F334 – The Steel Story:

Extracting and Purifying metals:

Most metals are found as compounds in the earth’s crust. Metal ores are mined and the metal extracted by reduction. Method depends on reactivity of metal. Reactive metals such as those in group 1 and 2 as well as aluminium are extracted by

electrolysis Less reactive metals such as iron and zinc are obtained from their ores by reduction with

carbon or carbon monoxide. Unreactive metals can be found as element in the Earth’s crust.

Preventing corrosion:

Barrier Protection: prevents oxygen and/or water coming into contact with iron and steel. Examples including painting, greasing, oiling and using polymer coatings.

Galvanising: Covering the steel in a thin layer of protective zinc oxides. Stainless steel contains chromium that oxidises, thus also leaving a protective oxide layer.

Sacrificial protection: attaching blocks of a more reactive metal such as zinc to large iron structures such as ships. An electrochemical cell is formed and the reactive metal corrodes preferentially.

The d block: transition metals (focusing mainly on first ten elements)

Electronic arrangements:

Properties of first ten elements are remarkably similar Each additional electron enters the 3d sub-shell as the atomic number increases. The 4s sub-shell has slightly lower energy than the inner 3d sub-shell and is filled first. However Cr and Cu have unusual electron arrangements due to the stability associated with

a half-full and completely full sub-shell arrangement. A transition metal is defined as a d-block element that forms at least one ion with a partially

filled sub-shell of d electrons. When d-block elements react to form ions the 4s electrons are the first to be lost, for

example:Fe[Ar]3d64s2 → Fe2+[Ar]3d64s0 → Fe3+[Ar]3d54s0

Scandium and Zinc don’t display the chemical properties associated with the transition metals because their ions have electron arrangements 3d0 and 3d10 respectively.

Physical properties:

Good conductors of heat and electricity They are denser and have higher melting and boiling points than s-block elements. They are hard and durable with high tensile strength and good mechanical properties. This makes them ideal for a wide range of uses, both as pure metals and in alloys.

Chemical Properties:

Four important chemical properties with directly relate to electronic arrangement of the elements or their ions

Variable oxidation states -This happens because the differences between successive ionisation enthalpies in the 3d and 4s sub-shell are relatively small, so multiple electron loss is possible.In the lower oxidation states, the elements exist as simple ions, but in the higher oxidation states they are covalently bonded to electronegative elements, such as oxygen or fluorine forming anions.Redox reactions are a key feature. Compounds containing transition metals in high oxidation states tend to be oxidising agents, whereas compounds in low oxidation states are often reducing agents.Oxidation is Loss of electrons or increases in oxidation number. Vice versa for reduction.Example:MnO4

- + 8H+ + 5Fe2+ → Mn2+ + 4H2O + 5Fe3+

As half equations – MnO4- + 8H+ +5e- → Mn2

5Fe2+ → 5Fe3+ + 5e-

Manganate(VII) ions have gained electrons as the oxidation state has decreased from +7 to +2, this is reductions. Iron(III) ions have lost electrons and the oxidation state has increased from +2 to +3, this is oxidation.

Formation of coloured ions -The metal compounds show many different colours. Electron transitions occur within the 3d sub shell when visible light is absorbed.This can only happen in ions that have a partially filled 3d sub shell. A fuller explanation is further on in the notes.

Formation of complexes-the 3d orbitals can accommodate the electrons donates by ligands. Further details will be given later in the notes.

Catalytic activity-the metals can act as heterogeneous catalysts, providing a surface onto which gaseous reactant molecules are adsorbed. Weak interactions between these and the 3d and 4s electrons keep the molecule in place while bonds are broken and formed.Can also act as homogeneous catalysts as they are able to change from one oxidation state to another during the reaction before they return to their original oxidation state.

Complex Formation:

In a complex a central metal atom or ion is surrounded by ligands. Ligands are molecules or anions with one or more lone pairs of electrons. They form dative covalent bonds with the central metal atom or ion. The number of bonds between the central metal and the ligands is called the coordination

number of the central metal.

