Redox Reactions

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REDOX Reaction

Both oxidation and Reduction going on side-by-side

Combustion

Oxidation that occurs so rapidly that noticeable heat and light are produced; burning.

Incomplete combustion

- Combustion that occurs in such a way that fuel is not completely oxidized ("burned up").
- The incomplete combustion of carbon-containing fuels (such as coal and oil) always results in the formation of some carbon monoxide.
- The relative amounts of carbon monoxide or carbon dioxide that form during combustion depend on two factors
- The amount of oxygen present
- the combustion temperature

When a large supply of oxygen is present and when the combustion temperature is high, carbon dioxide is more likely to be formed.
- With limited supplies of oxygen and at lower temperatures, carbon monoxide is produced.

Carbon monoxide

Used in industry primarily as a source of energy and as a reducing agent.
As a reducing agent, carbon monoxide is used to convert the naturally occurring oxide of a metal to the pure metal.
- When carbon monoxide is passed over hot iron oxides, for example, the oxides are converted to metallic iron.

Carbon

Essentially carbon acts as a reducing agent as well as the carbon monoxide that is inevitably formed by its heating in air.

DEFINITIONS OF OXIDATION AND REDUCTION (REDOX)

Oxidation and reduction in terms of :

A. Oxygen transfer

Definition

· Oxidation is gain of oxygen.

· Reduction is loss of oxygen.

For example, in the extraction of iron from its ore:

Oxidising agent

Reducing agent

- An oxidising agent is substance which oxidises something else.

- Oxidising agents give oxygen to another substance.

- In the above example, the iron(III) oxide is the oxidising agent.

- A reducing agent reduces something else.

- Reducing agents remove oxygen from another substance.

- In the equation, the carbon monoxide is the reducing agent.

B. Hydrogen transfer

Definition

For example, ethanol can be oxidised to ethanal:

Oxidizing agent

· Oxidation is loss of hydrogen.

· Reduction is gain of hydrogen.

Reducing agent

- You would need to use an oxidising agent to remove the hydrogen from the ethanol.

- A commonly used oxidising agent is potassium dichromate(VI) solution acidified with dilute sulphuric acid.

- Ethanal can also be reduced back to ethanol again by adding hydrogen to it.

- A possible reducing agent is sodium tetrahydridoborate, NaBH₄.

C. Electron transfer

Definition

How to remember…

A simple example

· Oxidation is loss of electrons.

· Reduction is gain of electrons.

The equation shows a simple redox reaction which can obviously be described in terms of oxygen transfer.

Copper(II) oxide and magnesium oxide are both ionic. The metals obviously aren't. If you rewrite this as an ionic equation, it turns out that the oxide ions are spectator ions and you are left with:

Or you could think it out like this:

How to visualize…

Oxidizing and Reducing agents

- If you look at the equation above, the magnesium is reducing the copper(II) ions by giving them electrons to neutralise the charge.

- Magnesium is a reducing agent.

- Looking at it the other way round, the copper(II) ions are removing electrons from the magnesium to create the magnesium ions.

- The copper(II) ions are acting as an oxidising agent.

· An oxidising agent oxidises something else.

· Oxidation is loss of electrons (OIL RIG).

· That means that an oxidising agent takes electrons from that other substance.

· So an oxidising agent must gain electrons.

· An oxidising agent oxidises something else.

· That means that the oxidising agent must be being reduced.

· Reduction is gain of electrons (OIL RIG).

· So an oxidising agent must gain electrons.

Oxidation State in Reduction-Oxidation reaction

Oxidation State is the charge an atom would have if it existed as an ion

To work out oxidation state, the rules are:

- Free elements have oxidation state zero, e.g. Cu, Fe, N₂
- Oxidation of an ion is the charge of the ion
  - e.g. Na⁺ = +1, Cu²⁺ = +2, O^2- = -2
- The oxidation state of some elements in their compounds is fixed, e.g.
  - Group I Elements = +1
  - Group II Elements = +2
  - Hydrogen in most compounds = +1
  - Iron or copper can have either +1, +2, +3, so it’s not fixed
- Oxidation states of the elements in a compound adds up to zero, e.g.
  - NaCl: (+1) + (-1) = 0
  - K₂O: (+1) x 2 + (-2) = 0
  - Al₂O₃: (+3) x 2 + (+2) x 3 = 0
- Sum of oxidation states of elements in an ion is equal to charge on the ion,
  - e.g. OH^-: (-2) + (+1) = -1

