Electron Transfer Reactions (DP IB Chemistry: SL): Exam Questions

Common household cleaning agents are powerful redox chemicals. Their properties and reactions are explored below.

Common household bleach is a cleaning product which smells like chlorine gas and is therefore, also called chlorine bleach. It contains a mixture of sodium chlorate (NaOCl), sodium chloride and water and can be made by dissolving chlorine gas in a solution of sodium hydroxide.

i) Write a balanced equation with state symbols for this reaction.

[2]

ii) Deduce the oxidation number of chlorine in all of the chlorine-containing reactants and products

[1]

A common safety warning is to never mix bleach with ammonia-based cleaners. This is because a toxic gas, chlorine, and a toxic liquid, hydrazine (N₂H₄) are produced. 

The overall redox reaction for this reaction is shown below. 

2NH₃ (aq) + 2ClO^- (aq) → N₂H₄ (aq) + Cl₂ (g) + 2OH^- (aq)

i) Deduce the oxidation numbers of the nitrogen atom in NH₃ and in N₂H₄.

[1]

ii) Identify the oxidising agent in this reaction? Explain your answer.

[2]

iii) Why is the hazard of the toxic chlorine gas being produced greater than the hazard of hydrazine?

[1]

Due to the risks associated with chlorine-based bleach, alternative bleaches are often used instead. These bleaches are based on peroxides such as hydrogen peroxide.  

Manganate(VII) ions oxidise hydrogen peroxide to oxygen gas. The reaction is carried out with both species under acidic conditions. 

i) Identify the oxidising and reducing agents in this reaction.

[1]

ii) Write the half-equation for the oxidation of hydrogen peroxide to oxygen gas.

[1]

iii) The manganate(VII) ions themselves get reduced to manganese(II) ions. Write down the half-equation for the reduction of manganate(VII) ions.

[1]

iv) Deduce the overall redox equation for this reaction. 

[2]

Explain how the oxidation number of the oxygen atom in H₂O₂ is different from its oxidation state in other compounds.

A student is given three unknown solutions labelled as Solution 1, Solution 2 and Solution 3. They are known to be aqueous solutions of sodium chloride (NaCl), sodium sulfite (Na₂SO₃), and sulfur dioxide (SO₂), but not necessarily in that order. The student performs a series of tests to identify them.

To a sample of Solution 1, the student adds silver nitrate solution, producing a white precipitate of silver chloride.

Explain why this precipitation is not a redox reaction.

The student researches halogen displacement reactions and finds the following reaction:

Cl₂ (aq) + 2Br^- (aq) → 2Cl^- (aq) + Br₂ (aq)

State, with a reason, whether chlorine or bromine is the stronger oxidising agent.

To a sample of Solution 2, the student adds acidified potassium dichromate(VI), and the solution turns from orange to green. The reaction is:

Cr₂O₇^2- (aq) + SO₃^2- (aq) → Cr³⁺ (aq) + SO₄^2- (aq)

i) Deduce the oxidation state of sulfur in the sulfite ion (SO₃^2-) and the sulfate ion (SO₄^2-).

[1]

ii) Hence, identify the reducing agent in this reaction.

[1]

i) The half-equation for the oxidation of sulfur dioxide is:

• SO₂ (aq) + 2H₂O (l) → SO₄^2- (aq) + 4H⁺ (aq) + 2e^-; [1]

ii) The half-equation for the reduction of chlorine is:

• Cl₂ (aq) + 2e^- → 2Cl^- (aq)
  OR
  ½Cl₂ (aq) + e^- → Cl^- (aq); [1] 

The overall redox equation is:

• Cl₂ (aq) + SO₂ (aq) + 2H₂O (l) → 2Cl^- (aq) + SO₄^2- (aq) + 4H⁺ (aq)
  OR
  Cl₂ (aq) + SO₂ (aq) + 2H₂O (l)→ 2HCl (aq) + H₂SO₄ (aq) ; [1] 

[Total: 3 marks]

Part (i):

• Oxidation half-equation for oxidation of SO₂: 

  • SO₂ →  SO₄^2-

• First, balance the O atoms by adding H₂O, and then balance H atoms by adding H⁺ ions

  • SO₂ + 2H₂O →  SO₄^2- + 4H⁺

• Balance the charge by adding two electrons to the right hand-side

  • SO₂ + 2H₂O →  SO₄^2- + 4H⁺ + 2e^-

Part (ii):

• Reduction half-equation for reduction of Cl₂:   

  • Cl₂ → 2Cl^-  

• Balance the charge by adding 2 electrons on the left hand-side 

  • Cl₂ + 2e^- → 2Cl^-  

Part (iii):

• Neither half-equation needs to be multiplied when combining them as they both have 2e

• Cancel the electrons (and anything else that appears on both sides, e.g., H₂O or H⁺)

• Note that two H⁺ ions can combine with SO₄^2- to form H₂SO₄

The rusting of iron is an electrochemical process that involves the oxidation of iron by oxygen in the presence of water. The overall equation can be represented as:

4Fe (s) + 3O₂ (g) + 6H₂O (g) → 4Fe(OH)₃ (s) 

Define reduction in terms of oxidation state.

State, with a reason, the oxidising agent in this reaction in part (a).

A student set up three test tubes to investigate the conditions required for rusting:

Predict the result in each test tube, explaining your reasoning for test tubes B and C.

The electrolysis of molten lead(II) bromide, PbBr₂, is a common laboratory demonstration for extracting a reactive metal.

State two different ways in which electrical charge is carried in the operating electrolytic cell.

Explain, in terms of ions and electrons, the processes occurring at the positive electrode (anode) and the negative electrode (cathode).

Write the half-equations, including state symbols, for the reaction occurring at each electrode.