Shapes of complexes:

Complexes with a coordination number of 6 are generally octahedral in shape.

Those with a coordination number of 4 are usually tetrahedral but can be square planar. Those with a coordination number of 2 are linear. If a complex has an overall charge it is called a complex ion. Remember to always draw a square bracket round a complex ion and write the charge

outside the bracket.

Types of ligands:

Ligands can form one bond to the central metal – monodentate. Can form two bonds to the central metal – bidentate Can form many bonds to the central metal – polydentate. Table below demonstrates names and formulae for common ligands:

Molecule/ion Formula Name of ligand

Type of ligand

water H2O aqua monodentateammonia NH3 ammine monodentatechloride ion Cl- chloro monodentatecyanide ion CN- cyano monodentatehydroxide ion OH- hydroxo monodentateethanedioate ion (COO-)2 ethanedioate bidentateethylenediaminetetracetate ion

edta4- (do not need to remember full formula

edta polydentate

Naming complexes:

Use the following rules to figure out the name of a complex:- Write the number of each type of ligand using the prefixes mono-, di-, tri-, tetra-, penta-, hexa-.- Write the name of each ligand in alphabetical order.- Write the name of the central metal. If there is an overall charge which is positive or neutral use the English name. If the overall charge is negative use the Latinised name.- Write the oxidation number of the central metal in brackets.

Below is the table for Latinised names of metals.

Metal Latinised nameCu CuprateV VanadateTi TitanateAg ArgentateZn ZincatePb PlumbateCr Chromate

Colour in complexes:

When light falls on a complex some frequencies are absorbed and some are reflected or transmitted.

If visible light is absorbed the complex will appear coloured. The presence of ligands causes the five d orbitals of the central metal to ‘split’. Some orbitals become slightly higher in energy and some slightly lower in energy. The small energy gap between these orbitals allows visible light to be absorbed. The colour of the complex (which we see) is the complementary colour to that absorbed.

Ligand substitution reactions:

These are reactions in which one ligand displaces another. For example, if concentrated hydrochloric acid is added to a solution of copper(II)

sulphate(VI), chlorine ligands replace water ligands. The solution changes colour from blue to yellow. Ligand substitution, also called a ligand exchange reaction, occurs if the new complex

formed is more stable than the previous complex.[Cu(H2O)6]2+ + 4Cl-

(aq) → [CuCl4]2-(aq) + 6H2O(aq)

Precipitation reactions of Cu2+(aq), Fe2+

(aq) and Fe3+(aq).

Solutions containing copper(II), iron(II) or iron(III) ions form coloured precipitates with sodium hydroxide solution.

Fe2+(aq) + 2OH-

(aq) → Fe(OH)2(s)

Green gelatinous solid. Fe3+

(aq) + 2OH-(aq) → Fe(OH)3(s)

Orange gelatinous solid. Cu2+

(aq) + 2OH-(aq) → Cu(OH)2(s)

Pale blue solid. If ammonia solution is added to a solution copper(II) ions, a dark blue/violet solution forms:

[Cu(H2O)6]2+(aq) + 4NH3(aq) → [Cu(NH3)4(H2O)2]2+

(aq) + 4H2O(l)

The larger the value of its stability constant, Kstab, the more stable a complex is.

Redox and redox titrations:

All redox reactions involve electron transfer. OILRIG – oxidation is loss, reduction is gain. A reducing agent donates electrons. An oxidising agent accepts electrons.

Potassium manganate(VII) titrations:

Potassium manganate(VII) solution is a strong oxidising agent and can be used in redox titrations to find the concentration of solutions containing iron(II) ions or hydrogen peroxide.

No indicator is required because the distinctive purple colour of potassium manganate(VII) disappears as it reacts.

A typical procedure is as followed: Use a pipette to transfer a known volume of the test solution into a conical flask.(containing

Fe2+ ions for example)

Acidify this solution with dilute sulfuric acid. Not hydrochloric acid as the chloride ions would be oxidised by the manganate(VII) ion.

