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Many chemical reactions involve a movement of electrons from one reacting species to another. An example of this is the formation of aqueous metal ions when a metal reacts with an acid or water or another metal's salt solution. If a special electrochemical cell is made up which includes a voltmeter, then the movement of electrons can be measured as a potential difference,
The salt bridge here is simply a piece of paper soaked in a concentrated solution of potassium nitrate. This allows the charge, in the form of ions, in the circuit to move from one beaker to another, just as the wires allow the electrons to flow around the top half of the circuit.
The first example of this type of cell was the Daniell cell involving copper in copper(II) sulphate(aq) and zinc in zinc sulphate(aq). Because zinc is a more reactive metal than copper it prefers to form ions in solution, therefore the two reactions occurring are,
Zn(s) → Zn2+(aq) + 2e- - oxidation
Cu2+(aq) + 2e- → Cu(s) - reduction
The oxidation reaction gives out electrons and therefore occurs at the negative electrode (cathode) and the reduction occurs at the positive electrode (anode). These reactions are called half-cells or half-equations.
Electrons flow from the negative electrode to the positive electrode, whilst ions also flow from the negative electrode via the salt bridge to the positive electrode,
The cell reactions occurring may be written in a short hand form like this,
Reading from left to right this diagram states zinc metal forms zinc ions(aq) whilst copper ions(aq) form copper metal.
In order to calculate the potential for an electrochemical cell, without having to run all the thousand's of possible combinations, some sort of standard electrode is needed to provide a reference point.
Originally a calomel electrode involving saturated potassium chloride(aq), mercury(I) chloride(s) (Hg2Cl2) and mercury was used. Now the standard hydrogen electrode is used. This involves hydrogen gas being turned into hydrogen ions,
H2(g) ⇌ 2H+(aq) + 2e-
A platinum-black electrode is needed to provide a surface on which the hydrogen gas can be in contact with the hydrogen ions(aq),
The hydrogen electrode is always placed as the negative electrode of a cell and is written as,
The potential of this cell is given the value of 0 V. The potentials for other cells are then tabulated relative to this value. These are the standard electrode potentials, Eo, e.g.
Pt(s)|H2(g),2H+(aq)┆Zn2+(aq)|Zn(s) Eo=-0.76 V
Pt(s)|H2(g),2H+(aq)┆Cu2+(aq)|Cu(s) Eo=+0.34 V
A positive value for the Eo value means a favourable reaction is occurring and therefore the metal is less reactive than hydrogen. A negative value for Eo means an unfavourable reaction and therefore the metal is more reactive than the standard hydrogen electrode. This ties in relatively well with the metal reactivity series learnt at GCSE level; however, it should be noted that lithium is placed at the top of the electrochemical reactivity series but below potassium and sodium in the Reactivity Series in GCSE.
When the half-cell to be measured involves a metal in its metal salt solution the electrode wire may be attached to the metal strip with a crocodile clip.
When the half-cell involves an element's aqueous ions in two different oxidation states the platinum electrode has to be used again. The platinum surface provides the means by which the two different ions may interact with one another.
e.g. Fe3+(aq) + e- ⇌ Fe2+(aq)
The Fe2+(aq) and Fe3+(aq) ions may react together on the platinum surface in a similar fashion to the process in a standard hydrogen electrode. The half-cell would be written as,
N.B.: Both aqueous ions have to be 1 mol dm-3 for a standard cell potential to be measured.
The standard cell potential, Eocell, for the Daniell cell may be found by combining the standard electrode potentials, Eo, for the copper and zinc half-cells.
|Eocell||= EoR(ight) - EoL(eft)|
|= Eo(Cu2+|Cu) - Eo(Zn2+|Zn)|
|= (+0.34 V) - (-0.76 V)|
|= +1.10 V|
The value tabulated for the zinc electrode (Zn2+(aq)+2e- ⇌ Zn(s)) must be reversed as the actual zinc reaction occurring here is the opposite of this. Hence the negative sign in the mathematical equation above.
Both half-cell reactions are essentially equilibria and as such a change in the concentration of ions in solution will effect the Eo value.
e.g. Zn2+(aq) + 2e- ⇌ Zn(s)
If the concentration of zinc ions(aq) is increased then the equilibrium will move to the right to compensate for this increase. This will mean more electrons are needed which will mean the Eo value will increase or become more positive.
A decrease in the concentration leads to the opposite effect i.e. the value of Eo will decrease.
The concentration used for all electrochemical cell solutions to obtain a standard electrode potential is 1 mol dm-3.
A positive value for an Eocell means a favourable reaction will occur. However, other factors, such as a high activation energy barrier may prevent any reaction at all from occurring or may make it very slow.
