Ligands and redox

Question 1

what is a ligand substitution reaction

Answer 1

this is a reaction in which one ligand in a complex ion is replaced by another ligand
- usually water molecules in an aqueous solution of a complex ion are replaced by another ligand

Question 2

the reaction of aqueous copper(II) ions and ammonia

Answer 2

• in an aqueous solution of copper(II) ion contains [Cu(H2O)6]2+ which have a pale blue colour
• when ammonia is added in excess it turns from pale blue to a dark blue solution
• [Cu(H20)6]2+ + 4NH3 = [Cu(NH3)4(H2O)2]2+ + 4H2O
• reversible reaction
• 4 water ligands have been replaced by 4 ammonia ligands
• octahedral shape as formed
• copper oxygen bonds are longer than the copper nitrogen bonds so the product is described as having a distorted octahedral shape

Question 3

what are the two changes that happen when you add ammonia to cooper (II) ions

Answer 3

• on addition of a small amount of ammonia a pale blue precipitate of copper(II) hydroxide Cu(OH)2 forms because the ammonia acts as an alkali
• when ammonia is put into water it reacts to form ammonium hydroxide which can partially ionise releasing the hydroxide ions

Question 4

the reaction of copper(II) ions and hydrochloric acid

Answer 4

• when concentrated HCL is added to aqueous solution containing copper (II) ions the pale blue solution forms a green solution before turning yellow, this react is in equilibrium and can be reversed by adding water to the yellow solution to return it to the original blue colour
• [Cu(H2O)]2+ + 4Cl- = [CuCl4]2- + 6H2O
• the complex with the chloride ions only has 4 ligands, this happens because the chloride ligands are larger so fewer chloride ligands can fit around the central metal ion - has a tetrahedral shape

Question 5

reaction of chromium(III) with ammonia

Answer 5

an aqueous solution of chromium(III) contains [Cr(H2O)6]3+, complex ions and has a green grey or dark green colour, when a small amount of ammonia is added hydrogen ions are pulled off the water ligands and this produces a green preciptate of [Cr(H2O)3(OH)3]

• [Cr(H2O)6]3+ + 3NH3 = [Cr(H2O)3(OH)3] + 3NH4+
• when excess ammonia is added some of the green precipitate re-dissolves to form a dark green solution, this is more noticeable when concentrated ammonia is used
• occurs as the ligands are replaced by ammonia
• [Cr(H2O)6)]3+ + 6NH3 = [Cr(NH3)6]3+ + 6H2O

Question 6

describe what the haem group does in a polypeptide of haemoglobin

Answer 6

• each haem group has an Fe2+ ion at its centre, oxygen can reversibly bind to the Fe2+ ion, this allows hameoglobin to carry out its function of transporting oxygen around the body
• when Fe2+ binds to oxygen the haem group is red in colour
• there are 4 coordinate bonds between the Fe2+ ion and the nitrogen atoms in the haem structure
• a further coordinate bond is formed to the protein globin
• a final coordinate bond is formed to an oxygen molecule which is then transported

Question 7

describe how carbon monoxide can kill you

Answer 7

• carbon monoxide and oxygen can both bind to haemoglobin at the same place, carbon monoxide binds more strongly to the haemoglobin than oxygen therefore if they are both present fewer oxygen molecules are bound to haemoglobin
• people get starved of oxygen
• no reversible
• low levels can cause headaches, nausea and potential suffocation whereas high levels can be fatal
• formed during incomplete combustion,
• have sensors avaliable to give warning
• burning tobacco can also released carbon monoxide therefore smokers become short of breath

Question 8

what is a precipitation reaction

Answer 8

• precipitation reaction is one in which soluble ions in separate solutions are mixed together to produce an insoluble compound this is called a precipitate

