Quantitative redox reaction analysis

GCE Advanced A Level REDOX Volumetric Analysis Titration Revision Questions

Quantitative volumetric analysis – exam practice redox titration questions based potassium manganate(VII)–iron(II)/ethanedioate–ethanedioic acid (oxalate, oxalic acid)/hydrogen peroxide/sodium nitrite titrations, sodium thiosulfate/thiosulfate–iodine titrations and potassium dichromate(VI)–iron(II) titration. Any suggestions for additional types of A level redox titration questions?

Volumetric analysis worksheet of structured questions of REDOX VOLUMETRIC TITRATION CALCULATIONS

Titrations and calculations based on oxidation–reduction techniques–reactions – solved problems

Relative atomic masses that may be needed, in alphabetical order of symbol ...

C=12.0,  Cr = 52.0, Fe=55.9,  H=1.0,  I=126.9,  K=39.1,  Mn=54.9,  N=14.0,  Na=23.0,  O=16.0,  S=32.1,

NOTE: Half reactions are usually quoted as the half–cell reduction equation. Reuse half–cell or full equations in later questions from earlier questions. It is assumed you will work through them in numerical question order. If you cannot work out the redox equations, you can just download the equations so that you can at least practice the 'pure' volumetric calculation aspects of the questions.

Questions 1/4/6/7/8/9/10/11/12/13/15/16 based potassium manganate(VII)–iron(II)/ethanedioate–ethanedioic acid (oxalate, oxalic acid)/hydrogen peroxide/sodium nitrite titrations, Q2/14/17 on sodium thiosulfate–iodine titration, Q3/5 on potassium dichromate(VI)–iron(II) titration,  further Q's will be added – suggestions?

REDOX–ionic EQUATION CHECKS

1. Use of the correct 'species' (e.g. usually two given/chosen as/from half–cell equation data)

2. The 'species' direction change – which is oxidised or reduced? In other words get the half–cell equations the right way round! (If not indicated, might have to decide from EØ data supplied, the more +ve half–cell is the reduction (oxidising agent). More in Equilibria Part 7 (being written)

3. The correct ratio of half–cell equations – the 'balance' must be based on oxidation number analysis or number of electrons transferred. The total increase in oxidation states = the total decrease in oxidation states or total electrons gained = total electrons lost by the species involved.

4. Add up the ion charges, the totals should be the same on both sides of the equation (I find this a handy extra check especially with stray H2O's or H+'s!).

5. 'traditional' atom count – placed last because its not completely reliable with redox equations!

6. In some Q's the full equation may be given or the two half–cell equations to be put together the right way round and in the right ratio (see Redox Chemistry Part 2

I've tried to quote the data to the appropriate significant figures and associated 'trailing zeros'.

Note that it is standard convention to show half-cell reactions as reductions, i.e. atom/ion/molecule + electron(s) to give the reduction product. This means you have to judge whether the half-reaction needs to be reversed to derive the full ionic redox equation, and any multiples of it are needed, - so take care, and don't get confused by conventions! they are there help!

I DO MY BEST TO CHECK MY CALCULATIONS, as you yourself should do,  BUT I AM HUMAN! AND IF YOU THINK THERE IS A 'TYPO' or CALCULATION ERROR PLEASE EMAIL ME ASAP TO SORT IT OUT!

Question 1: Given the following two half–reactions: (Q1 can be done as an experimental 'word–fill' version)

Question 1 has many parts covering the titration of iron(II) ions with a standard solution of potassium manganate(VII) and the problems are solved.

The relative atomic mass of iron = 55.9

(i) MnO4(aq) + 8H+(aq)  + 5e==> Mn2+(aq) + 4H2O(l)

and (ii) Fe3+(aq) e ==> Fe2+(aq)

(a) Construct the fully balanced redox ionic equation for the manganate(VII) ion oxidising the iron(II) ion

(b) 24.3 cm3 of 0.0200 mol dm–3 KMnO4 reacted with 20.0 cm3 of an iron(II) solution acidified with dilute sulfuric acid.

(i) Calculate the molarity of the iron(II) ion.

(ii) How do recognise the end–point in the titration?

(c) Calculate the percentage of iron in a sample of steel wire if 1.51 g of the wire was dissolved in excess of dilute sulfuric acid and the solution made up to 250 cm3 in a standard graduated flask.

A 25.0 cm3 aliquot of this solution was pipetted into a conical flask and needed 25.45 cm3 of  O.0200 mol dm–3 KMnO4 for complete oxidation.