Examples:

Work out the oxidation states of the underlined elements in these compounds:

(a) CO₂

(oxidation state of C) + (-2) x 2 = 0

(oxidation state of C) + (-4) = 0

Oxidation state of C = +4

(b) KMnO₄

(+1) + (oxidation state of Mn) + (-2) x 4 = 0

(oxidation state of Mn) + (+1) + (-8) = 0

(oxidation state of Mn) + (-7) =0

Oxidation state of Mn = +7

(c) Fe(NO₃)₂

(oxidation state of Fe) + (-1) x 2 = 0

(oxidation state of Fe) + (-2) = 0

Oxidation state of Fe = +2

Examples of elements with variable oxidation states

Some compounds with possible variable oxidation states have roman numeral as a guide about their oxidation state, e.g.

- Iron(II) chloride has formula FeCl2 and iron oxidation state +2
- Potassium(VI) dichromate has formula K₂Cr₂O₇ and potassium oxidation state +6
- Manganese(IV) oxide has formula MnO2 and manganese oxidation state +4

Oxidation is the increase of oxidation state by a substance
- - Losing electrons means gain in oxidation state
Reduction is the decrease of oxidation state by a substance
- - Gaining electrons means loss in oxidation state

Example

A. Metals with acids

Cu (s) + HCl (aq) → CuCl₂(aq) + H₂(g)

- Cu is oxidized as it gains oxidation state from 0 to +2. Cu is reducing agent
- H⁺ions in HCl reduced as it loses oxidation state from +1 to 0. H⁺ions are oxidising agent

B. Halide (Halogen) Displacement Reactions

Cl₂(aq) + 2KI (aq) → 2KCl (aq) + I₂ (aq)

- I^- ions in KI oxidized as it gains oxidation state from -1 to 0. I-ions is reducing agent
- Cl₂ is reduced as it loses oxidation state from 0 to -1. Cl₂ is the oxidizing agent

Test for Oxidising/Reducing Agents

Non-Redox Reactions

- Decomposition of carbonates by heat: CaCO₃(s) → CO₂(g) + CaO (s)
  - The oxidation state of each element don’t change.
- Neutralization: NaOH (aq) + HCl (aq) → NaCl (aq) + H₂O (l)
  - The oxidation state of each element don’t change.
- Precipitation reactions: Ag⁺(aq) + Cl^-(aq) → AgCl (s)
  - The oxidation numbers of silver and chloride ions unchanged.

Reversible Reactions

- Reversible reactions are denoted by the sign “⇌” where the arrow --> denotes forward reaction, where reactants react to form products, and the arrow <-- denotes backward reaction where products decompose to reform reactants.
- The reactions occur at the same time.
  - E.g. N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

Effect of Temperature on Reversible Reactions

- With higher temperature, the condition is now favored to break the bonds of the product formed (The bonding of products requires low temperatures).
- Thus, the products decompose to its constituents, leading to backward reaction.

Effect of Pressure on Reversible Reactions

- Increase in pressure encourages forward reaction because the higher pressure the more reactants collide to react.

Dynamic equilibrium

- Dynamic Equilibrium is the state when the rate of forward reaction is the same as the rate of backward reaction.
- Both reactants are reacted and products decompose at the same rate.
- Hence, there is no overall change in the amounts of reactants and products.
- When we remove the products, it will also encourage forward reaction as the reaction would try to achieve equilibrium.
- Similar thing happens when we remove the reactants, that the decomposition of products is encouraged to reach the point

