5.1.1 The reactivity series of metals

Reactivity measures how readily a metal forms positive ions

1. Metals are placed in order of how readily they react, with the most reactive at the top.
2. The order used in this course runs potassium, sodium, calcium, magnesium, aluminium, (carbon), zinc, iron, (hydrogen), copper, silver, gold.
3. A metal reacts by losing electrons from its atoms, which leaves positively charged ions behind.
4. The more readily a metal's atoms form cations, the higher that metal sits in the order.
5. Three kinds of evidence fix a metal's place: its reaction with water, with dilute acid, and with salt solutions.

Reactions with cold water separate the top of the series

1. Potassium reacts violently with cold water and the hydrogen given off burns with a lilac flame.
2. Sodium reacts vigorously, melting into a ball that skims across the surface.
3. Calcium reacts steadily, giving a stream of bubbles and a cloudy suspension of calcium hydroxide.
4. Each of these three metals gives a hydroxide and hydrogen: $$metal+water\to metal\; hydroxide+hydrogen$$
5. Magnesium barely reacts with cold water, but reacts readily with steam to give magnesium oxide and hydrogen.
6. Zinc, iron and copper show no reaction with cold water at all.

Reactions with dilute acid rank the metals below calcium

1. A metal above hydrogen in the order reacts with dilute acid: $$metal+acid\to salt+hydrogen$$
2. Magnesium gives rapid, steady bubbling with dilute hydrochloric acid.
3. Zinc bubbles more slowly than magnesium, and iron more slowly still.
4. Copper, silver and gold give no reaction with dilute acid, which places them below hydrogen.
5. Since the products are the same in every case, it is the rate of bubbling that ranks the metals.

Displacement in salt solutions compares two metals directly

1. A more reactive metal takes the place of a less reactive one in a solution of its salt.
2. Zinc displaces copper from copper sulfate solution: $$Zn+{CuSO}_4\to {ZnSO}_4+Cu$$
3. Copper placed in zinc sulfate solution gives no change, and that one-way answer fixes the order of the pair.
4. A colour change in the solution and a coating on the metal are the two observations that show a reaction has happened.
5. Testing every metal against every salt solution builds the whole order from pairs.

Carbon and hydrogen sit in the series as reference points

1. Carbon and hydrogen are non-metals, so they are written in brackets within the order.
2. Their places come from the same evidence as any metal's: whether they displace a metal, or are displaced by one.
3. A metal below carbon can be extracted from its oxide by heating the ore with carbon.
4. A metal above hydrogen reacts with dilute acid, and one below hydrogen does not.
5. These two markers turn the order into a tool for predicting reactions rather than a list to learn.

5.1.2 Displacement reactions as redox reactions

Displacement moves electrons from one metal to another

1. In a displacement reaction the atoms of the more reactive metal lose electrons and become ions.
2. The ions of the less reactive metal gain those same electrons and become atoms.
3. Loss of electrons is oxidation and gain of electrons is reduction.
4. Both changes occur in the same reaction, so a displacement is always a redox change.
5. The electrons pass directly between the two species, with no external circuit involved.

Half equations show each electron change on its own

1. Each half equation shows only the loss or only the gain, with electrons written as $e^{-}$.
2. Magnesium atoms lose two electrons, which is oxidation: $$Mg\to {Mg}^{2+}+2e^{-}$$
3. Copper ions gain two electrons, which is reduction: $${Cu}^{2+}+2e^{-}\to Cu$$
4. The electrons lost in one half equation equal the electrons gained in the other.
5. Adding the two half equations together cancels the electrons and rebuilds the overall change.

The ionic equation removes the spectator ions

1. A salt in solution is present as separate ions, so copper sulfate solution holds ${Cu}^{2+}$ and ${SO}_4^{2-}$.
2. The sulfate ions are unchanged on both sides, so they take no part in the reaction.
3. Leaving them out gives the ionic equation for magnesium and copper sulfate: $$Mg+{Cu}^{2+}\to {Mg}^{2+}+Cu$$
4. Stripping the equation this way leaves only the species whose charge changes.
5. The same ionic equation describes the reaction whatever copper salt is used, provided it dissolves.

A worked example: zinc in copper sulfate solution

1. Zinc is above copper, so zinc atoms are the species that lose electrons.
2. The oxidation half equation is: $$Zn\to {Zn}^{2+}+2e^{-}$$
3. The reduction half equation is: $${Cu}^{2+}+2e^{-}\to Cu$$
4. Adding them and cancelling the two electrons gives: $$Zn+{Cu}^{2+}\to {Zn}^{2+}+Cu$$
5. The blue of the copper ions fades and a copper coating forms on the zinc, which is the visible evidence.

