Chemistry: Developing Metals

    a Redox titrations:

      Test yourself
    • Redox titrations are used to find the amount of oxidising agent needed to react with a known concentration of reducing agent or vice-versa
    • An example is where manganate(VII) ions, MnO4-(aq), are used as the oxidising agent to find the concentration of a reducing agent. I'll use Fe2+(aq) in this example, as this is used in the PAG 12 practical at most schools (finding the iron content of an iron tablet)
    • The method is below:
      1) Add a known mass of impure Fe2+(aq) to a conical flask
      2) Add dilute sulfuric acid in excess (because the equation of the reaction contains H+ ions)
      3) Add MnO4-(aq) to the flask with the burette until the solution in the conical flask turns slightly pink and remains this colour
      4) Repeat until you have two readings within 0.10
    • The MnO4-(aq) ions in potassium manganate(VII) (KMnO4) are purple. They are reduced to Mn2+ ions (colourless) when they react with the iron and sulfuric acid. Therefore, once all of the reducing agent has reacted, the next drop of MnO4- will not react and will colour the solution a light shade of purple
    • By working out the equation for the reaction, you can calculate the concentration of Fe2+ ions. To find the equation, write balanced half-equations and then combine them for a full redox equation (see c)
    • The equation for this reaction is:
      MnO4-(aq) + 8H+(aq) + 5Fe2+(aq) → Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)

    b Coordination numbers:

      Test yourself
      It may make sense to read through i and j first before reading this - I only put this section here because it's the order in the specification

    • The coordination number is the number of coordinate bonds which form around a transition metal. It is usually either:
      - 4 (forming a tetrahedral shape, 109.5°, and occasionally a square planar shape, 90°)
      - 6 (forming an octahedral shape, 90°)

    c Equations of redox reactions:

      Test yourself
    • You need to be able to balance redox reactions. This is mostly detailed in Elements from the Sea
    • For some reactions, the method from ES will not work. In this case, try adding H+ and/or H2O
    • You can also add e- to a half-equation for oxidation/reduction. But they must cancel in the overall equation

    Combining and balancing example:

    • We'll use reaction between iron and manganate ions for this example
    • Fe2+ is being oxidised, so its half-equation must be Fe2+ → Fe3+ + e-
    • MnO4- is being reduced, so MnO4- → Mn2+
      - However, this is not balanced - neither the species nor the charge is conserved
      - First, balance the elements by adding water: MnO4- → Mn2+ + 4H2O
        And then hydrogen: 8H+ + MnO4- → Mn2+ + 4H2O
      - The total oxidation state on the left of the → is +7. On the right, it's +2
      - These can be balanced by adding electrons: 8H+ + 5e- + MnO4- → Mn2+ + 4H2O
    • You now need to combine the two balanced half-equations
    • However, if you did that, the right-hand side would have 1 electron and the left-hand side would have 5. They must be equal so that they cancel
    • Therefore, you can multiply the oxidation reaction by 5 to get 5Fe2+ → 5Fe3+ + 5e-
    • Now combine: 5Fe2+ + MnO4- + 8H+ + 5e- → 5Fe3+ + Mn2+ + 4H2O + 5e-
    • The electrons cancel to give 5Fe2+ + MnO4- + 8H+→ 5Fe3+ + Mn2+ + 4H2O

    d Electrochemical cells:

      Test yourself
    • Electrochemical cells are usually made from two metal electrodes, each dipped in a solution of their own ions
    • Alternatively, if the two ions are different oxidation states of the same element, then inert electrodes can be used instead
    • If none of the reactants or products are solids, inert electrodes can also be used
    • The electrodes are connected by wires to the rest of the circuit, and the circuit is completed with a salt bridge (usually filter paper soaked in a salt)
    • There is an oxidation reaction at the anode and reduction happens at the cathode - making the reaction a redox reaction
    • Always draw the anode on the left and the cathode on the right

    Setting up an electrochemical cell with two different metals:

    • Take a strip of each metal and clean them with sandpaper
    • Then clean them with propanone to remove oil/grease
    • Partially dip each strip into a beaker containing its ions (e.g. if the strip was zinc, you could put it in ZnSO4(aq))
    • Soak some filter paper in a salt solution (e.g. KCl(aq)) and dip it into both beakers
    • Use crocodile clips and wires to connect each electrode to the circuit or a voltmeter
    • It doesn't really matter which way you connect the voltmeter since most digital ones will simply show a - sign before the reading if it's the wrong way, which can be ignored to get the true potential difference

    e How electrochemical cells work:

      Test yourself
    • The electrons flow from the most reactive metal to the least reactive metal
    • This is because more reactive metals will lose their electrons more easily, so they're oxidised more easily
    • The reduced metal's mass will slowly increase as the ions in the solution gain electrons to form the metal
    • The half-equations are usually reversible, but they'll only occur one way. If you're writing them with a ⇌ symbol, you should write the reduction reaction going in the forward direction by convention (i.e. the electrons are on the left-hand side)
    • The salt bridge allows ions to transfer, completing the circuit
    • This constant flow of electrons creates a difference in the two cell potentials, creating a potential difference
    • The cell potential is also called the cell emf (electromotive force, a measure of a power supply's electrical potential), and written as Eo or
    • A voltmeter (or multimeter) can be connected to the electrodes to measure the potential difference of the cells in volts (V)

    f Standard electrode potential:

