Chpt. 15 Volumetric Analysis - Oxidation-Reduction

We have already studied how volumetric analysis can be used to determine the concentrations of solutions of acids and bases (Chpt. 13) and also how many chemical reactions in chemistry involve oxidation-reduction reactions (Chpt. 14). So now, lets combine this knowledge and investigate how volumetric analysis can be used to determine the concentrations of oxidising and reducing agents.

Redox Titrations: titrations using oxidising and reducing agents are called redox titrations or oxidation-reduction titrations.

Potassium Permanganate KMnO 4 (Potassium Manganate(VII) ) The most important oxidising agent to be studied is potassium permanganate. Potassium permanganate is a very useful oxidising agent in redox titrations. (For this reason it is important to be very familiar with this substance)

Potassium Permanganate – Key Points: 1) Potassium permanganate is a purple coloured solid. It is NOT a primary standard as it cannot be obtained in a state of very high purity and it decomposes in the presence of sunlight. 2) Potassium permanganate acts as an oxidising agent by gaining 5e - in acidic solution: *Must Learn - Half Reaction: MnO H + + 5e - Mn H 2 O (+7) (+2) Purple Colourless

As acidified potassium permanganate is added to the reducing agent, the manganese atom changes from an oxidation state of +7 (purple) to an oxidation state of +2 (colourless)

3) In order to get potassium permanganate to undergo the above reaction, ie reduce it from Mn(+7) to Mn(+2), some dilute acid is added to supply the H + ions. It is usual to use sulphuric acid (H 2 SO 4 ) in acidifying KMnO 4 for several reasons: - SO 4 2- is very stable to oxidation - the Cl - ions in HCl would be oxidised to Cl 2 by KMnO 4 - Nitric acid (HNO 3 ) cannot be used as it is a powerful oxidising agent itself and therefore would interfere with the reaction.

4) There are two situations where the above reaction will NOT occur: a) in neutral solution i.e. no acid b) in basic solution i.e. not enough acid In the above cases instead of going from Mn(VII) Mn(II) Purple Colourless the manganese ion changes to an intermediate oxidation state – Mn(IV) formation of MnO 2 : Mn(VII) Mn(IV) Purple Brown

When KMnO 4 reacts with a reducing agent in neutral or basic solution, the manganese ion changes from an oxidation state of +7 (purple) to an oxidation state of +4(brown)

5) Potassium permanganate acts as its own indicator, giving a sharp end-point. It changes colour from purple to colourless when acting as an oxidising agent. 6) Because of the intense purple colour of potassium permanganate, read the top rather than the bottom of the meniscus. 7) Generally potassium permanganate is used in fairly weak solutions – 0.02M. This is due to its strong oxidising power and the fact that it is not very soluble (i.e. hard to get to dissolve).

Two types of redox titrations to be studied: - the rxn of MnO 4 - and Fe 2+ ions - the rxn of iodine and thiosulphate ions

Reaction of MnO 4 - ions and Fe 2+ ions KMnO 4 – is not a primary standard (not possible to make up solution of KMnO 4 of accurate concentration by weighing out a known amount of the solid and dissolving in a known volume of solution) However, it is possible to determine the concentration of a KMnO 4 solution by titrating it against a solution containing Fe 2+ (iron) ions: Standardising Potassium Permanganate using Iron

KMnO 4 – oxidising agent – when added to an acidified (to ensure Mn+7 is reduced to Mn+2) solution containing Fe 2+ ions it converts the Fe 2+ ions to Fe 3+ ions. Reactions involved: Oxidising Agent (itself reduced): Reducing Agent (itself oxidised): (B) Fe 2+ - e - Fe 3+ (A)MnO H + + 5e - Mn H 2 O

Upon inspection of equations A & B it is clear to see that every MnO 4 - ion must receive 5e - in order for the reaction to go to completion but every Fe 2+ ion will only give 1e -. So, the following ratio can be determined: MnO 4 - ion : Fe 2+ ion 1 : 5

Rewriting equations A & B: Adding these two half equations: (A)MnO H + + 5e - Mn H 2 O (B) 5Fe e - 5Fe 3+ MnO H + + 5Fe 2+ Mn Fe H 2 O

Ammonium Iron(II) Sulphate (Ferrous Ammonium Sulphate) Vs Iron(II) Sulphate Cannot use iron(II) sulphate(FeSO 4 ) crystals as a primary standard because: - the crystals are oxidised slightly by the air - the crystals tend to lose their water of crystallisation when exposed to the air (efflorescence)

Ammonium Iron(II) Sulphate (NH 4 ) 2 SO 4.FeSO 4.6H 2 O is used as a source of Fe 2+ ions because: - it is a primary standard (crystals not affected by air) - it is stable and available in a high state of purity Ammonium Iron(II) Sulphate Crystals

When preparing ammonium iron(II) sulphate it must be dissolved in water containing dilute sulphuric acid: - the sulphuric acid prevents the Fe 2+ ions reacting with water to form a new substance – prevents hydrolysis - the sulphuric acid helps to prevent the Fe 2+ ions being oxidised to Fe 3+ ions by the oxygen in the air or dissolved in the water

Redox Reactions Volumetric Problems A formula similar to that used in acid-base titrations is used in these calculations: V O x M O = V r x M red n O n r V O = vol. of oxidising agent (cm 3 ) M O = molarity of oxidising agent n O = no. of moles of oxidising agent (from balanced eqn.) V r = vol. of reducing agent (cm 3 ) M red = molarity of reducing agent n r = no. of moles of reducing agent (from balanced eqn.)