A negative value for Eocell means that the reaction is happening in the opposite direction to the one written. For example, for the reaction between bromine and chloride ions -
Br2(aq) + 2Cl-(aq) → 2Br-(aq) + Cl2(aq)
Eocell = (+1.09) - (+1.36) = -0.27 V
The negative value for Eocell indicates the above reaction does not occur. The reaction that can occur is -
Cl2(aq) + 2Br-(aq) → 2Cl-(aq) + Br2(aq)
Eocell = (+1.36) - (+1.09) = +0.27 V
There is a quick method of determining whether a reaction will occur between two species called the anti-clockwise rule.
The half-cell equations for the species to be considered must be drawn out and placed in order from most negative potential at the top to most positive electrode at the bottom,
The reactions that will occur, i.e. those that will have positive Ecell values, are those that follow anti-clockwise paths.
So in the diagram above ,reactions will occur between chlorine and bromide ions and chlorine and iodide ions because these reactions follow an anti-clockwise path (green arrow). A reaction will also occur between bromine and iodide ions (red arrow). However, no reactions will occur between chloride ions and bromine or iodine and between iodine and chloride or bromide ions as these would be clockwise paths.
|Reduction half-equation||Reduction electrode notation||Eo (V)|
|Li+(aq) + e- ⇌ Li(s)||Li+(aq)|Li(s)||-3.03|
|Rb+(aq) + e- ⇌ Rb(s)||Rb+(aq)|Rb(s)||-2.93|
|K+(aq) + e- ⇌ K(s)||K+(aq)|K(s)||-2.92|
|Ca2+(aq) + 2e- ⇌ Ca(s)||Ca2+(aq)|Ca(s)||-2.87|
|Na+(aq) + e- ⇌ Na(s)||Na+(aq)|Na(s)||-2.71|
|Mg2+(aq) + 2e- ⇌ Mg(s)||Mg2+(aq)|Mg(s)||-2.37|
|Al3+(aq) + 3e- ⇌ Al(s)||Al3+(aq)|Al(s)||-1.66|
|Mn2+(aq) + 2e- ⇌ Mn(s)||Mn2+(aq)|Mn(s)||-1.19|
|V2+(aq) + 2e- ⇌ V(s)||V2+(aq)|V(s)||-1.18|
|Cd(OH)2(aq) + 2e- ⇌ Cd(s) + 2-OH(aq)||Cd(OH)2(aq),Cd(s) + 2-OH(aq)|Pt||-0.88|
|2H2O(l) + 2e- ⇌ 2-OH(aq) + H2(g)||2H2O(l),2-OH(aq) + H2(g)|Pt||-0.83|
|Zn2+(aq) + 2e- ⇌ Zn(s)||Zn2+(aq)|Zn(s)||-0.76|
|Cr3+(aq) + 3e- ⇌ Cr(s)||Cr3+(aq)|Cr(s)||-0.74|
|Fe2+(aq) + 2e- ⇌ Fe(s)||Fe2+(aq)|Fe(s)||-0.44|
|Cr3+(aq) + e- ⇌ Cr2+(aq)||Cr3+(aq),Cr2+(aq)|Pt||-0.41|
|Ti3+(aq) + e- ⇌ Ti2+(aq)||Ti3+(aq),Ti2+(aq)|Pt||-0.37|
|V3+(aq) + e- ⇌ V2+(aq)||V3+(aq),V2+(aq)|Pt||-0.26|
|Ni2+(aq) + 2e- ⇌ Ni(s)||Ni2+(aq)|Ni(s)||-0.25|
|Sn2+(aq) + 2e- ⇌ Sn(s)||Sn2+(aq)|Sn(s)||-0.14|
|Pb2+(aq) + 2e- ⇌ Pb(s)||Pb2+(aq)|Pb(s)||-0.13|
|2H+(aq) + 2e- ⇌ H2(g)||2H+(aq),H2(g)|Pt||0|
|S4O62-(aq) + 2e- ⇌ 2S2O32-(aq)||S4O62-(aq),2S2O32-(aq)|Pt||+0.09|
|Cu2+(aq) + e- ⇌ Cu+(aq)||Cu2+(aq),Cu+(aq)|Pt||+0.15|
|Hg2Cl2(s) + 2e- ⇌ 2Hg(s) + 2Cl-(aq)||Hg2Cl2(s),2Hg(s) + 2Cl-(aq)|Pt||+0.27|