      Test yourself
    • Every type of half-cell has a standard electrode potential (o). This is the potential measured under standard conditions when paired with a standard hydrogen electrode which has a potential of 0 V
    • The reaction at the hydrogen electrode is 2H+(aq) + 2e- ⇌ H2(g)
    • The half-reaction with the most positive standard electrode potential goes forwards and vice-versa for the other half-reaction
    • You can calculate the cell potential (o) from the two standard electrode potentials by subtracting the most negative one from the most positive one
    • More reactive metals are trying to form a positive ion by losing electrons. So they will usually have a more negative standard electrode potential
    • More reactive non-metals are trying to form a negative ion by gaining electrons. So they will usually have a more positive standard electrode potential

    Feasibility prediction:

    • You can predict whether a redox reaction between a solid and an aqueous metal is feasible with their electrode potentials. First, determine which direction each one goes in. Then combine them. If the reaction looks possible, it's most likely feasible
    • However, sometimes the rate of reaction is too low or the activation enthalpy is too high, making your prediction incorrect
    • Also, the electrode potentials may be slightly different if not under standard conditions
    • For example, consider the half-equation Zn(s) ⇌ Zn2+(aq) + 2e- (o = -0.76 V). If the concentration of Zn2+ is increased (rather than being at 1 as described by standard conditions), the equilibrium will move left. This increases the difficulty of electron loss, so the electrode potential will become less negative

    Rusting:

    • The first equation for rusting has the half-equations:
      - Fe2+(aq) + 2e- ⇌ Fe(s) (negative o)
      - 2H2O(l) + O2(g) + 4e- ⇌ 4OH-(aq) (positive o)
    • The two products from this react (the first half-equation will go backwards since it's the most negative): Fe2+(aq) + 2OH-(aq) → Fe(OH)2(s)
    • The iron(II) hydroxide (Fe(OH)2) is then oxidised again by oxygen: 2H2O(l) + O2(g) + 4Fe(OH)2(s) → 4Fe(OH)3(s)
    • This iron(III) hydroxide is then hydrated to rust (Fe2O3.xH2O)
    • The rate of rusting is lower in alkaline conditions because the equilibrium in the second half-equation in the first step (2H2O(l) + O2(g) + 4e- ⇌ 4OH-(aq)) will shift left with additional OH-. This means that more electrons will be produced in that reaction. So the equilibrium in the other half-equation (Fe2+(aq) + 2e- ⇌ Fe(s)) must shift right to counteract this. This reduces the number of Fe2+ ions available to form iron(II) hydroxide

    Prevention:

    • Iron can be coated with paint, a polymer, oil or grease to prevent water and oxygen from coming into contact with it
    • Alternatively, a more reactive metal than iron can be used as a coating - this is known as a sacrificial coating, or galvanisation
    • For example, the standard electrode potential of Zn/Zn2+ is more negative than Fe/Fe2+. Therefore, zinc will be oxidised instead of iron

    g, h Transition metals:

      Test yourself
    • A transition metal is a d-block element which can form at least one stable ion with at least one incompletely filled d-orbital
    • Since a d sub-shell contains up to 10 electrons, a transition metal ion must have between 1 and 9
    • Therefore, all d-block elements are transition metals, except scandium (Sc3+ has an empty d sub-shell) and zinc (Zn2+ has a full d sub-shell), and these metals can only form one ion each
    • The 4s orbitals usually fill before 3d, except from in chromium and copper, which are more stable with only one electron in 4s
    • When forming a positive ion, the s electrons are removed before the d electrons
    • Many transition metals can form more than one ion, each with a different oxidation state
    • Transition metals can exist in many oxidation states because the energy levels of 4s and 3d are very close to each other. Therefore, different numbers of electrons can be lost with similar amounts of energy
    • Different oxidation states of a metal usually have different colours in solution, for example (you need to memorise these four):
      - Fe2+ is green
      - Fe3+ is yellow/orange/brown
      - Cu+ is unstable in solution
      - Cu2+ is blue

    i Ligands:

      Test yourself
    • A coordinate bond is another name for a dative bond (a covalent bond where both electrons are from the same atom)
    • A ligand is an atom/ion/molecule which donates an electron pair to a transition metal ion, forming a coordinate bond
    • Because they are electron pair donors, all ligands have a lone pair of electrons
    • A complex ion is a metal atom/ion surrounded by coordinately bonded ligands

    Ligand substitution:

    • Often, one ligand can replace another, usually changing the colour of the metal
    • There's sometimes a simple substitution if the coordination numbers are the same and the shape is the same
    • The coordination number will usually be the same if the two ligands are of similar size (e.g. H2O and NH3)
    • A ligand substitution equation can be written like this:
      [Cu(H2O)6]2+ + 4Cl- ⇌ [CuCl4]2- + 6H2O
      Here, notice that the coordination number is 6 on the left and 4 on the right

