ASEI Lesson Plan and Note Template on Chemistry

Learning Objectives:

1. Identify the ion and oxidation number of an element/ compound by figuring out the protons, neutrons, and electrons of the element in the periodic table of element:








2. Explain the redox reactions in terms of:

Oxidation	Reduction
Gaining oxygen	Losing oxygen
Losing hydrogen	Gaining hydrogen
Addition of electronegative elements & removal of electropositive elements	Removal of electronegative elements & addition of electropositive elements
Losing electron	Gaining electron

1. REDUCTION is Losing (Removal) of Oxygen and OXIDATION is Gaining (Addition) of Oxygen
Reduction Reaction is defined as the loss of oxygen by a compound. E.g.

• the reduction of copper (II) oxide to copper by heating with charcoal (carbon).
  2CuO(s) + C → 2Cu + CO₂
• the reaction of copper (II) oxide and hydrogen, copper (II) oxide is reduced to copper after oxygen has been removed.
  CuO + H₂ → Cu + H₂O
• the reaction of magnesium with steam whereby steam is reduced to hydrogen by the removal of oxygen.
  Mg + H₂O(g) → MgO + H₂

Oxidation Reaction is defined as the reaction where substances are combined with the element oxygen. E.g.

• burning carbon to produce carbon dioxide is oxidation.
  C + O₂ → CO₂
• the reaction of a copper atom with oxygen, the copper atom is oxidised to copper (II) oxide.
  2Cu + O₂ → 2CuO
• the reaction of hydrogen with oxygen to form the compound water.
  2H₂ + O₂→ 2H₂O


2. REDUCTION is Gaining (Addition) Hydrogen and OXIDATION is Losing (Removal) of Hydrogen
A reduction reaction is defined as the addition of hydrogen to a substance. E.g.

• the reaction of hydrogen with oxygen to form the compound water.
  2H₂ + O₂→ 2H₂O
• the reaction of hydrogen sulphide with chlorine, the addition of hydrogen to chlorine is a reduction reaction.
  H₂S(g) + Cl₂(g) → 2HCl(g) + S(s)
• the reaction of hydrogen with magnesium oxide, the magnesium (II) oxide is reduced to magnesium by the addition of hydrogen.
  MgO + H₂(g) → Mg + H₂O



Oxidation Reaction is defined as the removal of hydrogen from a substance. E.g.




3. REDUCTION is the Removal of an Electronegative element as well as the addition of an Electropositive element and OXIDATION is the addition of electronegative elements and the removal of electropositive elements.




Electronegative elements tend to gain electrons, e.g. non-metals like chlorine.
Electropositive elements tend to lose electrons and become positive ions, e.g. metals of sodium.




4. REDUCTION is the gaining of electrons and OXIDATION is the loss of electrons.



2KI(aq) + Fe₂(SO₄)₃(aq) → I₂(s) + K₂SO₄(aq) + 2Fe(aq)
Ionically,
2I⁻(aq) + 2Fe³⁺(aq) → I₂(s) + 2Fe²⁺(aq)
In this reaction, iodide ions are oxidized to iodine molecules and iron(III) ions are reduced to iron(II) ions

Similarly, reactions where copper(I) changes to copper(II), lead(II) to lead(IV) etc, are all regarded as oxidation reactions. In all these changes, the charge on a positive ion is increased, e.g.
Cu⁺ → Cu²⁺,
while the charge on a negative ion is decreased, e.g.
I⁻¹ → I°.
A substance loses electrons when it is oxidized and gains electrons when it is reduced, while a redox reaction is a transfer of electrons from the reducing agent (the electron donor) to the oxidizing agent (the electron acceptor).


3. Identify oxidizing and reducing agents.

1. Oxidizing agents: are substances or chemical species containing the element or species that accepts electrons allowing another element or species to be oxidized
   By accepting electrons, the element or species in the oxidizing agent are reduced.
2. Reducing agents: are substances or chemical species containing the element or species that donate electrons, allowing another element or species to be reduced.
   By giving up electrons, the element or species in the reducing agent are oxidised.

Reducing agent	Oxidizing agent
Causes reduction by donating electron(s)	Causes oxidation by removing electron(s) from others
Loses one or more electrons	Gains one or more electrons
Undergoes oxidation; i.e. oxidation number of atoms increases	Undergoes reduction, i.e.oxidation number of atoms decreases

In general, metals give up electrons and act as reducing agents, while reactive nonmetals such as O₂ and halogens accept electrons and act as oxidizing agents.