Slowly add potassium manganate(VII) to the solution in the conical flask from a burette, swirly gently, until a pale pink colour persists.

Repeat the titration until you have two titres within 0.1cm3. During the titration, iron(II) ions are oxidised to iron(III) ions, and manganate(VII) ions are

reduced to manganese(II) ions. The ionic equation for the reaction is as followed:

5Fe2+(aq) + MnO4

-(aq) + 8H+

(aq) → 5Fe3+(aq) + Mn2+

(aq) + 4H2O(l)

Iodine-thiosulfate titrations:

Often used to find the concentration of solutions of oxidising agents. A known amount of the oxidising agent reacts with an excess of acidified potassium iodide

solution. The iodine produced is then titrated against a standard solution of sodium thiosulfate. Near the end point of the titration, a few drops of starch solution are added, giving an

intense blue/black colour. This disappears at the end point. The equation for the reaction is as followed:

2S2O32-

(aq) + I2(aq) → S4O62-

(aq) + 2I-(aq)

Electrode potentials:

Redox reactions can be considered to be two different reactions occurring simultaneously. One is a reduction, the other an oxidation. The equation for each is called a half-equation and involves ions and electrons.

Redox reactions:

Redox reactions involve electron transfer. They can be split up into two half reactions, one producing electrons and one accepting

electrons. For example, when zinc is added to copper(II) sulphate solution, a redox reaction takes

place. The blue colour of the solution becomes paler. Copper metal deposits on the zinc. The temperature rises because it is an exothermic reaction. Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s) This is an example of a displacement reaction. The sulphate ions play no part in the reaction and are spectator ions. On removing them from the overall equation we obtain the ionic equation: Zn(s) + Cu2+(aq) → Cu(s) + Zn2+(aq) Zn is grey as a solid and yet the Zn2+ is a colourless solution. Cu2+ is a blue solution and Cu(s) is an orange solid. What the reaction amounts to is Zn atoms transferring electrons to Cu2+ ions. The ionic

equation can be written as two half equations:

Zn(s) → Zn2+(aq) + 2e-

Cu2+(aq) + 2e- → Cu(s) Zinc provides the electrons which reduce Cu2+, therefore zinc is the reducing agent, while

Cu2+ is the oxidising agent. If copper is added to zinc sulphate solution no change is observed. However if copper is added to silver nitrate(V) solution the copper does react. A grey precipitate forms and the solution turns from colourless to blue. Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s) The half equations are: Cu(s) → Cu2+(aq) + 2e- oxidation 2Ag+(aq) + 2e- → 2Ag(s) reduction Individual half equations are reversible. The actual direction they take depends on what they are reacting with. As shown in the

example above.

Combining half equations:

Once we know the direction in which the half equation is going we can add the half equations together to get an equation for the overall reaction. For example if you add zinc to the silver ions, the atoms supply electrons.

Zn(s) → Zn2+(aq) + 2e-

Ag+(aq) + e- → Ag(s) To combine the two half equations we need to make sure the number of electrons is the

same in each half equation as every electron released by a zinc atom must be accepted by a silver ion.

This means we have to multiply the silver equation by 2 so there are 2e- in each equation. Now you can add the two half equations together to give the overall equation:

Zn(s) + 2Ag+(aq) → 2Ag(s) + Zn2+(aq) The 2e- disappears because they are on both sides of the equation. Overall:

STEP 1 – Write the half-equations for the oxidation and reduction reactions.STEP 2 – Make the number of electrons the same in each half-equation.STEP 3 – Add the two half-equations together.

Electrochemical cells:

Something must control the direction of electron transfer in a redox reaction. To find out more about redox reactions and what makes them go in a particular direction, we need to be able to study the half reactions.

We can arrange for the two half reactions to occur separately with electrons flowing through an external wire from one half reaction to the other.