    Ligands you need to know:

    • [Fe(H2O)6]2+, [Fe(H2O)6]3+ and [Cu(H2O)6]2+ all have 90° angles, e.g.
    • [CuCl4]2- and [Cu(NH3)4]2+ have 109.5° bond angles with a tetrahedral shape
    • The charge outside of the brackets is calculated by summing the charges of the ligands and the central metal ion charge. For example, [CuCl4]2- has a -2 charge because
      -1 [Cl- charge] × 4 [coordination number] + +2 [Cu2+ charge] = -2 [the charge of the complex]

    j Complex ions:

      Test yourself
    • Monodentate ligands have one lone pair (e.g. NH3)
    • Bidentate ligands have two lone pairs, so they can form two coordinate bonds with the transition metal ion
    • An example of a bidentate ligand which you need to know is ethanedioate which is shown below. Two lone pairs in the single-bonded oxygens will interact with the metal ion. So the coordination number will be 2 × the number of ethanedioate ligands in the complex
    • Polydentate ligands have three or more lone pairs and form three or more coordinate bonds with transition metal ions

    k Precipitates:

      Test yourself
    • If aqueous transition metal ions (written as [M(H2O)6]+(aq) or more simply M+(aq)) are mixed with aqueous sodium hydroxide or aqueous ammonia, a coloured hydroxide precipitate is formed
    • You need to know the following equations and colours:
      - Cu2+(aq) + 2OH-(aq) → Cu(OH)2(s) (pale blue solution → blue precipitate)

      - Fe2+(aq) + 2OH-(aq) → Fe(OH)2(s) (pale green solution → green precipitate)
        The precipitate colours darken on standing

      - Fe3+(aq) + 3OH-(aq) → Fe(OH)3(s) (yellow solution → orange precipitate)
        The precipitate colours darken on standing

    • The reactions are the same when dilute aqueous ammonia solution is added. This is because ammonia slightly dissociates in water to give sufficient OH- ions: NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH–(aq)
    • In excess NH3, the product Cu(OH)2(s) will react again to form [Cu(NH3)4(H2O)2]2+(aq) (a dark blue ammonia complex). This does not happen to iron(II) hydroxide or iron(III) hydroxide

    l Catalysts:

      Test yourself
    • Heterogeneous catalysts have different states to the reactants. Transition metal catalysts are usually in a solid phase and the reactants are liquids or solids
      - Weak bonds are formed between the reactants and the 3d and 4s electrons on the catalyst
      - These bonds break after the product is formed
      - Transition metals are ideal because the bonds are strong enough to attract reactants, but weak enough for products to desorb
    • Homogeneous catalysts have the same state as the reactants. Usually, this is aqueous
      - The catalyst usually forms an intermediate compound with one or more reactants
      - Later this will break down and the catalyst will be released again
      - Transition metal ions can move between different oxidation states easily
         - Because this enables them to be both reducing agents and oxidising agents, they are often good homogeneous catalysts
         - This is especially true for redox reactions

    m Transition metal complex colour:

      Test yourself
    • With the exception of those with a full or empty 3d sub-shell, most transition metal ions in solution (i.e. complexes) are coloured
    • 3d orbitals usually all have the same energy in transition metal ions
    • However, ligands can cause some orbitals to gain energy by splitting the sub-shell into two energy levels
    • As many electrons as possible will usually occupy the lower (ground) energy level. But when light is absorbed, some electrons move to the upper level. The amount of visible light absorbed depends on:
      - the metal
      - its oxidation state
      - the ligand
      - the coordination number
      - the size of the energy gap, Δ E
    • Since white light contains many frequencies, only some will be absorbed. The rest are transmitted or reflected, which makes the metal appear the complementary colour to the one that is absorbed

    n Colorimetry:

      Test yourself
    • Colorimeters measure the absorbance of a solution
    • Absorbance of a solution increases with concentration
    • A cuvette is a clear container which doesn't absorb light, used to hold the samples
    • Below is a method for using a colorimeter:
      1) Pick a filter with a similar colour to the one that's absorbed by the solution (i.e. its complimentary colour)
      2) Put a sample of the solvent (usually water) into the colorimeter and set it to zero
      3) Put a sample in a clean cuvette. The absorbance is measured relative to the blank sample from step 2
    • Note: the complementary colours don't need to be memorised for the exams. Writing 'use a filter of the complementary colour' will gain you the mark

    Finding the concentration of a transition metal solution sample:

    • If you measure the absorbance of several solutions of the metal with a known concentration, you can plot these on a straight line graph (calibration graph)
    • Now you can measure the absorbance of the sample and read its concentration off the graph using the trendline

    Visible spectroscopy:

    • This method uses a visible spectrophotometer, which passes a beam of light with a specified wavelength through a sample, measuring how much is absorbed
    • The data can be used to plot a concentration calibration curve, as with colorimetry above