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You have likely performed a titration before. Perhaps you used it to find out how much of an alkali you need to neutralise a certain amount of acid. Redox titrations help us find the exact amount of an oxidising agent needed to react with a reducing agent.Redox titrations with transition metals are exciting because of their colourful variable oxidation states.…
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Jetzt kostenlos anmeldenYou have likely performed a titration before. Perhaps you used it to find out how much of an alkali you need to neutralise a certain amount of acid. Redox titrations help us find the exact amount of an oxidising agent needed to react with a reducing agent.
Redox titrations with transition metals are exciting because of their colourful variable oxidation states. Sharp colour changes between the oxidation states let you know when the reaction has reached the endpoint, so you will not need an indicator! Let us use potassium manganate(VII) as an example to find out how this works!
Titration is a way of analysing chemicals to find an unknown concentration by using a substance with a known concentration.
Essentially, we slowly add a standard solution of a titrant from a burette to the analyte in the conical flask. We can use a colour indicator in order to know that the reaction has reached its endpoint.
Performing a titration will help you understand how it works. Let us next examine the steps involved in a titration.
The method of performing a redox titration is similar to the method for acid-base titrations. You may read about it in pH Curves and Titrations.
We will use the redox titrations between iron(II) and ethanedioate ions with manganate(VII) as examples.
People who suffer from anaemia - low iron concentration in the blood - may be prescribed iron tablets by their doctor or pharmacist. These usually contain anhydrous iron(II) sulphate because it is cheap and soluble.
You can estimate the amount of iron(II) sulphate in each tablet by titrating it against a standard solution of potassium manganate(VII). You will have to dissolve each tablet in diluted sulfuric acid first!
In a laboratory, you may carry out the following experiment with iron tablets from the pharmacist.
A diagram of the equipment you will need is shown below.
In this reaction, Fe2+ gets oxidised to Fe3+ while Mn7+ gets reduced to Mn2+. You write the half equations for the process as follows:
You must use diluted sulphuric acid because potassium permanganate works best as an oxidiser in acidic conditions. Remember, transition metal ions require strongly acidic conditions when going from a higher to a lower oxidation state. However, you cannot use just any acid!
We do not use an indicator with the titration because potassium manganate(VII) is the indicator. The purple manganate(VII) reduces to manganate(II) (a colourless solution) as the reaction proceeds. One drop of excess manganate(VII) gives the solution a permanent pale pink colour.
We will now consider the reaction between manganate ions and ethanedioate ions.
The reaction between manganate and ethanedioate ions (C2O42-) is intriguing because it is autocatalytic. Chemists use ethanedioic acid (also called oxalic acid) to standardise or determine the strength of permanganate solution.
Read about autocatalysts in Catalysts.
Ethanedioic acid, also called oxalic acid, can be found in plants such as spinach and rhubarb. Oxalic acid salts contain the ethanedioate ion (C2O42-). We can learn the concentration of free oxalate ions in solution by titrating against potassium permanganate. This reaction is used to analyse the ethanedioate content of spinach leaves, for example.
The redox reaction between manganate(VII) and ethanedioate ions takes place as follows:
MnO4- is reduced to Mn2+ and C2O42- is oxidised to CO2.
Here are the steps to perform the titration:
We heat the ethanedioate solution to about 60-70ºC to speed up the reaction with potassium permanganate. Be careful not to heat the solution past 70ºC, as ethanedioate begins to decompose at 70ºC and above.
Permanganate's oxidising power works best in an acidic environment. So we use dilute sulfuric acid in this experiment. Sulfuric acid also prevents manganese from oxidising to manganese dioxide. We cannot carry out the titration in the presence of acids such as hydrochloric acid or nitric acid. Hydrochloric acid is an oxidising agent that reacts with manganate(VII) to form chlorine.
As with the previous titration, permanganate acts as a self indicator. Purple MnO4- ions reduce to colourless Mn2+ ions. One drop of excess MnO4- ions presents a pale pink colour.
Have you got that? Okay, let us do some calculations!
After you have completed a titration, you will need to do some calculations to determine the concentration of the analyte. Let us try a few together!
