The oxidation states of transition element ions can change during?redox reactions
A species will either be oxidised or reduced, depending on what reaction is occurring
To find the concentration of specific ions in solution, a titration can be performed
There are three particular redox titrations that need to be learnt:
Iron (II) (Fe2+) and permanganate (MnO4-) in acid solution given suitable data
Permanganate (MnO4-) and ethanedioate (C2O42-) in acid solution given suitable data
Copper (II) (Cu2+) and iodide (I-) given suitable data
Reaction of MnO4-?& Fe2+?in acid
The concentration of Fe2+?ions can be determined by titrating a known?volume?of Fe(II) ions with a known?concentration?of MnO4-?ions
During the reaction of MnO4-?with Fe2+, the?purple?colour of the manganate(VII) ions disappears
The end-point is when all of the Fe2+?ions have reacted with the MnO4-?ions, and the first trace of a permanent?pink?colour appears in the flask
At this point, the MnO4-?is very slightly in?excess
The two half-reactions that are involved in this redox reaction are as following:
The half equations are combined to get the overall equation
Redox Systems: Permanganate & Oxalate
The second redox titration involving transition element ions that needs to be learned, is the titration of permanganate (MnO4-) and ethanedioate, sometimes known as oxalate (C2O42-) in acid solution given suitable data
Reaction of MnO4-?& C2O42-?in acid
The reaction of MnO4-?with ethanedioate, C2O42-?is an example of a redox reaction in which the ethanedioate ions (C2O42-) get?oxidised?by manganate(VII) (MnO4-) ions
A titration reaction can be carried out to find the concentration of the?toxic?ethanedioate ions
As before, the end point is when all of the ethanedioate ions have reacted with the MnO4-?ions, and the first permanent pink colour appears in the flask
At this point, the MnO4-?is very slightly in?excess
The two half-reactions that are involved in this redox reaction are as following:
The half equations are combined to get the overall equation
This is an example of an?autocatalysis reaction
This means that the reaction is catalysed by one of the products as it forms
In this reaction, the Mn2+?ions formed act as the?autocatalyst
The more Mn2+?formed, the faster the reaction gets, which then forms even more Mn2+?ions and speeds the reaction up even further
Transition element ions can act as autocalysts as they can change their oxidation states during a reaction
Redox Systems: Cupric & Iodide
The third redox titration involving transition metal ions that needs to be learnt is the titration between copper (II) ions (Cu2+) - sometimes known as cupric ions - and iodide ions (I-)
Reaction of Cu2+?& I-
The reaction of Cu2+?with I-?is an example of a redox reaction in which the copper ions (Cu2+)?oxidise?the iodide ions (I-) and as a result are themselves?reduced
The two half-reactions that are involved in this redox reaction are as follows:
The half equations could be combined to get this overall equation
When?excess?iodide ions are reacted with Cu(II), a precipitate of copper(I) iodide and?iodine?is formed
2Cu2+?(aq) + 4I-(aq) →? I2(aq) + 2CuI (s) ?????????Reaction 1
A titration reaction can be carried out to find an unknown concentration of the copper(II) solution
This is done by finding the amount of iodine which is liberated during the reaction, through a titration
Step 1:?A known concentration of?sodium thiosulfate solution?is added to the mixture formed in Reaction 1 from a burette
Step 2:?The I2?formed in Reaction 1 will react with the thiosulfate ions
Step 3:?As the iodine is used, up the?brownish?colour of the solution gets?lighter
Step 4:?When most of the iodine colour is gone, starch is added, which turns deep?blue/black?with the remaining I2?(aq)
Step 5:?Titrate further until the blue/black colour disappears, i.e. when all of the iodine has reacted
By knowing the number of moles of thiosulfate ions added in the titration, you can use the molar ratios from the reaction equations and work backwards to calculate the number of moles of Cu(II)
Step 6:?Look at Reaction 2: it can be concluded that half the number of moles of I2?reacts when compared to the moles of thiosulfate that react
Step 7:?Now look at Reaction 1: the number of moles of I2?which reacts in Reaction 2, is the moles formed in Reaction 1. The number of moles of Cu(II) is twice that of I2?(aq) (i.e. the same number of moles as thiosulfate ions added in the titration)
Step 8:?Divide the number of moles of Cu(II) by the volume in dm3?to get the concentration of Cu(II)