* Advanced Level Volumetric Analysis - Redox Titration Revision QUESTIONS *

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Basic Chemistry Calculations Index Doc Brown's Chemistry  Quantitative redox reaction analysis

REDOX VOLUMETRIC TITRATION CALCULATIONS

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

Revision notes for GCE Advanced Subsidiary Level AS Advanced Level A2 IB Revise AQA GCE Chemistry OCR GCE Chemistry Edexcel GCE Chemistry Salters Chemistry CIE Chemistry revising courses for pre-university students (equal to US grade 11 and grade 12 and Honours/honors level courses)

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,

Equations needed * FULL ANSWERS and WORKING * REDOX REACTION THEORY * Qualitative Analysis

and also acid-base and other non-redox titrations Questions * EMAIL query?comment

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-ethanedioc acid (oxalate, oxalic acid)/hydrogen peroxide/sodium nitrite titrations, Q2/14/17 on sodium thiosulphate-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  (being written))

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

(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 0.02 mol dm-3 KMnO4 reacted with 20.0 cm3 of an iron(II) solution.

(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 sulphuric acid and the solution made up to 250 cm3 in a standard graduated flask.  25.0 cm3 of this solution was pipetted into a conical flask and needed 25.45 cm3 of  O.02 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.

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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 thiosulphate ion S2O32-,and iodine.

(b) what mass of iodine reacts with 23.5 cm3 of 0.012 mol dm-3 sodium thiosulphate solution.

(c) 25cm3 of a solution of iodine in potassium iodide solution required 26.5 cm3 of 0.095 mol dm-3 sodium thiosulphate solution to titrate the iodine.

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

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Question 3: 2.83 g of a sample of haematite iron ore [iron (III) oxide, Fe203] were dissolved in concentrated hydrochloric acid and the solution diluted to 250 cm3.

25.00 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 required 26.4 cm3 of 0.02 mol dm-3 potassium dichromate(VI) for oxidation.

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

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

(c) Why isn't potassium manganate(VII) used for this titration? (as in Q1)

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Question 4: An approximately 0.02 mol dm-3 potassium manganate(VII) solution was standardized against precisely 0.1 mol dm-3 iron(II) ammonium sulphate solution. 25.00 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 ?

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Question 5: 10.0 g of iron(II) ammonium sulphate crystals were made up to 250 cm3 of acidified aqueous solution. 25 cm3 of this solution required 21.25 cm3 of 0.02 mol dm-3 potassium dichromate(VI) for oxidation.

Calculate x in the formula FeSO4.(NH4)2S04.xH2O

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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 ethane-dioic acid).

(b) 1.520 g of ethanedioic acid crystals, H2C2O4.2H2O, was made up to 250 cm3 of aqueous solution and 25.0 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.

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

0.15 g of potassium tetroxalate (tetraoxalate?), KHC2O4.H2C2O4.2H2O needed 23.2 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.

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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 cm3 of solution of hydrogen peroxide were diluted to 1 dm3 with water. 25.0 cm3 of this solution, when acidified with dilute sulphuric acid, reacted with 20.25 cm3 of 0.02 mol dm-3 KMnO4.

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

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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 sulphuric acid. The mixture was filtered and the filtrate and washings made up to 500 cm3 in a standard volumetric flask. 20.0 cm3 of this solution required 26.5  cm3 of 0.01 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.

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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 24.5 cm3 of a potassium manganate(VII) solution containing 0.05 mol dm-3.

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

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

25.0 cm3 of this solution required 43.85 cm3 of 0.01 mol dm-3 KMnO4 for complete oxidation.

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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 cm3 of an acidified 0.025 mol dm-3 KMnO4 solution.

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

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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 0.02 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.

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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 0.012 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 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 0.05 mol dm-3 sodium thiosulphate 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?

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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 sulphuric acid and required 22.5 cm3 of 0.02 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.

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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 sulphuric 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 0.020 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 25.0 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 0.10 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.

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Question 17: 25.0 cm3 of an iodine solution was titrated with 0.1 mol dm-3 sodium thiosulphate solution and the iodine reacted with 17.6 cm3 of the thiosulphate 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.

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Question 18: 1.01g of an impure sample of potassium dichromate(VI), K2Cr2O7, was dissolved in dil. sulphuric acid and made up to 250 cm3 in a calibrated volumetric flask. A 25 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 0.1 mol dm-3 sodium thiosulphate 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 thiosulphate ion and iodine.

(c) Calculate the moles of sodium thiosulphate 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 cm3 aliquot titrated.

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

 