MCQ Questions

1. Which one of the following is usually described as a reducing agent?

a. carbon monoxide

b. concentrated sulphuric acid

c. copper(II) oxide

d. Sodium oxide

2. Cu⁺ ions may be formed when

a. Cu²⁺ ions are oxidised by sulphur dioxide

b. Cu²⁺ ions are reduced by copper

c. Cu is oxidised by nitric acid

d. Zinc is placed in copper sulphate solution

3. An oxidisng agent was observed to turn from orange to green. It is most likely to be:

a. acidified potassium manganate (VII)

b. acidified potassium dichromate (VI)

c. maganese (IV) oxide

d. iron (III) chloride

4. When iron (II) oxide is heated strongly with carbon

a. iron ions gain 2 electrons

b. carbon is reduced

c. carbon is acting as the oxidising agent

d. only carbon monoxide is formed

5. When iron (III) sulphate is reduced to iron (II) sulphate, the colour of the solution turns from

a. brown to yellow

b. green to yellow

c. red to colourless

d. brown to green

6. In which of the following reactions is the underlined substance reduced?

a. C_l2 + 2I^- –> 2Cl^- + I₂

b. Zn + _H2SO₄ –> ZnSO₄ + H₂

c. Fe²⁺ + H₂O₂–> Fe³⁺ + H₂O

d. CuO + H₂ –> Cu + H₂O

7. Which conversion involves the smallest change in oxidation number of the underlined element?

a. C(s) –> CO₂

b. NO₃^-(aq) –> NO₂(g)

c. SO₃^2-(aq) –> SO₄^2-(aq)

d. MnO₄^-(aq) –> Mn²⁺(aq)

8. The oxidation states of chlorine in ClO4- and Cl2O are respectively

a. +1 and +3

b. +3 and +7

c. +7 and +1

d. +7 and +2

9. Which statement is true about oxidizing agents?

a. Their oxidation state is zero

b. They are easily oxidized

c. They never contain hydrogen

d. They readily accept electrons

10. What does an oxidizing agent do?

a. It turns acidified potassium dichromate (VI) green

b. It turns potassium manganate (VII) colorless

c. It turns Universal Indicator red

d. It turns aqueous potassium iodide brown

11. When zinc reacts with dilute sulphuric acid a gas is released. What happens to the zinc and what is the gas released?

12. Which compound, when added to aqueous iron(II) sulphate, takes part in a redox reaction?

a. ammonia

b. barium chloride

c. potassium manganate(VII)

d. sodium hydroxide

13. Which of the following is a property of aqueous potassium iodide?

a. it does not conduct electricity

b. it is decolorised by chlorine

c. it reacts with aqueous bromine to form iodine

d. it is a purple solution

14. What happens when a copper atom becomes a copper(II) ion?

a. it is oxidised by losing two electrons

b. it is oxidised by gaining two electrons

c. it is reduced by gaining two electrons

d. it is reduced by losing two electrons

15. Which reaction is an example of a redox reaction?

a. CuO + H₂SO₄--> CuSO₄ + H₂O

b. H⁺ + OH^- --> H₂O

c. Ag⁺ + Cl^- --> AgCl

d. 2SO₄ + O₂ --> 2SO₃

16. In which pair is the underlined element in the same oxidation state in both compounds?

a. CuCl₂ and NaCl

b. H₂S and SO₂

c. Fe₂O₃and FeSO₄

d. MnO₂ and MnCl₂

17. Solution X turns acidified potassium dichromate(VI) from orange to green. What must solution X contain?

a. an alkali

b an ammonium salt

c. an oxidising agent

d. a reducing agent

18. In which oxide does X have the same oxidation state as in the chloride, XCl₃?

a. X₃O

b. X₂O

c. XO₂

d. X₂O₃

19. In which of the following changes is the nitrogen reduced?

a. NH₃ to NO

b. NH₃ to NO₃^-

c. N₂ to NH₃

d. N^3- to N₂

20. When gas X is passed over heated copper(II) oxide the products are copper and steam only. What is gas X?

a. ammonia

b. carbon monoxide

c. hydrogen

d. methane

21. Which element in the reaction below is oxidised?

2FeSO₄ + Cl₂ + H₂SO₄ --> Fe₂(SO₄)₃ + 2HCl

a. chlorine

b. hydrogen

c. iron

d. sulphur

22. In which reaction does the oxidation state of iron remain unchanged?

a. 2Fe + 3Cl₂ --> 2FeCl₃

b. 2FeCl₂ + Cl₂ --> 2FeCl₃

c. Fe + 2FeCl₃ --> 3FeCl₂

d. Fe₂O₃ + 6HCl --> 2FeCl₃ + 3H₂O

23. Titanium is manufactured from a mixture of iron(II) titanate, FeTiO₃, and iron(III) titanate, Fe(TiO₃)₃. What are the oxidation numbers of titanium in these 2 compounds?