|Cu2+(aq) + 2e- ⇌ Cu(s)||Cu2+(aq)|Cu(s)||+0.34|
|VO2+(aq) + 2H+(aq) + e- ⇌ V3+(aq) + H2O(l)||VO2+(aq) + 2H+,V3+(aq) + H2O(l)|Pt||+0.34|
|O2(g) + 2H2O(l) + 4e- ⇌ 4-OH(aq)||O2(g) + 2H2O(l),4-OH(aq)|Pt||+0.40|
|Cu+(aq) + e- ⇌ Cu(s)||Cu+(aq)|Cu(s)||+0.52|
|NiO(OH)(aq) + H2O(l) + e- ⇌ Ni(OH)2(aq) + -OH(aq)||NiO(OH)(aq) + H2O(l),Ni(OH)2(aq) + -OH(aq)|Pt||+0.52|
|I2(aq) + 2e- ⇌ 2I-(aq)||I2(aq),2I-(aq)|Pt||+0.54|
|MnO42-(aq) + 2H2O(l) + 2e- ⇌ MnO2(s) + 4-OH(aq)||MnO42-(aq) + 2H2O(l),MnO2(s) + 4-OH(aq)|Pt||+0.59|
|2NH4+(aq) + 2e- ⇌ 2NH3(g) + H2(g)||2NH4+(aq),2NH3(g) + H2(g)|Pt||+0.70|
|Fe3+(aq) + e- ⇌ Fe2+(aq)||Fe3+(aq),Fe2+(aq)|Pt||+0.77|
|Ag+(aq) + e- ⇌ Ag(s)||Ag+(aq)|Ag(s)||+0.80|
|VO2+(aq) + 2H+(aq) + e- ⇌ VO2+(aq) + H2O(l)||VO2+(aq) + 2H+(aq),VO2+(aq) + H2O(l)|Pt||+1.00|
|Br2(l) + 2e- ⇌ 2Br-(aq)||Br2(l),2Br-(aq)|Pt||+1.07|
|MnO2(s) + 4H+(aq) + 2e- ⇌ Mn2+(aq) + 2H2O(l)||MnO2(s) + 4H+(aq),Mn2+(aq) + 2H2O(l)|Pt||+1.23|
|O2(g) + 4H+(aq) + 4e- ⇌ 2H2O(l)||O2(g) + 4H+(aq),2H2O(l)|Pt||+1.23|
|Cr2O72-(aq) + 14H+(aq) + 6e- ⇌ 2Cr3+(aq) + 7H2O(l)||Cr2O72-(aq) + 14H+(aq),2Cr3+(aq) + 7H2O(l)|Pt||+1.33|
|Cl2(aq) + 2e- ⇌ 2Cl-(aq)||Cl2(aq),2Cl-(aq)|Pt||+1.36|
|Mn3+(aq) + e- ⇌ Mn2+(aq)||Mn3+(aq),Mn2+(aq)|Pt||+1.49|
|MnO4-(aq) + 8H+(aq) + 5e- ⇌ Mn2+(aq) + 4H2O(l)||MnO4-(aq) + 8H+(aq),Mn2+(aq) + 4H2O(l)|Pt||+1.51|
|PbO2(s) + SO42-(aq) + 4H+(aq) + 2e- ⇌ PbSO4(s) + 2H2O(l)||PbO2(s) + SO42-(aq) + 4H+(aq),PbSO4(s) + 2H2O(l)|Pt||+1.69|
|MnO4-(aq) + 4H+(aq) + 3e- ⇌ MnO2(s) + 2H2O(l)||MnO4-(aq) + 4H+(aq),MnO2(s) + 2H2O(l)|Pt||+1.70|
|Co3+(aq) + e- ⇌ Co2+(aq)||Co3+(aq),Co2+(aq)|Pt||+1.81|
|F2(aq) + 2e- ⇌ 2F-(aq)||F2(aq),2F-(aq)|Pt||+2.87|
The co-ordination number for a metal ion in a complex is the number of ligand lone pairs that are attached to it.
For A level only co-ordination numbers of 4 and 6 are required, though co-ordination numbers can range up to 12 and beyond.
There are two shapes available for a metal ion connected to four lone pairs.
(i) tetrahedral, e.g. the tetrachloridecopper(II) ion, [CuCl4]2-,
the internal bond angles are all 109.5o.
(ii) square planar, e.g. the tetrachloridegold(III) ion, [AuCl4]-,
the internal bond angles are all 90o.
There is only one shape available for a metal ion surrounded by six lone pairs - octahedral.
e.g. the hexaaquacopper(II) ion, [Cu(H2O)6]2+,
In an octahedral complex there are two different positions that ligands can occupy relative to one another. In the two structures below it can be seen that the two ammonia ligands could be 90o or 180o relative to one another,
This gives a cis- isomer and a trans- isomer.