Basic Chemistry Calculations Index

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(spanish) Análisis cuantitativo reacción redox CÁLCULOS VALORACIÓN VOLUMÉTRICA Valoraciones y cálculos basados en técnicas de reducción- oxidación-reacciones - solved problemas Preguntas manganato de potasio 1/4/6/7/8/9/10/11/12/13/15/16 base (VII)-hierro (II) /-ethanedioc ácido ethanedioate (oxalato, ácido oxálico) / peróxido de hidrógeno / nitrito de sodio titulaciones, Q2/14/17 de sodio-yodo de valoración de tiosulfato, Q3 / 5 sobre el dicromato de potasio (VI) de hierro (II) la titulación, además de Q será añadido - sugerencias? * (thai) ปฏิกิริยาเชิงปริมาณการวิเคราะห์ปฏิกิริยาเคมี การคำนวณการไทเทรตรีดอกซ์ปริมาตร Titrations และคำนวณตามออกซิเดชันลดเทคนิคปฏิกิริยา -- solved ปัญหา หมายเหตุ : ครึ่ง ปฏิกิริยาที่ยกมามักจะเป็นเซลล์สมการลดลงครึ่ง . ครึ่งเซลล์ซ้ำหรือสมการในข้อต่อจากคำถามก่อนหน้านี้ จะถือว่าคุณจะทำงานผ่านพวกเขาในคำถามตัวเลข . ถ้าคุณไม่สามารถทำงานออกสมการรีดอกซ์คุณก็สามารถ ดาวน์ โหลดสมการ เพื่อให้คุณสามารถอย่างน้อยการที่'บริสุทธิ์'ด้านการคำนวณปริมาตรของคำถาม คำถามโพแทสเซียม manganate ตาม 1/4/6/7/8/9/10/11/12/13/15/16 (VII) เหล็ก (II) / - ethanedioc กรด ethanedioate (ออกซาเลต, กรดออกซาลิ) ไฮโดรเจนเพอร์ออกไซด์ / titrations ไนไตรท์โซเดียม Q2/14/17 ในโซเดียมไทเทรตไอโอดีน thiosulphate, Q3 / 5 บนโพแทสเซียมไดโครเม (VI) เหล็ก (II) ไทเทรต, เพิ่มเติม Q จะถูกบันทึก -- คำแนะนำ? รีดอกซ์ - ionic ตรวจสอบสมการ ของ'ชนิดถูกต้อง'(เช่นจะให้เลือกสอง / เป็น / จากเซลล์ครึ่งข้อมูลสม) T'ทิศทางชนิดเปลี่ยน'เขา -- ซึ่งเป็น oxidised หรือลดลง? ในคำอื่น ๆ ที่ได้รับสมครึ่งเซลล์ทางขวารอบ! อัตราส่วนที่ถูกต้องของ - สมเซลล์ครึ่ง -- ยอด'จะต้องขึ้นอยู่กับการวิเคราะห์เลขออกซิเดชันหรือจำนวนอิเล็กตรอนโอน เพิ่มขึ้นทั้งหมดในรัฐ oxidation = ลดลงทั้งหมดใน รัฐออกซิเดชันหรืออิเล็กตรอนรวม = ได้รับอิเล็กตรอนทั้งหมดหายไปโดยชนิดที่เกี่ยวข้อง dd ขึ้นค่า ion ที่ผลรวมจะเหมือนกันทั้งสองด้านของสมการ (I หานี้ตรวจสอบเพิ่มประโยชน์โดยเฉพาะอย่างยิ่งกับหลง H2O หรือ +'s H!) 'แบบนับอะตอม'-- วาง last เพราะสมบูรณ์ของสมการไม่น่าเชื่อถือกับปฏิกิริยา! Q ในบางสมการของการอาจได้รับหรือสองสมเซลล์ครึ่งจะใส่กันขวาตลอดทางและในอัตราส่วนขวา (ดูปฏิกิริยาเคมี Part 2 (การเขียน) * (swedish) Doc Browns kemi Kvantitativ redox reaktion analys Redox Volymetrisk TITRERING BERÄKNINGAR Titreringarna och beräkningar baserade på oxidation minska teknik-reaktionerna - solved problems OBS: Hälften reaktioner brukar anges som den halva cellen minskning ekvation. Återanvändning halv-cell eller hel ekvationer i senare frågor från tidigare frågor. Det antas att du kommer att arbeta igenom dem i numerisk fråga ordning. Om du inte kan räkna ut redox ekvationer, kan du bara ladda ner ekvationer så att du åtminstone kan utöva "ren" volymetriska beräkningar aspekter av frågorna. Questions 1/4/6/7/8/9/10/11/12/13/15/16 based potassium manganate(VII)-iron(II)/ethanedioate-ethanedioc acid (oxalate, oxalic acid )/hydrogen peroxide/sodium nitrite titrations, Q2/14/17 on sodium thiosulphate-iodine titration, Q3/5 on potassium dichromate(VI)-iron(II) titration, further Q's will be added - suggestions? Frågor 1/4/6/7/8/9/10/11/12/13/15/16 baserade kalium manganate (VII)-järn (II) / ethanedioate-ethanedioc syra (oxalsyra, oxalsyra) / väteperoxid / natriumnitrit titreringarna, Q2/14/17 på natriumtiosulfat-jod titrering, Q3 / 5 på kaliumdikromat (VI) järn (II) titrering ytterligare Q's kommer att läggas till - förslag? Redox-joniska JÄMSTÄLLANDE KONTROLLER Use de korrekta "art" (oftast två ges / valts / från halv-cell ekvation data) D et "arter" riktning förändring - som oxideras eller reduceras? Med andra ord får halv-cellen ekvationer åt rätt håll! (Om inte anges, kan ha att avgöra från E Ø uppgifter, desto mer + ve halv-cell är en minskning (oxiderande agent). Mer jämvikter del 7 (skrivs) Den korrekta förhållandet mellan halv-cell ekvationer - de "balans" måste grundas på oxidationstal analys eller antalet överförda elektroner. Den totala ökningen Oxidationstillstånd = den totala minskningen av oxidation stater eller total elektroner vunnits = totala elektroner förloras av de berörda arterna. En dd uppför jon avgifter, uppgår borde vara samma på båda sidor av ekvationen (tycker jag är en händig extra check särskilt med herrelösa H2O eller H+ har!)."Traditionella" atom räknas - placeras sist eftersom dess inte helt tillförlitlig med redox ekvationer? I vissa F: s hela ekvationen kan ges eller två halv-cells ekvationer som skall sätta ihop åt rätt håll och i rätt förhållande (se Redox kemi del 2 (skrivs)) *  
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