A system like this is used in all batteries and ‘dry’ cells. In one part of the cell an oxidation reaction occurs. Electrons are produced and transferred through an external circuit to the other part of the

cell where a reduction reaction takes place accepting the electrons. The two parts are called half cells, and when combined make an electrochemical cell.

The energy given out, instead of heating the surroundings, becomes available as electrical energy.

Cells are labelled with a positive and negative terminals and a voltage which measures the potential difference between the two cells.

As current flows voltage can drop. To compare we can measure the potential difference between the terminals when no

current flows. To measure this we use a high resistance voltmeter so that almost zero current flows. The potential difference is a measure of how much each electrode is tending to release or

accept electrons.

Metal ion – metal half cells:

You can set up a half cell by using a strip of metal dipping into a solution of metal ions. Each half cell has its own electrode potential.

Take a zinc one. The Zn atoms in the strip form Zn2+ ions by releasing electrons.Zn → Zn2+ + 2e-

The electrons released make the Zn strip negatively charged relative to the solution, therefore there is a potential difference between the zinc strip and the solution.

The Zn2+ ions in the solution accept electrons forming Zn atoms.Zn2+ + 2e- → Zn

When Zn2+ ions are turning back to Zn as fast as they are being formed, equilibrium is set up. For a general metal:

M2+ + 2e- ⇄ M Position of equilibrium determines the size of the potential difference. The further to the

right it lies, the greater the tendency of the electrode to accept electrons and the more positive the electrode potential.

When we put two half cells together, the one with the more positive potential will become the positive terminal of the cell and the other one will become the negative terminal.

Making a cell from two half-cells:

A connection is need between the two solutions, but the solutions should not mix together. A strip of filter paper soaked is saturated potassium nitrate(V) solution can be used as

junction, or salt bridge between the half cells. Sometimes this is called an ion bridge because the current is carried by the movement of

ions not electrons. The potassium ions and nitrate(V) ions carry the current in the salt bridge so that there is

electrical contact between the solutions but no mixing. The circuit is completed by a metal wire connecting the copper and zinc strips. A high resistance voltmeter can be included in the circuit to measure the measure voltage,

Ecell, produced by the cell.

Standard electrode potentials:

The standard hydrogen half-cell is chosen as the reference electrode against which all other electrode potentials are measured.

Its electrode potential under standard conditions is defined as 0.00V. The half reaction occurring in the cell is:

H+(aq) + e- → 1

2H2(g)

A diagram of the half-cell is shown below.

The glass tube around the platinum electrode has holes in to allow bubbles of H2(g) to escape. The standard conditions are:

- Temperature of 298K- Pressure of 1atm- Concentration of 1.00moldm-3 (all ions)

The standard electrode potential, Eθ, of a half-cell is defined as the potential difference between it and a standard hydrogen half-cell.

By convention, the half reactions are always written as reduction processes. (oxidised species and electrons on left-hand side)

Other half-cell reactions:

To measure a standard electrode potential the half-cell being investigated is connected to a standard hydrogen half-cell.

For half-cells involving molecules and ions (I2/2I-) or ions (Fe3+/Fe2+), an inert electrode such as platinum is dipped into a solution containing all the ions and molecules involved in the half reaction.

The solution into which the electrode is dipped contains equal concentrations of the ions/molecules.

Finding Eθcell

For example, what is Eθcell when the Fe2+/Fe and Cu2+/Cu half-cells are connected?

STEP 1: look up the standard electrode potentials for the two half-cell reactions - Fe2+

(aq) + 2e- → Fe(s) Eθ = -0.44VCu2+

(aq) + 2e- → Cu(s) Eθ = +0.34V STEP 2: construct an electrode potential chart. The half-cell with the most positive

electrode potential is at the bottom of the chart –

STEP 3: Find the difference between the two Eθ values.Eθ

cell = Eθ[most positive electrode] - Eθ[most negative electrode]Eθ

cell = +0.34V – (-0.44V) = +0.78V

Predicting the direction of a reaction:

The key idea is to remember that the half-cell with the more negative electrode potential supplies electrons to the half-cell with the more positive electrode potential.

