Redox titrations

​​In a nutshell

Transition elements with variable oxidation numbers are used as oxidising and reducing agents. Unknown concentrations are calculated through redox titration experiments. Indicators show a colour change when there is a chemical change, however some oxidising agents show a colour change when a species is oxidised. 

Titrations using transition elements

A titration is an analytical technique used to determine the concentration of a substance by measuring the volume added to a solution. Redox titrations allow you to determine the unknown quantity of a species by adding an oxidising agent to a reducing agent or vice versa.

An oxidising agent will be reduced by gaining electrons and it oxidises the other species. A reducing agent will be oxidised by losing electrons and it reduces the other species. Transition elements are used as oxidising/reducing agents in redox titrations as they have variable oxidation numbers, so readily oxidise or reduce.  

Example

Potassium dichromate ($$K_2Cr_2O_7$$) is a common oxidising agent which can be used to oxidise $$Fe^{2+}$$ to $$Fe^{3+}$$. $$Cr$$ is a transition element and will show a colour change, going from orange to green, when it is reduced from $$Cr^{6+}$$ as potassium dichromate to $$Cr^{3+}$$ ions. This colour change is used to determine the concentration of $$Fe^{2+}$$ ions.

Performing a redox titration:

You can now calculate the concentration of $$Fe^{2+}$$ ions as you know the volume of the $$Fe^{2+}$$ solution pipetted in, the concentration of the potassium dichromate oxidising agent and you have recorded the volume of the oxidising agent added.

Indicators and redox titrations

An indicator such as methyl orange is used in redox titrations as it changes colour in response to a chemical change. 

When using a potassium dichromate oxidising agent, no indicator is required as a colour change is observed in response to the oxidation of a species. Similarly, acidified potassium manganate ($$KMnO_4$$) is another oxidising agent that shows a colour change when a species is oxidised.

Calculating the concentration of a reactant

Example:

​$$40\ cm^3$$ of iron($$II$$​) nitrate was pipetted into a conical flask and $$0.15\ mol\ dm^{-3}$$ solution of potassium dichromate was added to a burette. $$32.5\ cm^3$$ of the oxidising agent was used to oxidise $$Fe^{2+}\rightarrow Fe^{3+}$$. 

What was the concentration of iron ($$II$$​) ions pipetted into the conical flask? Give you answer to $$2$$​ d.p.

How to calculate the concentration of a reactant:

To work out the half equations for this reaction, you know that $$2Cr$$ ions in $$Cr_2O_7^{2-}$$ are reduced from $$Cr^{6+}\rightarrow Cr^{3+}$$ so must gain six electrons:

$$Cr_2O_7^{2-}+6e^-\rightarrow 2Cr^{3+}$$​​

There is an excess of hydrogen ions coming from the sulfuric acid which will react with oxygen from the dichromate ion to form water:

$$Cr_2O_7^{2-}+xH^++6e^-\rightarrow 2Cr^{3+}+yH_2O$$

There are $$7O$$ which would produce $$7H_2O$$​, therefore you need $$14H^+$$:

$$Cr_2O_7^{2-}+14H^++6e^-\rightarrow 2Cr^{3+}+7H_2O$$

Now you can work out the oxidation half equation for iron:

$$Fe^{2+}\rightarrow Fe^{3+}+e^-$$​​

As six electrons have been gained in the reduction half equation, six electrons must be lost in this one:

$$6Fe^{2+}\rightarrow 6Fe^{3+}+6e^-$$​​

The two half equations balance out to give:

$$Cr_2O_7^{2-}+14H^++6Fe^{2+}\rightarrow 2Cr^{3+}+7H_2O+6Fe^{3+}$$​​

Now you can calculate the moles of the potassium dichromate oxidising agent:

​$$n=c\times V$$

$$n=0.15\times \frac{32.5}{1000}$$​​​

​$$n=4.875\times10^{-3}\ moles$$​​

You know from the reaction that one mole of dichromate ions reacts with six moles of $$Fe^{2+}$$ so you can work out the moles of iron:

$$n=4.875\times 10^{-3}\times 6$$​​

$$n=0.02925\ moles$$​​

 First you need to convert the volume from $$cm^3$$​ to $$dm^3$$​:

$$v=\frac{40}{1000}$$​​

$$v=0.04\ dm^3$$

Now you can calculate the concentration of $$Fe^{2+}$$ as you know the moles and volume added:

$$c=\frac{n}{V}$$

​​

$$c=\frac{0.02925}{0.04}$$​​

​$$c=0.73125\ mol\ dm^{-3}$$​​

​​

The concentration of $$Fe^{2+}$$ions in the iron($$II$$​) nitrate solution is $$\underline{0.73\ mol\ dm^{-3}}\ (2\ d.p.)$$.