(d) Suggest reasons why the presence of dil. sulfuric acid is essential for an accurate titration and why dil. hydrochloric and nitric acids are unsuitable to be used in this context.

(e) The analysis of a soluble iron(II) salt to obtain the percentage of iron in it.

8.25g of an iron(II) salt was dissolved in 250 cm3 of pure water. 25.0 cm3 aliquots were pipetted from this stock solution and titrated with 0.0200 mol dm–3 potassium manganate(VII) solution.

The titration values obtained were 23.95 cm3, 23.80 cm3 and 23.85 cm3.

(i) What titration value should be used in the calculation and why?

(ii) Calculate the moles of manganate(VII) used in the titration.

(iii) calculate the moles of iron(II) ion titrated

(iv) Calculate the mass of iron(II) titrated

(v) Calculate the total mass of iron in the original sample of the iron(II) salt.

(vi) calculate the % iron in the salt.

Question 2: Given the following two half–reactions

(a) Given (i) S4O62–(aq) + 2e ==> 2S2O32–(aq)

and (ii) I2(aq) + 2e ==> 2I(aq)

construct the full ionic redox equation for the reaction of the thiosulfate ion S2O32– and iodine I2.

(b) what mass of iodine reacts with 23.5 cm3 of 0.0120 mol dm–3 sodium thiosulfate solution.

(c) 25.0 cm3 of a solution of iodine in potassium iodide solution required 26.5 cm3 of 0.0950 mol dm–3 sodium thiosulfate solution to titrate the iodine.

What is the molarity of the iodine solution and the mass of iodine per dm3?

Question 3: 2.83 g of a sample of haematite iron ore [iron (III) oxide, Fe2O3] were dissolved in concentrated hydrochloric acid and the solution diluted to 250 cm3.

25.0 cm3 of this solution was reduced with tin(II) chloride (which is oxidised to Sn4+ in the process) to form a solution of iron(II) ions.

This solution of iron(II) ions required 26.4 cm3 of a 0.0200 mol dm–3 potassium dichromate(VI) solution for complete oxidation back to iron(III) ions.

(a) given the half–cell reactions

(i) Sn4+(aq) + 2e ==> Sn2+(aq)

and (ii) Cr2O72–(aq) + 14H+(aq) + 6e ==> 2Cr3+(aq) + 7H2O(l)

deduce the fully balanced redox equations for the reactions

(i) the reduction of iron(III) ions by tin(II) ions

(ii) the oxidation of iron(II) ions by the dichromate(VI) ion

(b) Calculate the percentage of iron(III) oxide in the ore.

(c) Suggest why potassium manganate(VII) isn't used for this titration? (though it was ok in Q1)

If you don't know, the following half-cell potential data will help!

Eθ = +1.33 for Cr2O72–(aq) + 14H+(aq) + 6e 2Cr3+(aq) + 7H2O(l)

Eθ = +1.36 for Cl2(aq) + 2e 2Cl(aq)

Eθ = +1.51 for MnO4(aq) + 8H+(aq) + 5e Mn2+(aq) + 4H2O(l)

Question 4: An approximately 0.02 mol dm–3 potassium manganate(VII) solution was standardized against precisely 0.100 mol dm–3 iron(II) ammonium sulfate solution. 25.0 cm3 of the solution of the iron(II) salt were oxidized by 24.15 cm3 of the manganate(VII) solution.

What is the molarity of the potassium manganate(VII) solution ?

Question 5: 10.0 g of iron(II) ammonium sulfate crystals were made up to 250 cm3 of acidified aqueous solution. 25.0 cm3 of this solution required 21.25 cm3 of 0.0200 mol dm–3 potassium dichromate(VI) for oxidation.

Calculate x in the formula FeSO4.(NH4)2SO4.xH2O

Question 6: Given the half–reaction C2O42–(aq) – 2e ==> 2CO2(g)

or H2C2O4(aq) – 2e ==> 2CO2(g) + 2H+(aq)

(a) write out the balanced redox equation for manganate(VII) ions oxidising the ethanedioate ion (or ethanedioic acid).

(b) 1.520 g of ethanedioic acid crystals, H2C2O4.2H2O, was made up to 250.0 cm3 of aqueous solution and 25.00 cm3 of this solution needed 24.55 cm3 of a potassium manganate(VII) solution for oxidation.

Calculate the molarity of the manganate(VII) solution and its concentration in g dm–3.

