Classwork Series and Exercises {Chemistry – SS2}: Oxidation – Reduction Reaction

Chemistry, SS 2 Week: 5

Topic: Oxidation-Reduction Reaction

Oxidation-reduction reactions involve the transfer of electrons between substances. They take place simultaneously, which makes sense because if one substance loses electrons, another must gain them. For example, all single-replacement reactions are redox reactions.

Electrochemistry: The study of the interchange of chemical and electrical energy.

Oxidation: The loss of electrons. Since electrons are negative, this will appear as an increase in the charge (e.g., Zn loses two electrons; its charge goes from 0 to +2). Metals are oxidized.

Oxidizing agent (OA): These are species that reduces and thus, causes oxidation.

Reduction: The gain of electrons. When an element gains electrons, the charge on the element appears to decrease, so we say it has a reduction of charge (e.g., Cl gains one electron and goes from an oxidation number of 0 to -1). Non-metals are reduced.

Reducing agent (RA): The species that is oxidized and thus causes reduction.

Oxidation number: The assigned charge on an atom. You’ve been using these numbers to balance formulas.

Half-reaction: An equation that shows either oxidation or reduction alone.

Rules for Assigning Oxidation States

A reaction is considered a redox reaction if the oxidation numbers of the elements in the reaction change in the course of the reaction. We can determine which elements undergo a change in oxidation state by keeping track of the oxidation numbers as the reaction progresses. You can use the following rules to assign oxidation states to the components of oxidation-reduction reactions:

1. The oxidation state of an element is zero, including all elemental forms of the elements (e.g., N₂, P₄, S₈, O₃).
2. The oxidation state of a monatomic ion is the same as its charge.
3. In compounds, fluorine is always assigned an oxidation state of -1.
4. Oxygen is usually assigned an oxidation state of -2 in its covalent compounds. Exceptions to this rule include peroxides (compounds containing the group), where each oxygen is assigned an oxidation state of -1, as in hydrogen peroxide (H₂O₂).
5. Hydrogen is assigned an oxidation state of +1. Metal hydrides are an exception: in metal hydrides, H has an oxidation state of -1.
6. The sum of the oxidation states must be zero for an electrically neutral compound.
7. For a polyatomic ion, the sum of the oxidation states must equal the charge of the ion.

Example

Assign oxidation numbers to each element in the following:

1. H₂S
2. MgF₂
3. 

Explanation

1. The sum of the oxidation numbers in this compound must be zero since the compound has no net charge. H has an oxidation state of +1, and since there are two H atoms, +1 times 2 atoms = +2 total charge on H. The sulfur S must have a charge of -2 since there is only one atom of sulfur, and +2 – 2 = 0, which equals no charge.
2. F is assigned an oxidation state of -1 (according to rule 3), and there are two atoms of F, so this gives F a total charge of -2. Mg must have a +2 oxidation state since +2 – 2 = 0 and the compound is electrically neutral.
3. This time the net charge is equal to -3 (the charge of the polyatomic ion—according to rule 7). Oxygen is assigned a -2 oxidation state (rule 4). Multiply the oxidation number by its subscript: -2 X 4 = -8. Since there is only 1 phosphorus, just use those algebra skills: P + -8 = -3. Phosphorus must have a +5 charge.

Oxidation and reduction in terms of oxygen transfer

Definitions;

• Oxidation is gain of oxygen.
• Reduction is loss of oxygen.

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

Because both reduction and oxidation are going on side-by-side, this is known as a redox reaction.

Oxidizing and Reducing Agents

An oxidizing agent is substance which oxidizes something else. In the above example, the iron (III) oxide is the oxidizing agent. A reducing agent reduces something else. In the equation, the carbon monoxide is the reducing agent.

• Oxidizing agents give oxygen to another substance.
• Reducing agents remove oxygen from another substance.

Oxidation and Reduction in Terms of Hydrogen Transfer

These are old definitions which aren’t used very much nowadays. The most likely place you will come across them is in organic chemistry.

Definitions

• Oxidation is loss of hydrogen.
• Reduction is gain of hydrogen.

Notice that these are exactly the opposite of the oxygen definitions.

For example, ethanol can be oxidized to ethanal:

You would need to use an oxidizing agent to remove the hydrogen from the ethanol. A commonly used oxidizing 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. Again the equation is too complicated to be worth bothering about at this point.

