LabMouse Chemistry A2 | Redox Calculations AQA 3.5.4 | Review Questions
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Kinetics AQA 3.4.1
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Thermodynamics AQA 3.5.1
Redox Equilibria AQA 3.5.3
Redox Calculations AQA 3.5.4
Redox Calculations
Redox Titrations
Practical: Standardisation of Potassium Manganate(VII) with Sodium Ethandioate
Practical: Determination of the Percentage of Iron in Iron Tablets
Practical: Standardisation of Sodium Thiosulfate with Potassium Manganate(VII)
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The following questions are a review section where you can test your understanding of the material covered. Try to answer each question for yourself before looking at the answers. If you are unsure of any answer, links are provided to the appropriate section in the study material - you can return to the question page through your browser's back button.
The reduction of Cr2O72- ions by zinc and hydrochloric acid goes through the colour changes from orange to green and finally to blue. Give the oxidation state of the chromium ions in each of these coloured solutions.
Molybdenum is the element below chromium in the Periodic Table. Cerium(IV) ions oxidise molybdenum(II) ions in aqueous solution to molybdenum(VI) ions according to the simplified equation below which shows the changes in oxidation states.
4Ce(IV) + Mo(II) → 4Ce(III) + Mo(VI)
1.300g of an ore of molybdenum were dissolved in dilute sulfuric acid to give a solution containing molybdenum(VI) ions. This solution was treated with an excess of zinc so that all the molybdenum was reduced to the molybdenum(II) state.
The excess of zinc was filtered off. The molybdenum(II) solution was then titrated with a 0.100M cerium(IV) solution when 25.0cm3 of the cerium(IV) solution were required to oxidise all the molybdenum(II) back to molybdenum(VI).
Oxidation state in orange solution: +6 Oxidation state in green solution: +3 Oxidation state in blue solution: +2
Click here to see the answer. Click here to go to the study material (Redox Calculations).
Given a solution of ammonium vanadate(V), NH4VO3, describe an experiment which demonstrates that vanadium has several oxidation states.
Potassium manganate(VII) solution oxidises all vanadium species to the vanadium(V) state.
When 0.234g of ammonium vanadate(V) was dissolved in dilute sulfuric acid and the solution treated with an excess of sulfur dioxide, a blue solution was formed. After the excess of sulfur dioxide was removed by boiling, the blue solution decolourised 20.0cm3 of 0.0200M KMnO4 solution.
Use the data to calculate the oxidation state of vanadium in the blue solution.
Dissolve solid in HCl(aq) ⇒ yellow solution, V(V). Add zinc and colour changes ⇒ blue, V(IV) ⇒ green, V(III) ⇒ lilac, V(II).
moles KMnO4 `= 20/1000 x 0.02 = 4.0x10^-4`
moles of e- `= 4.0x10^-4 x 5 = 2.0x10^-3`
moles NH4VO3 `= 0.234/117 = 2.0x10^-3`
change in oxidation state `= {2.0x10^-3}/{2.0x10^-3} = 1`
Hence oxidation state = +4
A 0.500g sample of steel was dissolved in hydrochloric acid to produce a solution of iron(II) ions. When this was titrated with 0.0500M potassium dichromate(VI) solution, 29.7cm3 were required to oxidise all the iron(II) to iron(III).
moles Cr2O72- `= 29.7/1000 x 0.05`
moles Fe2+ `= 29.7/1000 x 0.05 x 6 = 8.91x10^-3`
mass Fe `= 8.91x10^-3 x 56 = 0.499"g"`
% Fe `= 0.499/0.500 x 100 = 99.8%`
Potassium manganate(VII), KMnO4, can be used in the quantitative estimation of ethanedioate ions, C2O42-, in an acidified aqueous solution. In this reaction, ethanedioate ions are converted into carbon dioxide. Deduce half equations for the redox processes involved and hence derive an equation for the overall reaction.
A 1.93g sample of a crystalline ethanedioate salt was dissolved in water and made up to 250cm3. 25.0cm3 of this solution, after acidification, was found to react with 30.4cm3 of 0.0200M KMnO4. Calculate the percentage by mass of ethanedioate ions in the original salt.
C2O42- → 2CO2 + 2e- MnO4- + 8H+ + 5e- → Mn2+ + 4H2O
2MnO4-(aq) + 16H+(aq) + 5C2O42-(aq) → 2Mn2+(aq) + 8H2O(l) + 10CO2(g)
moles MnO4- `= 30.4/1000 x 0.02`
moles C2O42- `= 30.4/1000 x 0.02 x 5/2 = 1.52x10^-3`
In 250cm3 moles C2O42- `= 1.52x10^-2"mol."`
In 250cm3 mass C2O42- `= 1.52x10^-2 x 88 = 1.34"g"`
% C2O42- `= 1.34/1.93 x 100 = 69.4%`
Click here to see the answer. Click here to go to the study material (Practical: Standardisation of Potassium Manganate(VII) with Sodium Ethandioate).
In order to estimate the iron content of a ferrochrome alloy, a 0.300g sample of the alloy was dissolved in hydrochloric acid. The resulting solution was titrated with 0.0180M potassium dichromate(VI) solution when 28.0cm3 were required to oxidise all the iron(II).
Write half-equations to show the oxidation of iron(II) and the reduction of dichromate(VI) in this titration. Combine these to give an overall equation for the reaction occurring.
Calculate the number of moles of Cr2O72- used in this titration.
Calculate the number of moles of Fe2+ present in the solution.
Calculate the mass of Fe2+ present in the solution.
Calculate the percentage of iron in the alloy.
Fe2+ → Fe3+ + e- Cr2O72- + 14H+ + 6e- → 2Cr3+ + 7H2O
Cr2O72-(aq) + 14H+(aq) + 6Fe2+(aq) → 2Cr3+(aq) + 7H2O(l) + 6Fe3+(aq)
moles Cr2O72- `= 28.0/1000 x 0.018 = 5.04x10^-4`
moles Fe2+ `= 5.04x10^-4 x 6 = 3.024x10^-3`
mass Fe `= 3.024x10^-3 x 56 = 0.1693"g"`
% Fe `= 0.169/3 x 100 = 56.3%`