Question 7: A standardization of potassium manganate(VII) solution yielded the following data:

0.150 g of potassium tetraoxalate dihydrate, KHC2O4.H2C2O4.2H2O needed 23.20 cm3 of the manganate(VII) solution.

What is the molarity of the manganate(VII) solution? Use the equation and reasoning from Q6 to help you.

Question 8: Given the half–cell equation  O2(g) +2H+(aq) + 2e ==> H2O2(aq)

(a) construct the fully balanced redox ionic equation for the oxidation of hydrogen peroxide by potassium manganate(VII)

(b) 50.0 cm3 of solution of hydrogen peroxide were diluted to 1.00 dm3 with water.

25.0 cm3 of this solution, when acidified with dilute sulfuric acid, reacted with 20.25 cm3 of 0.0200 mol dm–3 KMnO4.

What is the concentration of the original hydrogen peroxide solution in mol dm–3?

Question 9: 13.2 g of iron(III) alum were dissolved in water and reduced to an iron(II) ion solution by zinc and dilute sulfuric acid. The mixture was filtered and the filtrate and washings made up to 500 cm3 in a standard volumetric flask.

If 20.0 cm3 of this solution required 26.5  cm3 of 0.0100 mol dm–3 KMnO4 for oxidation.

(a) give the ionic equation for the reduction of iron(III) ions by zinc metal.

(b) Calculate the percentage by mass of iron in iron alum.

Question 10: Calculate the concentration in mol dm–3 and g dm–3, of a sodium ethanedioate (Na2C2O4) solution, 5.00 cm3 of which were oxidized in acid solution by 24.50 cm3 of a potassium manganate(VII) solution containing 0.05 mol dm–3.

Question 11: Calculate x in the formula FeSO4.xH2O from the following data:

12.18 g of iron(II) sulfate crystals were made up to 500 cm3 acidified with sulfuric acid.

25.0 cm3 of this solution required 43.85 cm3 of 0.0100 mol dm–3 KMnO4 for complete oxidation.

Question 12:  Given the half–reaction NO3(aq) + 2H+(aq) + 2e ==> NO2(aq) + H2O(l)

(a) give the ionic equation for potassium manganate(VII) oxidising nitrate(III) to nitrate(V)

(b) 24.2 cm3 of sodium nitrate(III) [sodium nitrite] solution, added from a burette, were needed to discharge the colour of 25.0 cm3 of an acidified 0.0250 mol dm–3 KMnO4 solution.

What was the concentration of the nitrate(III)  solution in grammes of anhydrous salt per dm3?

Question 13: 2.68 g of iron(II) ethanedioate, FeC2O4, were made up to 500 cm3 of acidified aqueous solution. 25.0 cm3 of this solution reacted completely with 28.0 cm3 of 0.0200 mol dm–3 potassium manganate(VII) solution.

Calculate the mole ratio of KMnO4 to FeC2O4 taking part in this reaction. Give the full redox ionic equation for the reaction.

Question 14: Given the half–cell reaction IO3(aq) + 6H+(aq) + 5e ==> 1/2I2(aq) + 3H2O(l) (see also Q2)

(a) Deduce the redox equation for iodate(V) ions oxidising iodide ions.

(b) What volume of 0.0120 mol dm–3 iodate(V) solution reacts with 20.0 cm3 of 0.100 mol dm–3 iodide solution?

(c) 25.0 cm3 of the potassium iodate(V) solution were added to about 15 cm3 of a 15% solution of potassium iodide (ensures excess iodide ion). On acidification, the liberated iodine needed 24.1 cm3 of 0.0500 mol dm–3 sodium thiosulfate solution to titrate it.

(i) Calculate the concentration of potassium iodate(V) in g dm–3

(ii) What indicator is used for this titration and what is the colour change at the end–point?

Question 15: Calculate the molarities of iron(II) and iron(III) ions in a mixed solution from the following data.

(i) 25.0 cm3 of the original mixture was acidified with dilute sulfuric acid and required 22.5 cm3 of 0.0200 mol dm–3 KMnO4 for complete oxidation.

(ii) a further 25.0 cm3 of the original iron(II)/iron(III) mixture was reduced with zinc and acid and it then required 37.6 cm3 of the KMnO4 for complete oxidation.

Question 16: A piece of rusted iron was analysed to find out how much of the iron had been oxidised to rust [hydrated iron(III) oxide]. A small sample of the iron was dissolved in excess dilute sulfuric acid to give 250 cm3 of solution. The solution contains Fe2+ ions from the unrusted iron dissolving in the acid, and, Fe3+ ions from the rusted iron.