To predict the feasibility of the reaction between aqueous chlorine and potassium iodide solution, we calculate Eθ

cell. STEP 1: look up the half reactions and their standard electrode potentials –

I2(aq) + 2e- → 2I-(aq) Eθ = +0.54V

Cl2(g) + 2e- → 2Cl-(aq) Eθ = +1.36V

STEP 2: Identify which half reaction has the more negative electrode potential. Rewrite that react to show it supplying electrons. Half reaction 1 has the more negative Eθ so this half reaction is rewritten:2I-

(aq) → I2(aq) + 2e-

STEP 3: Balance the number of electrons, and then add the half reactions – 2I-

(aq) + Cl2(g) → I2(aq) + 2Cl-(aq)

Eθcell = +1.36V – (+0.54V) = +0.82V

Rusting:

An electrochemical process. Electrochemical cells are set up in the metal surface. Different areas act as sites of oxidation and reduction. The two half reactions involved in rusting are:

Fe2+(aq) + 2e- → Fe(s) Eθ = -0.44V

12

O2(g) + H2O(l) + 2e- → 2OH-(aq) Eθ = +0.44V

The reduction of oxygen to hydroxide ions occurs ay the more positive potential, and so electrons flow from the half-cell in which the iron is oxidised to iron(II) ions.

The concentration of dissolved oxygen in the water droplet determines which regions of the

metal surface are sites of oxidation or reduction. At the edges of the droplet, where the concentrations of dissolved oxygen is higher, oxygen

is reduced to hydroxide ions:12

O2(g) + H2O(l) + 2e- → 2OH-(aq) (Cathodic reaction)

The electrons needed to reduce the oxygen come from the oxidation of iron at the centre of the water droplet, where the concentration of dissolved oxygen is low. The Fe2+

(aq) pass into solution:Fe(s) → Fe2+

(aq) + 2e- (Anodic reaction) The electrons released flow in the metal surface to the edges of the droplet. This explains why corrosion is always greatest at the centre of a droplet of water or under a

layer of paint. These are the regions where the oxygen supply is limited. Pits are formed here where the iron has dissolved away. Rust forms in a series of secondary processes within the solution as Fe2+ and OH- ions diffuse

away from the metal surface. It does not from as a protective layer in contact with the iron surface:

Fe2+(aq) + 2OH-

(aq) → Fe(OH)2(s)

Fe(OH)2(s)

O2(aq)→

Fe2O3.xH2O(s)

Some ionic impurities such as NaCl from salt spray near the sea promote rusting by increasing the conductivity of the water.

Why recycle steel?

Saves resources and energy and helps to reduce waste. Steel is magnetic, and so can easily be separated from other waste.

Except for aerosols, all steel used for packaging is easily recycled. Scrap steel is important in the BOS process. It is added to the converter before the molten iron is poured in to help reduce ‘thermal

shock’.

Where does colour come from?

Why does an object appear coloured?

If visible light falls on a object some wavelengths are absorbed and others are reflected or transmitted.

What we see are the reflected or transmitted wavelengths. For example, copper(II) sulphate(VI) solution appears blue because it transmits blue light

and absorbs all colours other than blue. A colour wheel can help you work out which colour(s) are most easily absorbed by an object. A blue object absorbs most strongly in the orange region of the visible spectrum. Orange is

therefore called the complementary colour to blue.

Electronic transition:

When visible light falls on a coloured substance, the absorbed light is in the energy range that causes electronic transitions.

Electrons move to higher energy levels and the molecules become excited. Molecules do not remain excited for long, the electrons fall back to intermediate energy

levels. The energy is re-emitted in various forms, including vibrational energy. There are many different vibrational energy levels within each electronic energy level. An increase energy means there is an increase in vibrational energy.

Colorimetry:

An experimental technique used to find the concentration of a coloured solution. The amount of light absorbed by a solution, its absorbance, is proportional to the

concentration of the solution.