4. State the rules used to determine oxidation numbers of elements in chemical compounds and species and calculate/ assign the oxidation number of an element in a compound.

1. The oxidation number of an element in its elemental form is 0 (zero). For example,
   Na°, Cl₂°, Mg°, P₄°. The zero, however, is often omitted.
2. In compounds, the oxidation number of oxygen is almost always –2. The most common exception is in peroxides when the oxidation number is –1. Peroxides are compounds having two oxygen atoms bonded together. For example, hydrogen peroxide is H-O-O-H. In hydrogen peroxide, each oxygen atom has a ‐1 oxidation number. When Oxygen is bonded to fluorine, as in hypofluorous acid (HOF), the oxidation number of oxygen is 0. Oxygen‐fluorine compounds such as OF₂, the oxidation number of oxygen is +2
   In superoxide, O2-, we see that each oxygen atom bears a -1/2 oxidation state.
3. The oxidation number of a simple ion (i.e. an ion consisting of a single element) is equal to the charge on the ion.
   The oxidation number of Na⁺, Mg²⁺, Al³⁺, Cl⁻ and S²⁻ are +1, +2, +3, -1 and -2 respectively.
   Note that the negative or positive sign for an oxidation number always comes before the number, the “+” and “- “signs are just as important as the number (they indicate the nature of the charge).
4. The oxidation numbers of group 1 and group 2 elements are +1 and +2 respectively.
5. In compounds, the oxidation number of hydrogen is almost always +1. The most common exception occurs when hydrogen combines with metals; in this case, the oxidation number of hydrogen is typical –1.
6. In compounds, the oxidation number of fluorine is always –1. The oxidation number of other halogens (Cl, Br, I) is also –1, except when they are combined with oxygen. The oxidation state of halogens (except fluorine) combined with oxygen is typically positive. For example, in ClO⁻, chlorine’s oxidation number is +1.
7. For a complex ion, the sum of the positive and negative oxidation numbers of all elements in the ion equals the charge on the ion; e.g. SO₄²⁻ i.e [-2 = +6 + 4(-2)], OH⁻ i.e [-1 = -2 + (+1)], the oxidation numbers of NH₄⁺, NO₃⁻, PO₄³⁻ and MnO₄⁻ are +1, -1, -2, -3 and -1 respectively.
8. For an electrically neutral compound, the sum of the positive and negative oxidation numbers of all elements in the compound equals zero. In any compound, the more electronegative atom has the negative oxidation number while the less electronegative atom has the positive oxidation number. For example in SCl₂, sulfur has an oxidation number of +2 since it is less electronegative than chlorine but in sodium sulphide, Na₂S, its oxidation number is -2.
9. The sum of the oxidation numbers is 0 for a molecule and is equal to the net charge for a polyatomic ion.
   In the compound MgCl₂, which may also be written as (MgCl₂)° [0 = +2 + 2(-1)]
   This rule is particularly useful for finding the oxidation number of an atom in any chemical compound or species.
   The general procedure is to assign oxidation numbers to the “easy” or known atoms first and then find the unknown oxidation number by calculation.
   
   For example,
   1. Find the oxidation number of the sulfur atom in sulfuric acid (H₂SO₄).
   Solution
   Since each H atom is +1 and each O atom bears -2, the S atom must have an oxidation number of +6 for the compound to have no net charge: 2(+1) + (?) + 4(–2) = 0 net charge
   ? = 0 – 2(+1) – 4(–2) = + 6
   To find the oxidation number of the chlorine atom in the perchlorate anion (ClO₄⁻), we know that each oxygen has a charge of -2, so the Cl atom must have an oxidation number of +7 for there to be a net charge of –1 on the ion:
   ? + 4(–2 ) = –1 net charge
   ? = –1– 4(–2) = +7
   2. Find the oxidation number of the manganese atom in potassium tetraoxomanganate(VII), KMnO₄
   Solution
   (Oxidation number of K) + (Oxidation number of Mn) + 4(Oxidation number of O) = 0
   oxidation number of K = +1,
   oxidation number of Mn = X,
   oxidation number of O = -2,
   :. (+1) + X + [4 x (-2)] = 0
   1 + X - 8 = 0
   X = +7
   The oxidation number of the manganese atom in potassium tetraoxomanganate(VII) is +7.
   3. Find the oxidation number of the chromium atom in potassium heptaoxodichromate(VI), K₂Cr₂O₇.
   Solution
   2(Oxidation number of K) + 2(Oxidation number of Cr) + 7(Oxidation number of O) = 0
   oxidation number of K = +1,
   oxidation number of Cr = X,
   oxidation number of O = -2,
   :. 2(+1) + 2(X) [7 x (-2)] = 0
   2 + 2X - 14 = 0
   2X = +12
   X = +6
   The oxidation number of the chromium atom in potassium heptaoxodichromate(VI) is +6.
   4. Find the oxidation number of the chlorine atom in the perchlorate anion (ClO₄⁻).
   Solution
   each oxygen has a charge of -2, so Cl atom must have an oxidation number of +7 for there to be a net charge of –1 on the ion:
   ? + 4(–2 ) = –1 net charge
   ? = –1– 4(–2) = +7