Deciding which way a displacement will run

1. A displacement happens only when the free metal is above the metal in the compound.
2. The atoms of the higher metal give up electrons more readily, so they are the ones that are oxidised.
3. Reversing the pair gives no reaction, because the lower metal cannot force electrons onto the higher metal's ions.
4. The charge on the ion sets how many electrons move, so a $2+$ ion needs two per atom.
5. Balancing the electrons matters when the two charges differ, as with silver's ${Ag}^{+}$ and copper's ${Cu}^{2+}$.

5.1.3 Extracting metals from ores by reduction

Most metals are found in the Earth's crust as compounds

1. A rock is worth mining only when it holds enough of a metal compound to make extraction pay.
2. Most metals are found combined with other elements, very often as an oxide or a carbonate.
3. Gold and platinum are so unreactive that they occur as the uncombined elements.
4. An unreactive metal resists combining with oxygen, which is why it survives as the metal itself.
5. Obtaining a metal from a compound therefore means removing whatever it is joined to.

Extraction is a reduction, because the metal loses oxygen

1. Gain of oxygen is called oxidation and loss of oxygen is called reduction.
2. Heating a metal oxide with a substance that takes its oxygen reduces the ore: $$metal\; oxide+reducing\; agent\to metal+oxide\; of\; the\; reducing\; agent$$
3. The substance that takes the oxygen is itself oxidised in the same reaction.
4. The two descriptions of reduction agree: an oxide loses oxygen because the metal ions in it gain electrons.
5. Extracting a metal from a compound in its ore is a reduction of that metal.

Metals below carbon are reduced by heating with carbon

1. Carbon sits in the reactivity series between aluminium and zinc, placed there by experiment.
2. Carbon takes oxygen from the oxide of any metal below it, because those metals hold their oxygen less strongly.
3. Iron oxide is reduced by carbon on this basis: $$iron\; oxide+carbon\to iron+carbon\; dioxide$$
4. Zinc, iron, tin, lead and copper are all extracted this way.
5. Heating with carbon is cheap, since carbon comes from coke and the heat is supplied by burning fuel.

Metals above carbon are extracted by electrolysis

1. Potassium, sodium, calcium, magnesium and aluminium all sit above carbon, so carbon cannot reduce their ores.
2. Passing a direct current through the molten compound decomposes it and releases the metal.
3. Aluminium is obtained from aluminium oxide, which melts far too high to be liquefied on its own.
4. The oxide is instead dissolved in molten cryolite, which lowers the working temperature and cuts the energy used.
5. Aluminium forms at the negative electrode and oxygen is released at the positive electrode.

Reactivity and cost together decide the method

1. The position of the metal in the reactivity series decides whether carbon is strong enough to reduce its ore.
2. Where carbon works it is always chosen, because the process costs far less to run.
3. Electrolysis is reserved for the metals that nothing cheaper will reduce.
4. The price of a metal reflects this, so aluminium costs more to produce than iron of the same mass.
5. A metal found uncombined needs no reduction at all, only separation from the rock around it.

5.1.4 Biological methods of metal extraction

Low-grade ores make conventional extraction uneconomic

1. A low-grade ore holds only a small percentage of the metal compound.
2. Mining and smelting such an ore uses large amounts of energy for very little metal.
3. Traditional mining also scars the landscape and leaves large volumes of waste rock.
4. Rich ores are being used up, so low-grade deposits are an increasing share of what remains.
5. Two biological methods, bioleaching and phytoextraction, offer ways of working these deposits.

Bioleaching uses bacteria to produce a leachate

1. Certain bacteria feed on the sulfur in low-grade ores and break the ore down as they grow.
2. The liquid that drains away, the leachate, holds dissolved compounds of the metal.
3. For copper the leachate contains copper sulfate in solution, not copper metal.
4. Bioleaching needs no heat and no smelting, so it uses far less energy than mining and roasting.
5. The bacteria work slowly, so producing a useful quantity of metal can take months or years.

Phytoextraction uses plants to concentrate the compound

1. Plants are grown on soil or waste that holds a low concentration of metal compounds.
2. Their roots absorb the compounds, which build up in the leaves and stems as the plants grow.
3. The plants are harvested and burned, leaving an ash that is rich in the metal compound.
4. That ash is a far more concentrated source than the ground the plants grew on.
5. Phytoextraction can also clean up contaminated land while the metal is being collected.

The metal is recovered from the leachate or the ash

1. Both methods deliver a metal compound, so a reduction is still required.
2. Adding scrap iron to a copper leachate displaces copper, because iron is the more reactive metal.
3. The displacement that recovers the copper is: $$Fe+{CuSO}_4\to {FeSO}_4+Cu$$
4. Electrolysis of the solution is the alternative, and it gives copper of higher purity.
5. The same two routes are used on the solution made by dissolving the ash from phytoextraction.