(a) 25.0 cm3 of this solution required 16.9 cm3 of 0.0200 mol dm–3 KMnO4 for complete oxidation of the Fe2+ ions.

Calculate the moles of Fe2+ ions in the sample titrated.

(b) To a second 25.00 cm3 of the rusted iron  solution an oxidising agent was added to convert all the Fe2+ ions present to Fe3+ ions. The Fe3+ ions were titrated with a solution of EDTA4–(aq) ions and 17.6 cm3 of 0.100 mol dm–3 EDTA were required.

Assuming 1 mole of EDTA reacts with 1 mole of Fe3+ ions, calculate the moles of Fe3+ ions in the sample.

(c) From your calculations in (a) and (b) calculate the ratio of rusted iron to unrusted iron and hence the percentage of iron that had rusted.

Question 17: 25.0 cm3 of an iodine solution was titrated with 0.100 mol dm–3 sodium thiosulfate solution and the iodine reacted with 17.6 cm3 of the thiosulfate solution.

(a) give the reaction equation.

(b) what indicator is used? and describe the end–point in the titration.

(c) calculate the concentration of the iodine solution in mol dm–3 and g dm–3.

Question 18: 1.01g of an impure sample of potassium dichromate(VI), K2Cr2O7, was dissolved in dil. sulfuric acid and made up to 250 cm3 in a calibrated volumetric flask. A 25.0 cm3 aliquot of this solution pipetted into a conical flask and excess potassium iodide solution and starch indicator were added. The liberated iodine was titrated with 0.100 mol dm–3 sodium thiosulfate and the starch turned colourless after 20.0 cm3 was added.

(a) Using the half–equations from Q3(a)(ii) and Q2(a)(ii), construct the full balanced equation for the reaction between the dichromate(VI) ion and the iodide ion.

(b) Using the half–equations from Q2(a) construct the balanced redox equation for the reaction between the thiosulfate ion and iodine.

(c) Calculate the moles of sodium thiosulfate used in the titration and hence the number of moles of iodine titrated.

(d) Calculate the moles of dichromate(VI) ion that reacted to give the iodine titrated in the titration.

(e) Calculate the formula mass of potassium dichromate(VI) and the mass of it in the 25.0 cm3 aliquot titrated.

(f) Calculate the total mass of potassium dichromate(VI) in the original sample and hence its % purity.

Question 19: This question involves titrating ethanedioic acid (oxalic acid), H2C2O4 or (COOH)2 (i) with standard sodium hydroxide solution and then with potassium manganate(VII) solution (potassium permanganate, KMnO4).

The titration data is as follows:

10 cm3 of a H2C2O4 solution required 8.50 cm3 of a 0.20 mol dm-3 (0.20M) solution of sodium hydroxide for complete neutralisation using phenolphthalein indicator (first permanent pink end-point).

10 cm3 of the same H2C2O4 solution required 8.20 cm3 of a KMnO4 solution for complete oxidation to carbon dioxide and water in the presence of dilute sulfuric acid to further acidify the ethanedioic acid solution (first permanent pink end-point).

(a) Write an equation for the neutralisation reaction of ethanedioic acid with sodium hydroxide.

(b) Calculate the moles of H2C2O4 in the solution and the molarity of the ethanedioic acid solution.

(c) Given the following half-reactions:

(i)   MnO4(aq)  +  8H+(aq)  +  5e  ===>  Mn2+(aq)  +  4H2O(l)

(ii)   H2C2O4(aq)  – 2e   ===> 2CO2(g)   +   2H+(aq)

Deduce the full redox titration equation for the oxidation of ethanedioic acid by potassium manganate(VII).

(d) From the equation in (c) and the titration data, deduce the molarity of the potassium manganate(VII) solution.

Question 20: ?

I DO MY BEST TO CHECK MY CALCULATIONS, as you yourself should do,  BUT I AM HUMAN! AND IF YOU THINK THERE IS A 'TYPO' or CALCULATION ERROR PLEASE ASAP TO SORT IT OUT!