An update on oxidizing and reducing agents

• Oxidizing agents give oxygen to another substance or remove hydrogen from it.
• Reducing agents remove oxygen from another substance or give hydrogen to it.

Oxidation and Reduction in Terms of Electron Transfer

This is easily the most important use of the terms oxidation and reduction.

Definitions;

• Oxidation is loss of electrons.
• Reduction is gain of electrons.

It is essential that you remember these definitions. There is a very easy way to do this. As long as you remember that you are talking about electron transfer:

A simple example;

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

CuO + Mg → Cu + MgO

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:

If you look at the equation above, the magnesium is reducing the copper(II) ions by giving them electrons to neutralize 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 oxidizing agent. For example, what an oxidizing agent did in terms of electrons:

• An oxidizing agent oxidizes something else.
• Oxidation is loss of electrons (OIL RIG).
• That means that an oxidizing agent takes electrons from that other substance.
• So an oxidizing agent must gain electrons.

Example 1: Determine which element is oxidized and which element is reduced in the following reactions (be sure to include the oxidation state of each): 

1. Zn + 2H⁺ → Zn²⁺ + H₂
2. 2Al + 3Cu²⁺→2Al³⁺ +3Cu
3. CO₃^2- + 2H⁺→ CO₂ + H₂O

 Solutions:

1. Zn is oxidized (Oxidation number: 0 → +2) and acts as the reducing agent; H⁺ is reduced (Oxidation number: +1 → 0) and acts as the oxidizing agent.
2. Al is oxidized (Oxidation number: 0 → +3) and acts as the reducing agent; Cu²⁺ is reduced (+2 → 0) and acts as the oxidizing agent.
3. This is not a redox type because each element has the same oxidation number in both reactants and products: O= -2, H= +1, C= +4.

Example 2: The thermite reaction in which iron atoms of ferric oxide lose (or give up) O atoms to Al atoms, producing Al₂O₃.

Fe₂O_3(s) + 2Al_(s) → Al₂O_3(s) + 2Fe_(l)

Example 3: When powdered zinc metal is mixed with iodine crystals and a drop of water is added, the resulting reaction produces a great deal of energy. The mixture bursts into flames, and a purple smoke made up of I₂ vapor is produced from the excess iodine. The equation for the reaction is

Zn_(s) + I_2(s) →ZnI_2(s) + energy

Identify the elements that are oxidized and reduced, and determine the oxidizing and reducing agents.

Explanation

1. Assign oxidation numbers to each species. Zn and I₂ are both assigned values of 0 (rule 1). For zinc iodide, I has an oxidation number of -1 (group 7A—most common charge), which means that for zinc, the oxidation number is +2.
2. Evaluate the changes that are taking place. Zn goes from 0 to +2 (electrons are lost and Zn is oxidized). The half-reaction would look like this:

Zn⁰→Zn²⁺ + 2e^–

And I₂ goes from 0 to -1 (it gains electrons and so is reduced). This half-reaction would look like this: 

I⁰₂ + 2e^– → 2I^–

3. Here, zinc metal is the reducing agent—it causes the reduction to take place by donating electrons—while iodine solid is the oxidizing agent; iodine solid accepts electrons.

Balancing in Basic and Acidic Solution

• Balancing in acidic solution is similar to balancing in neutral solutions however, instead of three steps to follow, there are six. These rules are:
• Write and balance the half reactions.
• Balance oxygen, O, by adding with H₂O
• Balance hydrogen, H, by adding H⁺ (acidic)
• Balance charge by adding electrons (you should be adding the same number of electrons as H⁺ ions)
• Multiply both half reactions by some integer to cancel out electrons
• Add the half reactions together and cancel out what appears on both sides

Example 1: Balance the redox reaction in acidic solution: 

MnO₄^– + I⁻ →  Mn²⁺ +I_2(s) 

Solution

• Write and balance the half reactions:
  MnO₄^–+ I^––> Mn²⁺ + I_{2(s)
  O.S:  +7 -2      -1            +2             0        (Mn is reduced and I}^– _{is oxidized)}
  Oxidation Rx:2I^–_(aq) –> I_2(s) + 2e^–
  Reduction Rx: MnO₄^– + 5e^– –> Mn²⁺
• Balance oxygen, O, by adding H₂O
  Oxidation Rx:2I^–_(aq)–> I_2(s) + 2e^–
  Reduction Rx: MnO₄^– + 5e^– –> Mn²⁺ + 4H₂O
• Balance hydrogen, H, by adding H^+<