With certain ligands an octahedral complex ion can exhibit optical isomerism. This has nothing to do with the idea of a central carbon atom surrounded by four different groups, as in organic chemistry (see the stereoisomerism section in the nitrogen page).
It does have to do with the idea of two molecules that are not superimposable onto each other. Shown below are two different complex ions formed when 1,2-diaminoethane is used as the ligand with copper(II) ions,
The left hand molecule cannot be rotated so that it matches exactly the right hand molecule, so the complex exhibits stereoisomerism.
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There are five d-orbitals. Their shapes are -
In an octahedral complex the ligands are arranged around a central metal ion along the x, y and z axes as shown below -
As can be seen the dx2-y2 and dz2 orbitals overlap with the positions of the six ligands. This raises their energy relative to the other three d-orbitals
The amount of splitting (ΔE) produced in a transition metal ions d-orbitals is affected by the ligand used in the complex. e.g.
As can be seen from the equation above substitution of four water molecules by ammonia molecules in the complex causes the wavelengths of light absorbed to change, therefore changing the colour.
The wavelength of light absorbed by a transition metal ion can be recorded by taking a visible spectrum of the complex.
e.g. hexaaquacopper(II) ion -
As can be seen from the above spectrum the copper(II) complex absorbs wavelengths of light in the red/orange/yellow parts of the spectrum. Blue wavelengths are emitted therefore the complex appears blue.
Q/ From the following visible spectrum deduce the colour of the transition metal complex.
A/ Place the mouse pointer over the image to show a coloured line indicating the colour of the complex.
Titanium dioxide is used as the white pigment in paints (and polos!)
Monastral blue/other porphyrin complexes are used as pigments for paints.back to top
|Oxidation State = +2 (d3)||Oxidation State = +3 (d2)||Oxidation State = +4 (d1)||Oxidation State = +5 (d0)|
|VCl2, green||VCl3, pink||VO2+(aq), blue||VO3-(aq), yellow|
|V2O2, light grey||VBr3, dark grey||VO2, dark blue||VO2+(aq)|
|V2S2, black||VF3, green||VCl4, red||VF5, yellow|
|V2O3, black||V2O4, blue||V2O5, orange|
|V2S3, black||VOCl2, green||V2S5, black|
|VOCl, brown||VOBr2, yellow||VOCl3, yellow|
|VOBr, violet||VOBr3, red|
|[V(H2O)6]2+(aq), lavender||[V(H2O)6]3+(aq), green|
The different oxidation states of vanadium can be made by reacting solutions of metavanadate ions (VO3-) with different reducing agents.
|V3+(aq) + e-⇌ V2+(aq)||-0.26|
|VO2+(aq) + 2H+(aq) + e- ⇌ V3+(aq) + H2O(l)||+0.34|
|VO2+(aq) + 2H+(aq) + e- ⇌ VO2+(aq) + H2O(l)||+1.00|
|Zn2+(aq) + 2e-⇌Zn(s)||-0.76|
|SO42-(aq) + 4H+(aq) + 2e-⇌SO2(aq) + 2H2O(l)||+0.17|
As can be seen from the above data, reacting vanadium(V) ions with sulphur dioxide will only result in green vanadium(III) ions being produced (or more probably blue vanadium(IV) ions as the Ecell value for that reaction is larger). However, if the vanadium(V) ions are boiled with a mixture of zinc and dilute sulphuric acid the lavender vanadium(II) ions can be produced -
2VO3-(aq) + 12H+(aq) + 3Zn(s) → 3Zn2+(aq) + 6H2O(l) + 2V2+(aq)
The lower oxidation states of vanadium are unstable and easily oxidised. For example, vanadium(III) ions are oxidised in air to vanadium(IV) and vanadium(II) ions react with water to liberate hydrogen gas and give vanadium(IV) ions -
V2+(aq) + H2O(l) → VO2+(aq) + H2(g)
These low oxidation state ions can only be stabilised by keeping a reducing agent with them (e.g. zinc).
Vanadium pentoxide (V2O5) is used as the catalyst in the Contact Process.
|Oxidation State = +3 (d3)||Oxidation State = +6 (d0)|
|CrBr3, olive green||CrO3, red|
|CrCl3, pink||CrO42-(aq), yellow|
|CrF3, green||Cr2O72-(aq), orange|
|Cr2O3, dark green|
The least stable oxidation state for chromium is +6, which is a powerful oxidising agent, according to the half-equation below -
|Cr2O72-(aq) + 14H+(aq) + 6e- ⇌ 2Cr3+(aq) + 7H2O(l)||+1.33|
The stability of the two chromium(VI) ions - chromate(VI) (CrO42-) and dichromate(VI) (Cr2O72-) - is dependent on the pH of their solutions. The two ions exist in the following equilibrium -
2CrO42-(aq) + 2H+(aq) ⇌ Cr2O72-(aq) + H2O(l)
Therefore, at high pH the equilibrium lies to the left hand side and the yellow chromate(VI) ion dominates and at low pH the equilibrium lies to the right hand side and the orange dichromate(VI) ion dominates.