a. +2, +2

b. +4, +4

c. +2, +3

d. +4, +2

24. Which of the following does not represent a redox reaction?

a. Sn²⁺ (aq) + 2Fe³⁺ (aq) ---> Sn⁴⁺ (aq) + 2Fe²⁺ (aq)

b. Ca (s) + 2HCl (aq) ---> CaCl₂ (aq) + H₂ (g)

c. Ag⁺ (aq) + Cl^- (aq) ---> AgCl (s)

d. Zn (s) + 2H⁺ (aq) ---> Zn²⁺ (aq) + H₂ (g)

25. In which pair is the underlined element in the same oxidation state in both compounds?

a. CuCl₂ and NaCl

b. H₂S and SO₂

c. Fe₂O₃ and FeSO₄

d. MnO₂ and MnCl₂

26. Which compound, when added to aqueous iron(III) sulphate, takes part in a redox reaction?

a. ammonia

b. barium chloride

c. potassium manganate (VII)

d. sodium hydroxide

27. Solution X turns acidified potassium dichromate(VI) from orange to green. What must solution X contain?

a. alkali

b. ammonium salt

c. oxidising agent

d. reducing agent

Answers

1. a

2. b

3. b

4. a

5. d

6. d (CuO + H₂ –> Cu + H₂O)

7. b (NO₃^- (aq) –> NO₂(g))

8. c (+7 and +14)

9. d (They readily accept electrons)

10. d (It turns aqueous potassium iodide brown)

11. a

12. c

13. c

14. a

15. d (the oxidation number of S increased from +4 in SO₂ to +6 in SO₃. The oxidation number of oxygen decreased from 0 in O₂ to -2 in SO₃)

16. a (the oxidation state of chlorine in both is -1)

17. d

18. d

19. c

20. c (H₂ reduces CuO to Cu while itself is oxidised to H₂O)

21. c

22. d (the oxidation number of Fe in iron(III) oxide and iron(III) chloride is +3)

23. b

24. c

25. a (both have an oxidation state of -1)

26. c (this is a good oxidising agent in an acidic medium)

27. d

Structured Questions

1a. What is the meaning of (II) in copper(II) oxide

1b. In each of the following, underline the formula of the substance that is being oxidised. Give a reason in each case.

i. 2CO + O₂ ---> 2CO₂

ii. 2NH₃ + 3CuO ---> N₂ + 3Cu + 3H₂O

iii. 2Fe²⁺ + Cl₂ ---> 2Fe³⁺ + 2Cl^-

Solution

1a. It refers to the valency of the metal copper

1bi. 2CO + O₂ --> 2CO₂

reason: addition of oxygen to the compound

1bii.2NH₃ + 3CuO --> N₂ + 3Cu + 3H₂O

reason: removal of hydrogen from the compound

1biii. 2Fe²⁺ + Cl₂ --> 2Fe³⁺ + 2Cl^-

reason: increase in valency from +2 to +3 for the iron ion

2a. Define oxidation in terms of electron transfer

2b. Name a substance which is an oxidizing agent in aqueous solution. Explain how aqueous potassium iodide can be used to confirm that this named substance is an oxidizing agent.

Solution

2a. Oxidation is the transfer of electron from one particle to another. The particle that loses the electron is said to be oxidised.

2b. Chlorine is an oxiding agent. Aqueous chlorine is added to aqueous potassium iodide with starch added. A dark blue color is seen, indicating the presence of I₂. So I^- must have been oxidised to I₂.

3. Name 5 different salts which can be prepared using only the substances dilute sulphuric acid, dilute ethanoic acid, aqueous ammonia and potassium hydroxide.