5. Balancing redox reactions (equations) in aqueous/neutral solutions, acidic solutions and basic solutions using the Half Reaction Method.

   In redox reaction using the half-reaction method, it is useful to separate the reaction into two half-reactions one involving oxidation reaction and the other involving reduction reaction. Then after balancing those half-reactions, find the overall oxidation-reduction (redox) reaction by summing up the two half-reactions

   NOTE: In balancing redox reaction using the half-reaction method, you need to make sure that not only the particles, atoms and ions are balanced but also the chargers must be balanced on both sides.

   1. The Half-Reaction Method for Balancing Oxidation-Reduction Reactions in Aqueous Solutions.
   For example, Balance the following equation of the reaction which occurs in aqueous solutions.
   Al + Ni²⁺ → Al³⁺ + Ni
   Solution
   There is an equal (one) Aluminium atom on both sides.
   There is an equal Nickel particle on both sides, hence, the atoms are balanced.
   There is a net charge of +2 ion on the left and a net charge of +3 ion on the left, hence the charges are not balanced.
   Separate the equation into half-reaction.
   Al → Al³⁺ (net charge on the left = 0, net charge on the right = +3)
   Difference between charges +3 - (0) = 3, add 3 electrons to the side with the highest charges
   Al → Al³⁺ + 3e‾ This half-reaction is now balanced (Whenever we have the electron (e‾) on the right of the reaction, it represents the oxidation half-reaction)
   Ni²⁺ → Ni (net charge on the left = +2, net charge on the right = 0)
   Difference between charges +2 - (0) = 2, add 2 electron to the side with the highest charges
   Ni²⁺ + 2e‾ → Ni This half-reaction is now balanced (Whenever we have the electron (e‾) on the left of the reaction, it represents a reduction in half-reaction)
   Before adding both half-reactions, make sure the numbers of electrons are the same on both sides, the least common multiple of 2 and 3 is 6, hence, we need a total of 6 electrons on both sides of the half-reactions.
   2(Al → Al³⁺ + 3e‾)
   3(Ni²⁺ + 2e‾ → Ni)
   
   2Al → 2Al³⁺ + 6e‾
   3Ni²⁺ + 6e‾ → 3Ni
   Now add both half reactions (Note, the 6e‾ on both side cancel out)
   2Al + 3Ni²⁺ → 2Al³⁺ + 3Ni
   Check that the elements and charges are balanced.
   
   • The number of Nickel particles is the same on both sides.
   • The number of the Aluminium particle are the same on both side.
   • The net charge on the left is 3 x +2 = +6, and the net charge on the right is 2 x +3 = +6, hence they are the same on both sides.
   The number of atoms is the same on both sides and the number of charges is also the same on both side.
   