Example 1: (Please leave a page for answer) 24cm 3 of a potassium manganate(VII) solution required 25cm 3 of a M ammonium iron(II) sulphate solution for complete reaction. The equation for the reaction is: Calculate the concentration of the potassium manganate(VII) solution in (a) moles per litre (b) grams per litre MnO H + + 5Fe 2+ Mn Fe H 2 O

Mandatory Expt.: To prepare a standard solution of ammonium iron(II) sulphate and to use this solution to standardise a solution of KMnO 4 by titration Points to Note: KMnO 4 is its own indicator. Before the end point, all added KMnO 4 solution is decolourised by the reaction with Fe 2+ ions. At the end point, all of the Fe 2+ ions have reacted, and added KMnO 4 solution is no longer decolourised: Purple colourless pink Read from the top of the meniscus.

Example 2: (Please leave a page for answer) A sample of iron(II) sulphate crystals FeSO 4. xH 2 O was analysed in order to find out the value of x. 6.42g of the crystals were dissolved in deionised water to which some dilute sulphuric acid had been added and the resulting solution was made up to 250cm 3 in a volumetric flask. 25cm 3 samples of this solution were titrated against 0.018M KMnO 4 solution. The balanced equation for the reaction is: The average titration figure was 25.5cm 3. Calculate the value of x in the above formula. MnO H + + 5Fe 2+ Mn Fe H 2 O

Applications of Potassium Permanganate – Iron(II) Titrations Iron is an important element in our bodies as it forms part of the haemoglobin molecule in our blood. Haemoglobin is the molecule responsible for transporting oxygen in our blood. The oxygen actually becomes attached to the iron atom in the haemoglobin molecule. Our bodies can easily become deficient in iron resulting in anaemia. This condition can be remedied with a change in diet and a course of iron tablets (FeSO 4 ) In the next experiment we will attempt to calculate the amount of iron(II) sulphate in iron tablets

Haemoglobin Molecule *Do Not take Diagram

Mandatory Expt.: To determine the amount of iron in an iron tablet

Example 1: (Please leave a page for your answer) A student was asked to analyse an iron tablet to determine if the amount of iron sulphate stated on the packet was correct. She dissolved five iron tablets of total mass 1.8g in dilute sulphuric acid and made the solution up to 250cm 3 with deionised water. 25cm 3 of this solution required 16.5cm 3 of a 0.005M potassium manganate(VII) solution for complete reaction. Calculate (a) the mass of anhydrous FeSO 4 in each tablet (b) the mass of iron in each tablet (c) the percentage of FeSO 4 in each tablet and (d) the percentage of iron in each tablet

Sodium Thiosulphate (Na 2 S 2 O 3 ) Sodium thiosulphate (hypo) is an important reducing agent used in chemistry Used in developing photographs and in analytical chemistry

Most commonly encountered as a colourless crystalline solid of the form: Na 2 S 2 O 3.5H 2 O (pentahydrate)

Reaction of Iodine and Thiosulphate Ions Sodium thiosulphate is a reducing agent and reacts with iodine molecules (I 2 ), to convert them to iodide (I - ions): I 2(aq) + 2S 2 O 3 2- (aq) S 4 O 6 2- (aq) + 2I - (aq) Iodine Thiosulphate Ion Tetrathionate Ion Iodide Ion Oxidising Agent Reducing Agent

Ratio I 2 : S 2 O : 2

Thiosulphate Ion S 2 O 3 2- Tetrathionate Ion S 4 O 6 2- *Take Diagram

Sodium Thiosulphate is NOT usually a primary standard because it is difficult to obtain in the anhydrous form (no water) and the crystals lose water of crystallisation to the atmosphere and so are unstable. However, it is possible to determine the concentration of a sodium thiosulphate solution by titrating it against a standard iodine solution: Standardising Sodium Thiosulphate using Iodine

Iodine is NOT usually a primary standard. A standard solution of iodine cannot be made up by direct weighing because: - it is too volatile (vaporises slightly at room temp.) - it does NOT dissolve in water A STANDARD SOLUTION of IODINE (0.06M) is made up by reacting a standard solution of acidified potassium permanganate (0.02M) with excess potassium iodide.