Question 9

why can transition metals take part in oxidation and reduction reactions

Answer 9

• because they have variable oxidation states

Question 10

describe Fe2+ and Fe3+ in terms of redox reactions

Answer 10

• iron has two common oxidation states in its compounds Fe2+ and Fe3+
• the iron(II) oxidation state is less stable than the iron(III) oxidation state
• in the presence of air or when in contact with another oxidising agent iron (II) is readily oxidised to iron(III)
• Fe2+ = Fe3+ + e-

Question 11

describe Manganese in terms of reduction

Answer 11

• MnO4- is a strong oxidising agent and in acidic solution MnO4- can be reduced to form Mn2+
• MnO4- + 8H+ + 5e- = Mn2+ +4H20
• MnO4- is commonly used to oxidise solutions containing iron(II) but is also used as an oxidising agent in many other chemical reactions, half equations can be added together to generate an ionic equation
  MnO4- + 8H+ + 5Fe2+ = Mn2+ + 5Fe3+ + 4H20
• these can be used to calculate the concentration of a transition metal ion in an unknown solution or to calculate the percentage composition of a metal in a solid sample of a compound or alloy

Question 12

How do you carry out redox titrations

Answer 12

• involves the transfer of electrons from one spieces to another
• oxidising agent is titrated against a reducing agent
• indicators can be used in redox titrations but in some reaction there is a colour change that provides the end point without the need for an indicator
• for example aqueous potassium manganate (VII) contains Mn04- and is a common oxidising agent that is self indicating

MnO4- becomes Mn2+ ( goes from purple to colourless)

• Mn2+ ions are pale pink in colour, in manganate(vii) titrations the solutions used are dilute and the Mn2+ are in low concentrations which means that the pale pink colour cannot be seen and the solution appears colourless,
• MnO4- have a deep purple colour that masks any other colour present

Question 13

describe how potassium dichromate solutions are used

Answer 13

Cr2O72- ions will oxidise Fe2+ ions in solution and so can also be used in redox titrations

• colour change is subtle and cannot be seen with the naked eye therefore a redox indicator such as diphenylamine sulfonate is used which produces a violet blue coloured end point
• two half equations which are
• Fe2+ = Fe3+ + e- (oxidation)
• Cr2O72- + 14H+ + 6e- = 2Cr3+ 7H20

Question 14

redox titrations with iodine and thiosulfate

Answer 14

• iodine is reduced to iodide ions by aqueous sodium thiosulfate which forms the tetrahionate ion S4O62-
• the concentration of iodine in a solution can be determined by titration with a solution of sodium thiosulfate of known concentration
• this titration can be used to determine the concentration fo solution of an oxidising agent that reacts with the idoide ions I- to produce iodine I2
• example of oxidising agents that can be used include Copper(II), dicromate(VI) and chlorate(I)
• idodie ions are added to the oxidising agent
• redox reaction takes place
• the iodine is titrated against a sodium thiosulfate solution of known concentratio
• from the results you can calulate the amount of iodine and the concentration of the oxidising agent

Question 15

How do you estimate the copper content of solutions and alloys

Answer 15

• if a solution containing Cu2+ ions is mixed with aqueous iodidie ions I-
• iodide ions are oxidised to iodine 2I- = I2 + 2e-
• copper (II) ions are reduced to copper(I) ions Cu2+ + e- = cu+
• this produces a light brown/yellow solution and a white precipitate of copper(I) iodide, the precipitate appears to be light brown due to the iodine in the solution
• the mixture is then titrated against sodium thiosulfate of known concentration, as the iodine reacts the iodine colour gets paler during the titration, when the solution becomes a pale straw colour a small amount of starch is added to help with the identification of the end point - a blue black colour forms
• the blue black colour disappears sharply at the end point because all the iodine has reacted
• use the results to determine the concentration of iodine in the titration mixture and then the concentration of copper (II) ions in the original solution can be determined
• estimate teh concentration of copper ions in an unknown solution or the percetnage of copper in alloys such as brass and bronze
• the alloy is first reacted with nitric acid, brass and broxe can both be reacted to produce a solution containing copper(II) ions, this can then be reacted with potassium iodide solution, the iodine that forms is then titrated against sodium thiosulfate