting a more reactive metal which then becomes the anode and oxidizes in preference to the protected metal. Magnesium is used to protect underground iron pipes and iron ships, the magnesium oxidizing before the iron which it then protects.


An electrolytic cell is one which is operating in the non-spontaneous direction by the application of an electrical pressure (voltage) to push electrons in the opposite direction to the spontaneous direction. Thus chemical reactions can readily be made to occur to give high energy products.

Electrolysis of water produces H2(g) and O2(g) and this may be a way of collecting the sun's energy where and when it is most available and then transporting or storing the energy (as high energy hydrogen gas) for later use (perhaps in a fuel cell). This has been proposed as a basis for a "hydrogen" economy.

Commercially, electrolysis is used for the production of Cl2(g) and NaOH, by the electrolysis of NaCl(aq). This process also produces hydrogen gas. The chloralkali process is shown on page 991 of your text.

Electrolysis of molten sodium chloride is used to produce Cl2(g) and sodium metal. The process is similar to that shown on page 994 of your text for the production of fluorine.

And the electrolysis of bauxite is used to produce aluminum.

(see your text book, page 971 for this process independently discovered by two 22-year olds!)

Calculations of Amount of Reaction.

The quantity of charge on one mole of electrons is the Faraday, or 96 485 Coulombs. Also a coulomb is equal to amps x seconds:

C = A * s.

Thus in a reaction that involves 1.00 mol of electrons being transferred 96 485 C must be used. If the circuit is of 10.0 amps, then the time must be 9 649 s, or 2.68 days.

Example 1:

What mass of copper can be deposited at a cathode if a current of 1.00 amps is allowed to flow for 1.00 hours?

Quantity = A * s = 1.00 * 3600 = 3600 C

Mol e- = 3600 C / 96,485 C mol-1 = 0.0373 mol e-

Reaction: Cu2+ + 2 e- Ž Cu

Thus 2 mol e- required for each 1 mol of Cu deposited.

Mol Cu = mol e- / 2 = 0.0373 / 2 = 0.0187 mol Cu

Example 2.

What mass of aluminium can be deposited at the cathode of an electrolytic cell if a current of 100 amps flows for 1.00 days?

Quantity = A * s = 100 * 86,400 = 8.64 x 106 C

Mol e- = 8.64 x 106 C / 96485 C mol-1 = 89.5 mol e-

Reaction: Al3+ + 3 e- Ž Al

Thus 3 mol e- required for each mole of Al deposited.

Mol Al = mol e- / 3 = 89.5 / 3 = 29.8 mol Al.

Mass Al = 29.8 mol * 27.0 g mol-1 = 806 g.

Note the vast amount of electrical power needed to produce less than 1 kg of aluminium. This is why aluminium smelters must be located where there is an ample and cheap supply of electricity. This does not occur where the aluminium ore is found and consequently it is cheaper to move the ore to the smelters, often found in mountainous regions where cheap hydroelectricity can be found.


Electroplating is the deposition of a metal on to another surface. Whatever this other surface is made of, it must be capable of acting as a cathode. Thus iron can be electroplated with silver or chromium, both of which protect the iron from corrosion.

Your text, figure 17-24 shows the set-up for electroplating silver.

Calculations of how much material (silver, say) is deposited are done as above.

How long must electroplating be carried out if a mass of 0.100 g of silver is to be plated out using a current of 3.30 amps?

Moles of silver = 0.100 g / 107.9 g/mol = 9.27 x 10-4 mol

Reaction: Ag+ + e- Ž Ag

Moles e- required = 9.27 x 10-4 mol.

C required = 96,485 C / mol * 9.27 x 10-4 mol = 89.4 C

C = A * s

Time required = C / A = 89.4 C / 3.33 A = 26.9 s.