practice redox titration questions, how do you do redox titrations? how do you do redox titration calculations? what apparatus do you need to do redox titrations? describe redox titration methods and procedures, what is the definition of a redox titration? how to do redox titration experiments, describe examples of redox titrations, (potassium permanganate) potassium manganate(VII) - iron iron(II) sulfate titration calculation method, sodium thiosulfate Na2S2O3 - I2 iodine titration calculation method, potassium manganate(VII) - iron(II) ammonium sulfate (ferrous ammonium sulfate) standardising titration calculation method, potassium manganate(VII) - ethanedioic acid (oxalic acid) titration calculation method, KMnO4 - potassium tetraoxalate dihydrate titration calculation method, hydrogen peroxide H2O2 KMnO4 potassium manganate (VII) titration calculation method, potassium manganate(VII) KMnO4 - sodium ethanedioate Na2C2O4 titration calculation method, determining water of crystallisation in FeSO4 hydrated iron(II) sulfate, KMnO4 potassium manganate(VII) - nitrite nitrate(III) potassium nitrite titration calculation method, potassium manganate (VII) - iron(II) ethanedioate FeC2O4 titration calculation method, KIO3 potassium iodate iodate(V) - iodine - sodium thiosulfate titration calculation method, analysing iron content of rust or iron ore iron(III) oxide Fe2O3 with potassium manganate(VII) titration calculation method, analysing potassium dichromate dichromate(VI) by oxidising iodide to iodine and titrating iodine with Na2S2O3 sodium thiosulfate keywords–phrases etc.: MnO4–(aq) + 8H+(aq) + 5Fe2+(aq) ==> Mn2+(aq) + 5Fe3+(aq) + 4H2O(l) * 2S2O32–(aq) + I2(aq) ==>  S4O62–(aq) + 2I–(aq) * Cr2O72–(aq) + 14H+(aq) + 6Fe2+(aq) ==> 2Cr3+(aq) + 6Fe3+(aq) + 7H2O(l) * 2MnO4–(aq) + 16H+(aq) + 5C2O42–(aq) ==> 2Mn2+(aq) + 8H2O(l) + 10CO2(g) or  2MnO4–(aq) + 6H+(aq) + 5H2C2O4(aq) ==> 2Mn2+(aq) + 8H2O(l) + 10CO2(g) * 2MnO4–(aq) + 6H+(aq) + 5H2O2(aq) ==> 2Mn2+(aq) + 8H2O(l) + 5O2(g) * 2MnO4–(aq) + 6H+(aq) + 5NO2–(aq) ==> Mn2+(aq) + 5NO3–(aq) + 3H2O(l) * IO3–(aq) + 5I–(aq) + 6H+(aq) ==> 3I2(aq) + 3H2O(l) * Cr2O72–(aq) + 14H+(aq) + 6I–(aq) ==> 2Cr3+(aq) + 3I2(aq) + 7H2O(l) * MnO4– + 8H+ + 5Fe2+ ==> Mn2+ + 5Fe3+ + 4H2O * 2S2O32– + I2 ==>  S4O62– + 2I– * Cr2O72– + 14H+ + 6Fe2+ ==> 2Cr3+ + 6Fe3+ + 7H2O * 2MnO4– + 16H+ + 5C2O42– ==> 2Mn2+ + 8H2O + 10CO2 or  2MnO4– + 6H+ + 5H2C2O4 ==> 2Mn2+ + 8H2O + 10CO2 * 2MnO4– + 6H+ + 5H2O2 ==> 2Mn2+ + 8H2O + 5O2 * 2MnO4– + 6H+ + 5NO2– ==> Mn2+ + 5NO3– + 3H2O * IO3– + 5I– + 6H+ ==> 3I2 + 3H2O * Cr2O72– + 14H+ + 6I– ==> 2Cr3+ + 3I2 + 7H2O advanced level chemistry redox titration questions for AQA AS chemistry, redox titration questions for Edexcel A level AS chemistry, redox titration questions for A level OCR AS chemistry A, redox titration questions for OCR Salters AS chemistry B, redox titration questions for AQA A level chemistry, redox titration questions for A level Edexcel A level chemistry, redox titration questions for OCR A level chemistry A, redox titration questions for A level OCR Salters A level chemistry B redox titration questions for US Honours grade 11 grade 12 redox titration questions for pre-university chemistry courses pre-university A level revision notes for redox titration questions  A level guide notes on redox titration questions for schools colleges academies science course tutors images pictures diagrams for redox titration questions A level chemistry revision notes on redox titration questions for revising module topics notes to help on understanding of redox titration questions university courses in science careers in science jobs in the industry laboratory assistant apprenticeships technical internships USA US grade 11 grade 11 AQA A level chemistry notes on redox titration questions Edexcel A level chemistry notes on redox titration questions for OCR A level chemistry notes WJEC A level chemistry notes on redox titration questions CCEA/CEA A level chemistry notes on redox titration questions for university entrance examinations with advanced level chemistry

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