Chromium is used in stainless steel and in hardening steel.
|Oxidation State = +2 (d7)||Oxidation State = +3 (d6)|
|CoBr2, green||CoCl3, red|
|CoBr2.6H2O, red||Co(OH)3, black|
|CoCO3, red||Co2O3, black|
|CoCl2, blue||CoF3, brown|
|CoCl2.6H2O, pink||Co2(SO4)3, blue|
|Co(OH)2, blue||[Co(NH3)6]3+, yellow/orange|
|CoO, olive green|
The least stable oxidation state for cobalt is +3, except in the presence of strong complexing ligands.
|Co3+ + e- ⇌ Co2+||+1.81|
As can be seen from the above half-equation cobalt(III) ions are powerful oxidising agents, being reduced to cobalt(II) ions.
The colour of cobalt complex ions can be affected (a) by changing the ligand used
e.g. [Co(H2O)6]2+(aq) + 4Cl-(aq) ⇌ [CoCl4]2-(aq) + 6H2O(l)
or (b) by heating the complexes
e.g. [Co(NH3)6]2+(aq) ⇌ [Co(NH3)6]3+(aq) + e-
Strong complexing agents, such as ammonia (or nitrile ions) allow the cobalt(II) ions to be readily oxidised to cobalt(III).
|Oxidation State = +1 (d10)||Oxidation State = +2 (d9)|
|Cu2CO3, yellow||CuSO4.5H2O, blue|
|CuOH, yellow||CuCl2.2H2O, green|
|Cu2O, brick red||CuCO3, green|
|CuBr, white||CuO, black|
|CuCl, white||Cu(NO3)2.6H2O, blue|
|CuI, white||CuF2.2H2O, blue|
The least stable oxidation state for copper is +1. As can be seen from the electrode potentials below copper(I) ions readily undergo disproportionation in aqueous solutions forming copper(II) ions and copper metal (overall Ecell = +0.37 V).
|Cu2+ + e- ⇌ Cu+||+0.15|
|Cu+ + e- ⇌ Cu||+0.52|
|2Cu+⇌ Cu2+ + Cu||+0.37|
This disproportionation is also seen in the reaction of copper(I) oxide with sulphuric acid:
Cu2O(s) + H2SO4(aq) → CuSO4(aq) + Cu(s) + H2O(l)
Copper(I) ions can be stablised by either (i) adding anions, e.g. I-, iodide ions, that precipitate out an insoluble salt:
2Cu2+(aq) + 4I-(aq) → 2CuI(s) + I2(aq)
or (ii) by adding a ligand that forms a more stable complex with copper(I) ions than copper(II) ions, e.g. ammonia, NH3.
The mass of copper present in a sample of brass can be found by a multi stage process.
Stage 1 - A known mass of the sample of brass is dissolved in concentrated sulphuric acid. This changes the copper into copper(II) ions according to the equation:
Cu(s) + 2H2SO4(l) → CuSO4(aq) + 2H2O(l) + SO2(g)
Stage 2 - This solution of copper(II) ions is then reacted with an excess of iodide(aq) ions. This produces copper(I) iodide and iodine according to the equation:
2Cu2+(aq) + 4I-(aq) → 2CuI(s) + I2(aq)
Stage 3 - The iodine produced in stage 2 is then titrated with a standard solution of thiosulphate ions. Once the initial yellow colour has disappeared starch is added and the end point is reached when the new blue-black colour produced has disappeared.
I2(aq) + 2S2O32-(aq) → 2I-(aq) + S4O62-(aq)
From the equations above, the mass of copper present in the original sample of brass =
63.5x½xthe amount thiosulphate used in the titration
The percentage of copper in the sample of brass =
(mass of copper calculated/mass of original sample)x100
Copper is used to make electrical wiring and in various alloys such as brass (with zinc), bronze (with tin), duralumin (with aluminium and magnesium) and other coinage metal alloys.
Finely divided copper is also used in the industrial oxidation of methanol to methanal.back to top
written by Dr Richard Clarkson : © Saturday, 1 November 1997
Updated : Monday, 28th May, 2012
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