Solution

- potassium sulphate

- potassium ethanoate

- ammonium sulphate

- ammonium ethanoate

- potassium hydrogensulphate

4. By giving either the name or the formula, identify each of the lettered substances in the following:

a. when an excess of aqueous chlorine was added to aqueous potassium iodide, a black solid, P was formed. What is P?

b. Q is a sodium salt. When a solution of Q was added to aqueous silver nitrate, a white precipitate, R was formed. This precipitate did not dissolve in dilute nitric acid. What is Q and R?

c. S is a sodium salt. When a solution of S was added to aqueous barium nitrate, a white precipitate, T was formed. When dilute nitric acid was added to T, it dissolved and a gas, U which turned limewater milky was evolved. What is S, T, and U?

d. When aqueous sodium hydroxide was added to aqueous zinc chloride, a white precipitate, V was formed. This precipitate dissolved in an excess of sodium hydroxide to form a colorless solution in which zinc was present as a salt, W. What is V and W?

Solution

4a. P = iodine, I₂

Cl₂ oxidises I^-to I₂.

Cl₂ + 2I^- --> 2Cl^- + I₂

4b. Q = sodium chloride, NaCl

R = silver chloride, AgCl

Cl^- + Ag⁺ --> AgCl

4c.

S = sodium carbonate, Na₂CO₃

T = Barium carbonate, BaCO₃

U = carbon dioxide, CO₂

Since the white precipitate dissolves in dilute HNO₃, the anion cannot be SO₄^2- because BaSO₄ is insoluble in dilute HNO₃

Ba²⁺ + CO₃^2- --> BaCO₃

BaCO₃ + 2H⁺ --> Ba²⁺ + H₂O + CO₂

4d. V = zinc hydroxide, Zn(OH)₂

W = sodium zincate, Na₂Zn(OH)₄ or Na₂ZnO₂

Zn²⁺ + 2OH^- --> Zn(OH)₂

Zn(OH)₂ + 2OH^---> Zn(OH)₄^2- or ZnO₂^2- + 2H₂O

5. Aqueous copper(II) sulphate reacts with aqueous potassium iodide according to the equation below.

2Cu²⁺ + 4I^- --> 2CuI + I₂

a. Identify the reducing agent in this reaction. Explain your answer

b. Describe briefly how acidified potassium manganate(VII) can be used to test for a reducing agent

Solution

5a. Aqueous potassium iodide. Iodide ion gives away electrons and become oxidised to iodine molecules. The oxidation state of iodine increased from -1 in I^- to 0 in I₂ indicating that iodide is oxidised.

b. Addition of purple acidified potassium manganate(VII) solution to a reducing agent causes the purple solution to decolorise.

6. Aqueous iron(II) ions react with acidified potassium manganate(VII) according to the equation below

5Fe²⁺ + 8H⁺ + MnO₄^- --> 5Fe³⁺ +4H₂O +Mn²⁺

a. What is the reducing agent in this reaction? Explain your answer.

b Briefly describe how aqueous potassium iodide can be used to test for an oxidising agent

Solution

6a. Iron(II) irons, Fe²⁺ is the reducing agent. Each Fe²⁺ ion has lost 1 electron to form Fe³⁺ ion causing the oxidation number of iron to increase from +2 to +3

b. Add the oxidising agent to potassium iodide solution. A brown solution of iodine is obtained.

7. Potassium dichromate is an oxidizing agent. The oxidation number of chromium in potassium dichromate is +6.

ai. name one compound which can be oxidized by potassium dichromate

aii. state the condition(s) under which the compound reacts with potassium dichromate

aiii. what product is formed from the compound in the redox reaction?

b. In the presence of a dilute acid, chromium(II) ions react with atmospheric oxygen to from chromium(III) ions and water.

bi. write the half equations for the formation of chromium(III) ions

bii. write the half equation for the formation of water

biii. write the overall equation for the reaction

c. suggest 2 ways in which chromium can be used to prevent the corrosion of iron.

Solution

ai. potassium iodide OR iron(II) sulphate

aii. mix the potassium iodide with acidified potassium dichromate OR mix the iron(II) sulphate with acidified potassium dichromate

aiii. iodine OR iron(III) ions

bi. Cr²⁺ --> Cr³⁺ + e^-

bii. O₂ + 4H⁺ + 4e^- --> 2H₂O

biii. 4Cr²⁺ + O₂ + 4H⁺ --> 4Cr³⁺ + 2H₂O

c. chromium can be used in chromium-plating and making stainless steel.

iodide ions are oxidised to brown iodine molecules.