Draw a simple, labelled diagram of the apparatus required to carry out this electrolysis, showing the direction of electron flow.

The list below shows three metals from the activity series in order of reactivity.

Deduce which of the three metals is the strongest reducing agent.

A voltaic cell can be made by joining two half-cells together, such as  Zn/Zn²⁺ and Ni/Ni²⁺.

i) Write a balanced equation for the overall reaction taking place when the two half-cells are connected together.

[1]

ii) Explain which species is undergoing oxidation.

[1]

Cell diagrams are a way to represent the redox reactions taking place in voltaic cells.

Write the conventional representation for this cell in part (b).

A diagram of the apparatus for the voltaic cell in part (b) is shown below.

Complete the diagram by adding the essential components required for the cell to operate and by showing the direction of electron flow.

Explain the function of the salt bridge in this cell.

Ethene, C₂H₄, can be made into a number of useful compounds. A reaction sequence for this is shown below:

i) State the type of reaction occurring in step 1.

[1]

ii) Write an equation, using structural formulas, for the reaction in step 2 where C₂H₅Cl reacts with aqueous sodium hydroxide to form C₂H₆O.

[1]

The product of step 2 can undergo combustion.

i) Write a balanced equation for the complete combustion of the product of step 2.

[1]

ii) Write a balanced equation for the incomplete combustion of the product of step 2.

[1]

State the reagents and conditions required to carry out step 3.

The product of step 2 has a higher boiling point than the product of step 3.

i) State the names of the products of step 2 and 3.

[1]

ii) Explain, with reference to intermolecular forces, which product has a higher boiling point.

The oxidation of a primary alcohol, ethanol, using acidified potassium dichromate(VI) can produce two different organic products.

Name the two products and state the specific experimental condition required to maximize the yield of each.

Two structural isomers, A and B, with the formula C₅H₁₀O₃ are shown below.

i) State the IUPAC name for each isomer.

[2]

ii) Explain, by classifying the alcohol group in each isomer, why isomer B can be oxidised by acidified potassium dichromate(VI) but isomer A cannot.

[2]

The oxidation of isomer B produces a keto-acid, C₅H₈O₃. The reduction half-equation for the dichromate(VI) ion is:

Cr₂O₇^2- (aq) + 14H⁺ (aq) + 6e^- ⭢ 2Cr³⁺ (aq) + 7H₂O (l)

i) Deduce the half-equation for the oxidation of isomer B.

[1]

ii) Hence, deduce the overall balanced equation for the reaction.

[2]

A student sets up a titration to determine the amount of iron(II) sulfate in an iron tablet. They titrate the iron(II) sulfate solution with potassium manganate(VII) solution.

i) Write the balanced, ionic half equations to show the reduction of the manganate(VII) ion and the oxidation of the Fe²⁺.

[2]

ii) Use your answers to part (i) to write an overall redox equation for the titration of iron(II) sulfate with potassium manganate(VII) solution.

[1]

The iron(II) sulfate solution is acidified before titration to stop the manganate ion forming unwanted manganese dioxide. 

Explain the effect that not acidifying the iron(II) sulfate would have on the final calculation of the estimated mass of iron.

The student dissolved the iron tablet in excess sulfuric acid and made the solution up to 250 cm³ in a volumetric flask. 25.0 cm³ of this solution was titrated with 0.0100 mol dm^-3 potassium manganate(VII) solution. The average titre was found to be 26.65 cm³ of potassium manganate(VII) solution. 

i) Calculate the amount, in moles, of iron(II) ions in the 250 cm³ solution.

[3]

ii) Calculate the mass of iron, in mg, in the tablet. 

[2]

State the oxidation state of phosphorus in the following compounds.

• H₂PO₄^- 

• HPO₃ 

• H₃PO₃

The tetrathionate ion is shown below:

i) Determine the oxidation state of sulfur in the ion.

[1]

ii) Justify your answer to part (i).

[1]

Sodium tetrathionate can be formed by reacting sodium thiosulfate, Na₂S₂O₃, with iodine.

i) State the balanced symbol equation for this reaction.

[2]

ii) Identify the oxidising agent in this reaction.

[1]

State the expected observation as the reaction in part (c) proceeds to completion

A solution of ethanedioic acid, H₂C₂O₄ (aq), was analyzed by two different titration methods.

In an acid-base titration, a 15.00 cm³ sample of ethanedioic acid solution required 10.30 cm³ of a 0.250 mol dm^-3 NaOH (aq) for complete neutralisation.

The equation is:

H₂C₂O₄ (aq) + 2NaOH (aq) → Na₂C₂O₄ (aq)+ 2H₂O (l) 

Calculate the concentration of the ethanedioic acid solution.

In a redox titration, a separate 15.00 cm³ sample of the same ethanedioic acid solution was acidified and titrated with potassium manganate(VII) solution, KMnO₄ (aq). The products were Mn²⁺ (aq) and CO₂ (g).

i) Deduce the balanced half-equation for the oxidation of ethanedioic acid.

[1]

ii) Deduce the overall balanced ionic equation for the redox reaction.

[2]

The redox titration required 12.35 cm³ of the KMnO₄ solution.

Using your answers from parts (a) and (b), calculate the concentration of the potassium manganate(VII) solution.

Aqueous sodium tetrahydridoborate, NaBH₄, is a common reducing agent.

State the IUPAC name of the two isomers with the formula C₃H₆O that can be reduced by aqueous NaBH₄.

State the IUPAC name of two non-cyclic isomers with the formula C₃H₆O that cannot be reduced by aqueous NaBH₄.

A student states that when the following compound is reduced using NaBH₄, the carboxylic acid group will be reduced to a primary alcohol.

Explain whether the student is correct.