Reaction equations for formation of standard solution of iodine (A) MnO H + + 5e - Mn H 2 O (Each permanganate ion needs 5e - ) (B) 2I - I 2 + 2e - (Each iodide loses 1e - (2I - = 2e - and forms iodine as a result)

Upon inspection of equations A & B it is clear to see that every MnO 4 - ion must receive 5e - in order for the reaction to go to completion. But the 2I - only results in the loss of 2e -, so in order to balance the e - gain and loss we must: (A) x2: 2MnO H e - 2Mn H 2 O (B) x5: 10I - 5I e - Adding these two half equations: 2MnO I H + 2Mn I 2 + 8H 2 O

In the previous equation the I - ions (from potassium iodide) are kept in excess. As a result the amount of iodine formed in this reaction depends on the amount of permanganate present. So, by using a standard solution of KMnO 4 we can determine the amount of iodine formed. This known amount of iodine formed can then be used to standardise a sodium thiosulphate solution

**Note: Iodine is non-polar and water is a polar solvent. For this reason iodine will NOT dissolve in water but will dissolve in a solution of potassium iodide (KI). Adding I 2 to KI: - I 2 molecule reacts with I - to form I 3 - (tri-iodide ion) I 2 + I - I 3 - state of equilibrium - I 3 - is identical in its chemical behaviour to I 2 and because of its charge it dissolves easily in water - As a result a solution of iodine in potassium iodide may be treated as if it were a solution of iodine in water

Determining the end point of an iodine/thiosulphate titration As thiosulphate is added to the iodine the reddish/brown colour gradually turns yellow which gradually turns paler and paler until it turns colourless – very difficult to detect end point. Therefore need to use an indicator – starch indicator I 2(aq) + 2S 2 O 3 2- (aq) S 4 O 6 2- (aq) + 2I - (aq) Reddish Brown Colourless (Colourless)

Starch forms a blue-black colour with iodine therefore when ALL of the iodine has reacted with the thiosulphate the blue-black colour suddenly goes colourless. Note: starch is only added when most of iodine has disappeared i.e. when reddish/brown colour of iodine has gone pale(straw) yellow. Why??? If too much added or added too early starch participates in rxn and removes to much iodine When pale colour is observed it indicates end point is near

Colour Change Reddish/brown Straw Yellow Blue/Black Colourless Add Starch End Point Colour Change

Points to Note: 1. A standard solution of sodium thiosulphate can be used to standardise an iodine solution 2.In an iodine-thiosulphate titration: - iodine in conical flask - sodium thiosulphate in burette 3.Starch indicator must be prepared fresh as it deteriorates quickly on standing. 4.A 0.02M standard solution of potassium iodate may be used in place of potassium permanganate in preparing the iodine solution.

Mandatory Expt.: To prepare a solution of sodium thiosulphate and standardise it by titration against a solution of iodine Two reactions: 1.Formation of iodine in conical flask: 2. Main rxn – Reaction of Iodine with Sodium Thiosulphate So, 2MnO 4 - : 5I 2 : 10S 2 O 3 2- Ratio 1 : 2 2MnO I H + 2Mn I 2 + 8H 2 O I 2(aq) + 2S 2 O 3 2- (aq) S 4 O 6 2- (aq) + 2I - (aq)

Example 1: (Please leave a page for answer) 25cm 3 of 0.018M potassium permanganate is pipetted into a conical flask. Some dilute sulphuric acid and excess potassium iodide solution are then added and the liberated iodine is titrated against a solution of sodium thiosulphate of unknown concentration. The average titration figure is 15.6cm 3. The reactions may be represented as: Calculate the concentration of the sodium thiosulphate solution in a) moles per litre and b) grams per litre of crystalline Na 2 S 2 O 3.5H 2 O 2MnO I H + 2Mn I 2 + 8H 2 O I 2(aq) + 2S 2 O 3 2- (aq) S 4 O 6 2- (aq) + 2I - (aq)

Applications of Iodine –Thiosulphate Titrations Most household bleaches contain chlorine in the form of sodium hypochlorite, NaClO. NaClO contains the hypochlorite ion (chlorate(I) ion) – ClO - Hypochlorite ion is an oxidising agent and converts iodide ions to iodine: ClO - + 2I - + 2H + Cl - + I 2 + H 2 O I 2(aq) + 2S 2 O 3 2- (aq) S 4 O 6 2- (aq) + 2I - (aq)

Ratio: ClO - : I 2 : 2S 2 O : 1 : 2 Where, ClO - is the oxidising agent and 2S 2 O 3 2- is the reducing agent

Experiment: To determine the percentage (w/v) of sodium hypochlorite in household bleach To determine the strength of household bleach a sample of the bleach is reacted with a solution of iodide ions. The liberated iodine is then titrated against a solution of sodium thiosulphate of known concentration. Results obtained will allow calculation of the concentration of sodium hypochlorite in bleach. *Note: As household bleach is fairly concentrated it must be diluted first otherwise an excessive amount of reagents would be used in titration

Example 1: (Please leave a page for answer) 25cm 3 of household bleach was diluted to 250cm 3. A 25cm 3 portion of the diluted solution was added to an excess of acidified potassium iodide solution and titrated against 0.12M sodium thiosulphate solution (standardised). The average titration figure was 32.1cm 3. Calculate the concentration of sodium hypochlorite in the bleach in (a) moles per litre (b) grams per litre and (c) % w/v. The equations for the reaction are: ClO - + 2I - + 2H + Cl - + I 2 + H 2 O I 2(aq) + 2S 2 O 3 2- (aq) S 4 O 6 2- (aq) + 2I - (aq)