Weighing the biological methods against traditional mining

1. Both methods use less energy and cause less damage to the landscape than opening a mine.
2. Both work ores that would otherwise be left as waste, which extends the supply of the metal.
3. Both are much slower, which is the main reason they have not replaced conventional extraction.
4. Phytoextraction takes up land that could be used for growing food.
5. Whether a method is worth using depends on the price of the metal and how little of it the ore holds.

5.1.5 Resistance to oxidation and recycling metals

Resistance to oxidation follows the reactivity series

1. A metal reacts with the oxygen in air to form its oxide, which is an oxidation.
2. A metal high in the reactivity series is oxidised quickly, because its atoms lose electrons readily.
3. Potassium and sodium tarnish within seconds of being cut, so they are stored under oil.
4. Iron and zinc are oxidised slowly in moist air over weeks and months.
5. Gold and platinum are not oxidised in air at all, which is why they keep their shine.

Aluminium behaves better than its position suggests

1. Aluminium sits above zinc and iron, so it would be expected to corrode faster than either.
2. In practice aluminium objects last for years without visibly corroding.
3. The metal reacts at once with air to form a thin layer of aluminium oxide.
4. That layer sticks firmly to the surface and seals the metal underneath from further attack.
5. The oxide layer, not low reactivity, is what makes aluminium useful outdoors.

Recycling returns a used metal to service

1. A used metal object is collected, sorted, melted and cast into new stock.
2. The metal is an element, so it is not used up and can be recycled many times over.
3. Recycling skips mining, transport and reduction of the ore entirely.
4. Recycling aluminium uses only a small fraction of the energy that electrolysis of its ore demands.
5. Each tonne recycled leaves a tonne of ore in the ground for later use.

The case for recycling is economic as well as environmental

1. Less energy is bought, so the running cost per tonne of metal is usually lower.
2. Less ore is mined, which limits habitat damage and the waste rock left behind.
3. Fewer fossil fuels are burned, so less carbon dioxide reaches the atmosphere.
4. Less metal is sent to landfill, where it takes up space and can leach into the ground.
5. A reserve of ore is a finite raw material, so using less of it protects future supply.

Judging the evidence on recycling a particular metal

1. The saving is largest where extraction is most expensive, which is why aluminium is the clearest case.
2. For a cheaply extracted metal the saving is smaller, though it still exists.
3. The quality of the recycled metal matters, since some scrap is only fit for lower-grade uses.
4. Distance counts, because transporting scrap a long way uses fuel that eats into the saving.
5. A sound judgement compares energy, cost and supply together rather than any one of them alone.

5.1.6 Life cycle assessment

A life cycle assessment follows a product from cradle to grave

1. The assessment looks at the effect a product has on the environment across its whole life.
2. Four stages are considered: obtaining the raw materials, manufacturing, using the product, and disposing of it.
3. Each stage is examined for the energy it uses, the resources it consumes and the waste it produces.
4. Putting the four together prevents a saving at one stage hiding a cost at another.
5. The result is used to compare two products that do the same job.

Obtaining the raw materials

1. Raw materials are mined, quarried, drilled or grown, and every route disturbs land.
2. Extracting a metal from its ore uses energy and leaves waste rock behind.
3. A material made from crude oil ties the product to a finite resource.
4. Transporting the raw material to the factory burns fuel and releases carbon dioxide.
5. A recycled input cuts this stage sharply, because the extraction has already been paid for.

Manufacture, use and disposal

1. Manufacturing uses energy for heating, shaping and assembly, and can release waste into air or water.
2. The use stage covers everything the product consumes while it is doing its job.
3. A product that uses electricity in service may dominate its own assessment over many years.
4. A long-lived product spreads the cost of making it across more years of use.
5. Disposal covers whether the product is recycled, burned or buried, and how long it persists.

Reading the data from an assessment

1. Data are usually given per stage, so the four figures are compared before they are added.
2. The totals decide the comparison, and the largest stage shows where an improvement would count most.
3. Figures must be quoted for the same quantity of product, or the comparison means nothing.
4. A difference of a few percent is too small to separate two products reliably.
5. A conclusion is stated with the stage that drives it, not as a bare preference.

Where an assessment involves judgement

1. Energy use can be measured, but the damage done by habitat loss cannot be put on the same scale.
2. Deciding how much weight to give each effect is a judgement, not a measurement.
3. Some figures are estimates, particularly for how long a product will be kept and used.
4. An assessment paid for by a manufacturer may be selective about which stages it reports.
5. For these reasons two honest assessments of the same product can reach different conclusions.