   :. 2Al + 3Ni²⁺ → 2Al³⁺ + 3Ni
   2. The Half-Reaction Method for Balancing Oxidation-Reduction Reactions in Acidic Solutions.
   STEPS
   1. Write separate equations for the oxidation and reduction half-reactions.
   2. For each half-reaction,
      • balance all the elements except hydrogen and oxygen.
      • then balance oxygen using H₂O.
      • then balance hydrogen using H⁺.
      • then balance the charge using electrons.
   3. If necessary, multiply one or both balanced half-reactions by an integer to equalize the number of electrons transferred in the two half-reactions.
   4. Add the half-reactions, and cancel identical species.
   5. Checking that the elements and charges are balanced.
   For example, Balance the following equation of the reaction which occurs in acidic solutions.
   Zn + BrO₃‾ → Zn²⁺ + Br¹‾
   Solution
   There is an equal (one) Zinc atom on both sides.
   There is an equal Bromine particle on both side.
   There is an unequal Oxygen element on both sides.
   Hence, the atoms are not balanced.
   There is a net charge of -1 ion on the left and a net charge of +1 (i.e +2 - (-1) = +1) ion on the left, hence the charges are not balance
   Separate the equation into half-reaction.
   Zn → Zn²⁺ (net charge on the left = 0, net charge on the right = +2)
   Difference between charges +2 - (0) = 2, add 2 electron to the side with the highest charges
   Zn → Zn²⁺ + 2e‾ This half-reaction is now balanced (oxidation half-reaction).
   BrO₃‾ → Br¹‾ Net charge on the left = -1, net charge on the right = -1, they are balanced.
   We have equal bromine atoms on both sides but unequal Oxygen atoms (3 molecules of Oxygen on the left null on the right), to balance the oxygen atom add 3 water molecule to the right side of the reaction.
   BrO₃‾ → Br¹‾ + 3H₂O There are 3 Oxygen atoms on the left and 3 Oxygen atoms on the right, there are null Hydrogen atoms on the right but 6 Hydrogen atoms on the left, add 6 Hydrogen atoms to the left.
   BrO₃‾ 6H⁺ → Br¹‾ + 3H₂O All atoms on both sides are now balanced. Net charge on the left is -1 + 6(+1) = +5, net charge on the right is -1. Difference between the net charges +5 - (-1) = 6 :. add 6 electrons to the side with the highest charge.
   BrO₃‾ 6H⁺ + 6e‾ → Br¹‾ + 3H₂O Now make the number of electrons equal on both sides, the least common factor of 2 that will give us 6 is 3, hence, we need a total of 6 electrons on both sides of the half-reactions.
   3(Zn → Zn²⁺ + 2e‾)
   3Zn → 3Zn²⁺ + 6e‾
   BrO₃‾ 6H⁺ + 6e‾ → Br¹‾ + 3H₂O Now add both half reactions (Note, the 6e‾ on both side cancel out)
   3Zn + BrO₃‾ + 6H⁺ → 3Zn²⁺ + Br¹‾ + 3H₂O
   Check that the elements and charges are balanced.
   
   • one bromine atom on both side
   • Three zinc particles on both side
   • Six hydrogen atoms on both side
   • The net charge on the left is -1 + 6(+1) = +5, and the net charge on the right is 3(+2) + (-1) = +5, hence they are the same on both sides.
   The number of particles is the same on both side and the number of charges is also the same on both side.
   
   :. 3Zn + BrO₃‾ + 6H⁺ → 3Zn²⁺ + Br¹‾ + 3H₂O
   3. The half-reaction method for balancing equations for oxidation-reduction reactions occurring in basic solution.
   STEPS
   1. Write separate equations for the oxidation and reduction half-reactions.
   2. For each half-reaction,
      • balance all the elements except hydrogen and oxygen.
      • then balance oxygen using H₂O.
      • then balance hydrogen using H⁺.
      • addition of some OH- ions equal to the number of H+ ions to each side of each half-equation.
      • then balance the charge using electrons.
   3. If necessary, multiply one or both balanced half-reactions by an integer to equalize the number of electrons transferred in the two half-reactions.
   4. Add the half-reactions, and cancel identical species.
   5. Checking that the elements and charges are balanced.