24.55cm3 of 0.020M aqueous potassium manganate(VII) reacted with 25.0cm3 of acidified iron(II) sulfate solution. Find the concentration of Fe2+ ions in the solution.
Step 1: Write out the balanced equation
5Fe2+ (aq) + MnO4- (aq) + 8H+ (aq) ➔ 5Fe3+ (aq) + Mn2+ (aq) + 4H2O (l)
Step 2: Work out the number of moles of MnO4- ions added to the flask.
Moles of MnO4- =
We divide by 1000 to convert the volume from cm3 to dm3.
Moles of MnO4- =
Step 3: The equation tells you that 1 mole of MnO4- reacts with 5 moles of Fe2+.
5Fe2+ (aq) + MnO4- (aq) + 8H+ (aq) ➔ 5Fe3+ (aq) + Mn2+ (aq) + 4H2O (l)
Step 4: Multiply moles of MnO4- by 5.
0.000419 x 5 = 0.002455 moles of Fe2+
Step 5: Work out the concentration of Fe2+
moles of Fe2+ =
Rearrange so that concentration =
Concentration = 0.0982 mol dm-3
Titration calculations generally follow the same principles as you will see in the next example.
Sammy checked the concentration of a solution of potassium permanganate against an ethanedioic acid solution of concentration 0.04 mol dm-3.
He placed 25cm3 of the ethanedioic acid solution in a flask with excess dilute sulphuric acid. After warming the solution, he carried out a titration. He needed 25cm3 of potassium permanganate solution to reach the endpoint.
Calculate the actual concentration of the permanganate solution.
Step 1: Write the balanced equation for the reaction
2MnO4– + 16H+ + 5C2O42- → 2Mn2+ + 10CO2 + 8H2O
Step 2: Find the number of moles of ethanedioic acid
No. of moles of 5C2O42- =
Step 3: Find the number of moles of potassium manganate(VII).
The balanced equation tells us that we need as many moles of permanganate ions as ethanedioate ions.
2MnO4- (aq) + 16H+ (aq) + 5C2O42- (aq) ➔ 2Mn2+ (aq) + 10CO2 (aq) + 8H2O (l)
Step 4: Multiply the no. of moles of ethanedioate ions by
0.001 x in 25cm3 = 0.0004 moles of manganate (VIII)
Step 5: Find the concentration by rearranging the formula
Rearrange the formula so that
Concentration = 0.0004 x
Concentration = 0.016 mol dm-3
It takes practice to get the hang of titration calculations. Try the examples in the exercises section to improve your skills!
Titration is a way of analysing chemicals to find an unknown concentration by using a substance with known concentration.
Universal indicator gives a different colour for different pH ranges. That makes it hard to titrate to a specific pH value. On the other hand, specialised indicators like phenolphthalein change from colourless to deep red at pH above 9.0.
Let us try an example together.
23.9cm3 of 0.040 mol dm-3 of aqueous potassium permanganate reacted with 25cm3 of acidified iron(II) sulphate solution. What was the concentration of Fe2+ ions in the solution?
First things first, write down the equation for the reaction. The redox process between manganate(VII) and iron(II) takes place as follows:
5Fe2+ (aq) + MnO4- (aq) + 8H+ (aq) ➔ 5Fe3+ (aq) + Mn2+ (aq) + 4H2O (l)
Next, use the values provided to find the number of moles of MnO4- ions added to the flask.
Use the formula: no. of moles = concentration x volume / 1000
0.04 x 23.9/1000 = 0.000956 or 9.56x10-4 moles of MnO4-
Now we can figure out the number of moles of Fe2+ in the flask!
Use the reaction equation to find the proportion of the reaction between the titrant and analyte.
From the equation, we can see that 1 mole of manganate(VII) reacts with 5 moles of iron(II).
So multiply 9.56x10-4 by 5.
9.56x10-4 x 5 = 0.00478 or 4.78x10-3 of Fe2+ ions
Use the previous formula to calculate the concentration of Fe2+ ions.
0.00478 = Conc. x 25 / 1000
Conc. = 0.00478 x 1000 / 25
Conc. = 0.1912 mol dm-3
Congratulations, you have completed a titration calculation! Getting the hang of titration calculations takes practice. Have a go at the examples in the exercises section.
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