A2-Level Transition Elements
Titrations can be used to find the unknown concentration of a solution.
Redox reactions consist of a species that is oxidised and a species that is reduced.
If the moles of one species is known, using the ratios of the oxidised to species to reduced species, the moles of the other species can be found and the concentration calculated.
Redox reactions describe reactions where one species gains electrons (reduction) and another species loses electrons (oxidation). As transition elements have multiple oxidation states, they can form different charged ions and often take part in redox reactions.
Redox titrations use redox reactions to find the concentration of a species or molecule – in the same way titrations are used to find the find the concentration of an acid or alkali. An oxidising agent and reducing agent react, and when a colour change occurs, a set amount of either one has reacted or been produced.
Redox titrations can look complicated, but if approached in the right way they can be answered confidently. Determining the redox reaction(s) taking place is always the first step when dealing with redox titrations. This is the part that frequently causes students alarm.
Half equations are used to show the species being oxidised and reduced.
In the reaction between ethanoic acid and potassium manganate (VII) solution, the following reaction takes place.
This is a redox reaction, the ethanoic acid is oxidised and the manganate ions are reduced. Now, we can write out the half equations that show this oxidation and reduction.
By using these half equations and the overall equation for the reaction, we can find the concentration of one of the reactants with a titration.
In a titration, 24.5 cm of 0.02 moldm potassium manganate (VII) solution was required to react completely with 25.0cm of ethanoic acid solution. Calculate the concentration of the ethanoic acid solution.
The first thing to do is find the number of moles of manganate ions that reacted.
From the overall equation, the ratio of manganate ions to ethanoic acid is 2:5. This means that for every two moles of manganate ions, five moles of ethanoic acid are required.
We can use this ratio to calculate the moles of ethanoic acid that must have reacted with the 4.9x10 moles of manganate ions.
We know the volume of ethanoic acid that was used in the titration was 25.0cm. By re-arranging the equation used previously, we can find the concentration of this solution.