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   For example, Balance the following equation of the reaction which occurs in the basic solution.
   Al + ClO₄‾ → Al(OH)₄‾ + Cl‾
   Solution
   There is an equal Aluminium atom on both sides
   There is an equal Chlorine atom on both sides
   There is an equal Oxygen atom on both sides
   There is an unequal Hydrogen element on both sides
   Hence, the atoms are not balanced
   There is a net charge of -1 ion on the left and a net charge of 0 [i.e -1 + (-1) = 0] ion on the right, hence the charges are not balance
   Separate the equation into half-reaction
   Al → Al(OH)₄‾There is an unequal number of hydroxide molecules (4 molecules of Oxygen and 4 molecules of Hydrogen on the right, there are null of Hydrogen and Oxygen on the left, add 4 Hydroxide molecules to the right).
   Al + 4OH‾ → Al(OH)₄‾ (net charge on the left = -4 i.e 4 x -1 = -4, net charge on the right = -1) Difference between charges -4 - (-1) = -3, add 3 electron to the side with the highest charges
   Al + 4OH‾ → Al(OH)₄‾ + 3e‾ This half-reaction is now balanced (oxidation half-reaction).
   ClO₄‾ → Cl‾ There are an unequal number of Oxygen atoms on the right (4 molecules of Oxygen on the left, null on the left). First, we right this reaction under acidic conditions by adding four molecules of water.
   ClO₄‾ → Cl‾ + 4H₂O There is 4 Oxygen atom on the left and 4 Oxygen atom on the right, there are null Hydrogen atom on the right but 8 Hydrogen atom on the right, add 8 moles of Hydrogen ion to the left to balance the hydrogen atoms on both sides.
   ClO₄‾ + 8H⁺ → Cl‾ + 4H₂O On the basic condition the H⁺ does not exist so we have to get rid of it by adding 8OH‾ to both sides.
   ClO₄‾ + 8H⁺ + 8OH‾ → Cl‾ + 4H₂O + 8OH‾ The 8H⁺ + 8OH‾ molecule will combine to form 8 water molecules.
   ClO₄‾ + 8H₂O → Cl‾ + 4H₂O + 8OH‾ The 8H₂O on the right and the 4H₂O on the left can be simplified by subtracting 4H₂O molecule from both sides.
   ClO₄‾ + 8H₂O - 4H₂O→ Cl‾ + 4H₂O - 4H₂O + 8OH‾
   ClO₄‾ + 4H₂O→ Cl‾ + 8OH‾ The number of atoms on both side are now balanced, now check the net charges on both sides. The the left the net charges is -1, on the right the net charges is -9 [i.e -1 + (-8) = -9], difference between the net charges on the right and left -1 - (-9) = 8, add 8 electron to the side with the highest charges
   ClO₄‾ + 4H₂O + 8e‾ → Cl‾ + 8OH‾ Reduction half-reaction
   Al + 4OH‾ → Al(OH)₄‾ + 3e‾ Oxidation half-reaction
   Make the number of electrons on the reduction half-reaction and the number of electrons on the oxidation half-reaction equal. The least common multiple of 8e‾ and 3e‾ is 24, hence, add 8 to the oxidation reaction and 3 to the reduction reaction.
   3(ClO₄‾ + 4H₂O + 8e‾ → Cl‾ + 8OH‾)
   8(Al + 4OH‾ → Al(OH)₄‾ + 3e‾)
   
   8Al + 32OH‾ → 8Al(OH)₄‾ + 24e‾
   3ClO₄‾ + 12H₂O + 24e‾ → 3Cl‾ + 24OH‾ Now add both half reactions (Note, the 24e‾ on both side cancel out)
   8Al + 32OH‾ + 3ClO₄‾ + 12H₂O → 8Al(OH)₄‾ + 3Cl‾ + 24OH‾
   Simplify 32OH‾ on the left and 24OH‾ on the right by subtracting 24OH‾ from both side
   8Al + 32OH‾ + 24OH‾ + 3ClO₄‾ + 12H₂O → 8Al(OH)₄‾ + 3Cl‾ + 24OH‾ + 24OH‾
   8Al + 8OH‾ + 3ClO₄‾ + 12H₂O → 3Cl‾ + 8Al(OH)₄‾
   Check that the elements and charges are balanced.
   
   • 8 aluminium atoms on both sides
   • 3 chlorine atoms on both sides
   • 32 hydrogen atoms on both sides
   • 32 oxygen atoms on both sides
   • The net charge on the left is 8(-1) + 3(-1) = -11, and the net charge on the right is 3(-1) + 8(-1) = -11, hence they are the same on both sides.
   The number of particles is the same on both side and the number of charges is also the same on both side.
   Therefore, 8Al + 8OH‾ + 3ClO₄‾ + 12H₂O → 3Cl‾ + 8Al(OH)₄‾

Rationale:

When we talk about chemical reactions, we usually discuss the breakage and formation of bonds, gain and loss of electrons, and conversion from one state of matter to another. If we look closely, we might observe hundreds of chemical reactions taking place in our vicinity. You may find it quite surprising that almost one-third of the chemical reactions taking place in the surroundings fall under the category of redox reactions. E.g. The rusting of iron. The food that we consume is converted into energy by redox reactions. In photosynthesis, water is oxidised and carbon dioxide is reduced. The process of developing a photographic film also employs redox reactions, a positive image is obtained by the exposure of the negative to light. Following exposure to light, silver cations are reduced. The living matter in nature, primarily, is made up of carbon, hydrogen, and oxygen. Upon the death of any living organism, the organic compounds present in the organism start reacting with oxygen. This process commonly referred to as “decay” or “decomposition” forms another example of redox reactions.

Prerequisite/ Previous knowledge:

Learning Materials:

Iron powder, 2M sulphuric acid, Bunsen burner, hydrogen peroxide (20 cm3), filter paper, test-tube and test-tube racks, 2M sodium hydroxide, filter funnel, 2M ammonia solution. The periodic table of metals.

Reference Materials:

New School Chemistry. By Ababio
Lamlad's S.S.C.E and UTME by F. O. Ayinde and F. O. I. Asubiojo

Lesson Development: