ASEI Lesson Plan and Note Template on Chemistry

Lesson Note and Plan on Reduction and Oxidation Reaction


Subject: Chemistry
Theme: Types of Reaction
Topic: Reduction & Oxidation Reaction
Sub Topic: Redox Reaction
Date: dd/mm/yyyy
Class: S.S.S 2
Duration: 35 Minutes
No of Learners: 30

Learning Objectives:

By the end of the lesson learners should be able to:
  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:

  3. OxidationReduction
    Gaining oxygenLosing oxygen
    Losing hydrogenGaining hydrogen
    Addition of electronegative elements & removal of electropositive elementsRemoval of electronegative elements & addition of electropositive elements
    Losing electronGaining electron
    1. REDUCTION is Losing (Removal) of Oxygen and OXIDATION is Gaining (Addition) of Oxygen
    2. 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

    3. REDUCTION is Gaining (Addition) Hydrogen and OXIDATION is Losing (Removal) of Hydrogen
    4. 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.


    5. 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.



    6. 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).

  4. 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 agentOxidizing agent
    Causes reduction by donating electron(s)Causes oxidation by removing electron(s) from others
    Loses one or more electronsGains one or more electrons
    Undergoes oxidation; i.e. oxidation number of atoms increasesUndergoes 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.

  5. 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).
      2. 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
      3. Find the oxidation number of the manganese atom in potassium tetraoxomanganate(VII), KMnO₄
      4. 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.
      5. Find the oxidation number of the chromium atom in potassium heptaoxodichromate(VI), K₂Cr₂O₇.
      6. 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.
      7. Find the oxidation number of the chlorine atom in the perchlorate anion (ClO₄⁻).
      8. 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
  6. 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.
    2. 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
    3. The Half-Reaction Method for Balancing Oxidation-Reduction Reactions in Acidic Solutions.
      1. STEPS
      2. Write separate equations for the oxidation and reduction half-reactions.
      3. 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.
      4. If necessary, multiply one or both balanced half-reactions by an integer to equalize the number of electrons transferred in the two half-reactions.
      5. Add the half-reactions, and cancel identical species.
      6. 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
    4. The half-reaction method for balancing equations for oxidation-reduction reactions occurring in basic solution.
      1. STEPS
      2. Write separate equations for the oxidation and reduction half-reactions.
      3. 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.
      4. If necessary, multiply one or both balanced half-reactions by an integer to equalize the number of electrons transferred in the two half-reactions.
      5. Add the half-reactions, and cancel identical species.
      6. Checking that the elements and charges are balanced.

      We provide educational resources/materials, curriculum guide, syllabus, scheme of work, lesson note & plan, waec, jamb, O-level & advance level GCE lessons/tutorial classes, on various topics, subjects, career, disciplines & department etc. for all the Class of Learners

      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:

Types of chemical reactions.

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:

STAGETEACHER'S ACTIVITYLEARNER'S ACTIVITYLEARNING POINTS
INTRODUCTION
full class session (5mins)
The teacher begins the lesson by reviewing the types of chemical reactions.
Chemical reactions are changes in which some new chemical substances are formed. The substances which undergo the chemical change are known as reactants, while the new substances formed are called products. The following are types of chemical reactions;
  1. Combination reaction: This occurs when two or more substances combine to form one single compound.
    1. Fe(s) + S(s) → FeS(s)
    2. PbO₂(s) + SO₂(g) → PbSO₄(s)
  2. Decomposition reaction: This reaction is characterized by a single compound splitting up into two or simpler substances. When heat is needed to bring about decomposition, the reaction is known as thermal decomposition. In a displacement reaction, one element (or radical) replaces another element (or radical) in a compound. The ability of an element (or radical) to displace another is determined by its relative positions in the electrochemical series. Thus, the more electropositive copper from an aqueous solution of a copper(II) salt.
    1. Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s)
    2. Cl₂(g) + 2KBr(aq) → 2KCl(aq) Br₂(l)
  3. Double Decomposition reaction: In double decomposition, the reactants decompose to form new substances by an exchange of radicals. The two reactants are soluble, and one of the products is soluble while the other is insoluble, volatile or gaseous.
    1. AgNO₃(aq) + NaCl(s) → AgCl(s)[precipitate] + NaNO₃(aq)
    2. KNO₃(aq) + H₂SO₄(aq) → HNO₃(g)[volatile] + KHSO₄(aq)
    3. NaCl(aq) + H₂SO₄(aq) → HCl(g) + NaHSO₄(aq)
  4. Catalytic reaction: These are substances used to alter the rates of chemical reactions. These substances remain chemically and quantitatively unchanged at the end of the reactions. The acceleration or retardation of reactions by catalysts is known as catalysis. There are two types of catalyzed reactions;
    1. Homogeneous catalysis: In this reaction, the catalyst, reactants and products are all in the same phase. E.g. The oxidation of sulphur(IV) oxide using nitrogen(II) oxide as a catalyst occurs in the gaseous phase.

    2. Heterogeneous catalysis: The reactants, products and catalysts are in different phases in heterogeneous catalysis. The formation of margarine from vegetable oil in the presence of a nickel catalyst is an example of heterogeneous catalysis.

      Characteristics of a catalyst
      1. A catalyst alters the rate of a chemical reaction.
      2. A catalyst remains unchanged in chemical nature and mass at the end of a reaction. Its physical features such as colour and texture, however, may be changed.
      3. A catalyst is specific in action, i.e. a given catalyst will act on only one particular reaction, although many inorganic catalysts will catalyze more than one reaction.
      4. A catalyst cannot start a reaction. It is effective only in a reaction which is already in progress.
      5. A catalyst does not affect the types of product formed in a reaction.
      6. A catalyst does not affect the equilibrium of a reversible reaction.
      7. A catalyst will affect the rate of a reaction even though it is present in very small amounts. An increase in the amount of catalyst up to a certain limit, however, would usually increase the reaction rate.
      8. The effect of a solid catalyst is improved by increasing its surface area.
    Types of catalysts
    1. Inorganic catalyst: Inorganic catalysts can be grouped into positive and negative catalysts.

    2. A catalyst which increases the rate of a reaction is a positive catalyst. E.g, manganese(IV) oxide which speeds up the decomposition of potassium trioxochlorate(V).

      A catalyst which decreases the rate of a reaction is a negative catalyst. E.g, tetraoxophophate(V) acid decreases the rate of decomposition of hydrogen peroxide, while a 2% ethanol solution suppresses the oxidation of trichloromethane by air.

    3. Organic catalysts: Are known as enzymes and are produced by living cells.
    4. Enzymes are organic catalysts which control the rate of biochemical reactions in living organisms.
      Enzymes help to digest the food we eat. They cause blood to clot and fruit juices to ferment. The industrial manufacture of alcohol from sugars and starches uses enzymes.
  5. Promoters and inhibitors:
    Promoters improve the efficiency of a catalyst while inhibitors or catalyst poisons inhibit a catalyst. E.g, Hydrogen sulphide, arsenic(III) oxide, Hydrogen cyanide and Mercury salts are catalyst poisons which harm us by inhibiting important enzymes in our bodies.
  6. Reversible reactions:In some chemical reactions, the products of the reaction can react together to produce the original reactants. These reactions are called reversible reactions.
    The symbol for the reversible reaction has two half arrowheads, one pointing in each direction:
    • the forward reaction is the one that goes to the right
    • the backward reaction is the one that goes to the left
    Example:
    1. Ammonium chloride is a white solid. It breaks down when heated, forming ammonia and hydrogen chloride. When these two gases are cool enough, they react together to form ammonium chloride again.

    2. Blue copper sulfate is described as hydrated. The copper ions in its crystal lattice structure are surrounded by water molecules. This water is driven off when blue hydrated copper sulfate is heated, leaving white anhydrous copper sulfate. This reaction is reversible.

    3. At 20⁰C and 140⁰C, dinitrogen(IV) oxide and nitrogen(IV) oxide exist together in an equilibrium mixture.

The learners listen to the teacher's review and ask questions and make a contribution where necessary.Reviewing the previous lesson on chemical reactions.
STEP 1
5 mins.
Development
The teacher guides learners to explain how a chemical reaction occurs in the following areas:
  1. Cellular Respiration
  2. Application in Combustion
  3. Photosynthesis
  4. Corrosion
  5. Decomposition
  6. Bleaching agent
Learners follow teacher guidelines and list and explain the various application of chemical reactions in the environment both in living and non-living things.

1. Application in Cellular Respiration:

Cellular Respiration which is the ultimate source of energy in human beings encompasses a series of reactions. The food that we consume is converted into energy. During the process of respiration, the carbon dioxide is reduced whereas the water is oxidised to form oxygen.
6CO₂ + 12H₂O + Light energy → C₆H₁₂O₆ + 6H₂O + 6O₂

2. Application in Combustion

In the burning of organic material and combustion of hydrocarbons in fossil fuels, the oxygen present in the atmosphere bonds with carbon and hydrogen present in the compound being burned. During the process of combustion, the oxygen present in the atmosphere is reduced whereas the compound which is being burned is undergoing oxidation.
CₓHᵧ + O₂ → Heat and light + H₂O + CO₂

3. Application in Photosynthesis

The process of photosynthesis takes place in the leaves of plants, the carbon dioxide and water combine in the presence of sunlight to release oxygen and glucose. The glucose which is formed in the whole process of photosynthesis is used to fuel the metabolic reactions of the plants. In photosynthesis, water is oxidised and carbon dioxide is reduced.
CO₂ + 6H₂O + Energy from light + Chlorophyl in leaves → Heat and light + C₆H₁₂O₆ + 6O₂

4. Application in Corrosion

Oxygen atoms present in water oxidise iron (or metal) and, thereby, lead to the generation of free hydrogen ions. The hydrogen ions are generated to combine with oxygen to yield water, and the whole cycle begins once again.
Fe + 2H⁺ → Fe²⁺ + H₂
O₂ + 2H₂O → 4OH
Fe²⁺ + 2OH‾ → Fe(OH)₂
Fe + CO₂ + H₂O → FeCO₃ + H₂
Fe + H₂S → FeS + H₂

5. Application in Decomposition

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. The aforesaid reaction is a prolonged process. This process is commonly referred to as "decay" or "decomposition".
CH₂O (Organic matter) + O₂ → H₂O + CO₂ + Nutrients (NO₃, PO₄, Fe, Si, etc)

6. Application in Bleaching agent

A bleaching agent is a substance which can whiten or decolourize other substances. Decolourization of a substance occurs because the electrons move between different energy levels. Any sort of decolourization can be removed by the oxidation of the electrons.
Ca(ClO)₂ + 2H₂O → Ca⁺ + 2OH‾ + 2HClO
2HClO + OH‾ → ClO‾ + H₂O
HClO + ClO‾ → ClO + Cl‾ + OH
ClO + ClO‾ + OH‾ → 2Cl‾ + 2O + OH
ClO‾ + OH → ClO + OH‾

Developing the concept of the topic, oxidation and reduction.
The teacher ask learners to identify and group the compounds/ element on learners activity in a tabular form in
  1. GROUP A
  2. if there are
    • Gaining oxygen
    • Losing hydrogen
    • Adition of electronegative elements & removal of electropositive elements
    • Losing electron
  3. or GROUP B
  4. if there are
    • Losing oxygen
    • Gaining hydrogen
    • Removal of electronegative elements & addition of electropositive elements
    • Gaining electron
  5. Reasons for the answer.



























The teacher explains to the learners that chemical reactions in which the following occurs are regarded as OXIDATION REACTION
  1. There is an addition of Oxygen
  2. There is a lose of hydrogen
  3. Addition of electronegative elements & removal of electropositive elements.
  4. There is a loss of electron


While chemical reactions in which the following occurs are regarded as REDUCTION REACTION
  1. There is a loss of Oxygen
  2. There is a gain in hydrogen
  3. Removal of electronegative elements & addition of electropositive elements.
  4. There is a gain of electron


The teacher gives learners A useful mnemonic for redox:- OILRIG:
Oxidation Is Loss (of electrons) and
Reduction Is Gain (of electrons).

Oxidizing agents are always reduced and reducing agents are always oxidized.
  1. the reaction of Copper(II) oxide with Hydrogen to form copper and Water.
    CuO(s) + H₂(g)→ Cu(s) + H₂O(l)
  2. the reaction of Iron and steam to form Iron Iron(III) oxide and Hydrogen.
    3Fe(s) + 4H₂O(g) → Fe₃O₄(s) + 4H₂(g)
  3. the reaction of hydrogen sulphide with chlorine to form Hydrogen chloride and Sulphur.
    H₂S(g) + Cl₂(g) → 2HCl(g) + S(s)
  4. the reaction of Calcium and Chlorine to form Calcium chloride.
    Ca(s) + Cl₂(g) → CaCl₂(s)
  5. the reaction of Sodium and Chlorine to form Sodium chloride.
    2Na(s) + Cl₂(g) → 2NaCl(s)
  6. the reaction of Mercuric oxide to form Mercury and Oxygen.
    2HgO(s) → 2Hg(l) + O₂(g)
  7. the reaction of Mercuric chloride and Tin chloride to form Mercuric(II) chloride and Tin chloride.
    2HgCl₂(aq) + SnCl₂(aq) → Hg₂Cl₂(s) + SnCl₄(aq)
  8. the reaction of Ethene with Hydrogen to form Ethane.
    CH₂=CH₂(g) + H₂(g) → H₃C-CH₃(g)
  9. the reaction of Ferric Chloride and Hydrogento form Iron(II) chloride
    2FeCl₃(aq) + H₂(g) → FeCl₂(aq) + 2HCl(aq)
Learners respond to the teacher's questions.
(Hands and Minds-on activity)
GROUP AReasons for answerGROUP BReasons for answer
addition oxygen in H₂O(l)
CuO(s) + H₂(g)→ Cu(s) + H₂O(l)
Oxygen is added to Hydrogen to form waterloss of oxygen in Cu(s)
CuO(s) + H₂(g)→ Cu(s) + H₂O(l)
Oxygen is removed from the Copper(II) oxide to form metallic copper
addition oxygen in Fe₃O₄(s)
3Fe(s) + 4H₂O(g) → Fe₃O₄(s) + 4H₂(g)
Oxygen is added to Iron(II) to form Iron(III)loss of oxygen in H₂(g)
3Fe(s) + 4H₂O(g) → Fe₃O₄(s) + 4H₂(g)
Oxygen is removed from the steam to form hydrogen gas
removal of hydrogen
H₂S(g) + Cl₂(g) → 2HCl(g) + S(s)
Hydrogen is removed from the Hydrogen sulphide to form yellow sulphuraddition of hydrogen
H₂S(g) + Cl₂(g) → 2HCl(g) + S(s)
Hydrogen has added the chlorine to form Hydrogen-chloride
addition of electronegative elements and removal of electropositive elements
Ca(s) + Cl₂(g) → CaCl₂(s)
chlorine (electronegative elements) receive electron from calcium to form calcium chloride.

calcium (electropositive elements) lost electron to chlorine to form calcium chloride.
removal of electronegative elements and addition of electropositive elements
2HgO(s) → 2Hg(l) + O₂(g)
Oxygen (electronegative elements) is removed from Mercuric Oxide to form Oxygen gas.
addition of electronegative elements and removal of electropositive elements
2Na(s) + Cl₂(g) → 2NaCl(s)
chlorine (electronegative elements) gain electron from sodium to form sodium chloride.

sodium (electropositive elements) lose electron to chlorine to form sodium chloride.
removal of electronegative elements and addition of electropositive elements
2HgCl₂(aq) + SnCl₂(aq) → Hg₂Cl₂(s) + SnCl₄(aq)
Mercury (electropositive elements) is added to Mercuric Chloride
CH₂=CH₂(g) + H₂(g) → H₃C-CH₃(g)Hydrogen (electropositive elements) is added to Ethene
2FeCl₃(aq) + H₂(g) → FeCl₂(aq) + 2HCl(aq)Chlorine (electronegative elements) is removed from Ferric Chloride and Hydrogen (electropositive elements) is added to chloride to form Iron(II) chloride and Hydrochloric acid respectively.
STEP 2
5 mins.
Full class discussion

The teacher asks the learners to explain oxidizing and reducing agents.
HINT: using the mnemonic for redox. OILRIG. Oxidation Is Loss (of electrons) and Reduction Is Gain (of electrons).

The learners' response to the teacher's question.
An element/compound that act as a reducing agent is one that:
  1. undergoes oxidation.
  2. loses one or more electrons. i.e. donates electrons.
  3. causes reduction by donating electron(s).
An element/compound that acts as an oxidizing agent is one that:
  1. undergoes reduction.
  2. gain one or more electrons. i.e. accepts electrons.
  3. causes oxidation by removing electron(s) from others.
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.
STEP 3
(5 mins)
Group discussion
The teacher asks learners in group discussions to write a chemical equation for the formation of hydrogen chloride (HCl) and sulfur dioxide (SO₂).


The teacher explains to learners that there is a partial transfer of electrons (from H to Cl in HCl and from S to O in SO 2 ). and asks learners to write the transferred electron of the elements at the superscript on the elements.

NOTE: Oxidation numbers enable us to identify elements that are oxidized and reduced at a glance. The elements that show an increase in oxidation number—hydrogen and sulfur in the preceding examples—are oxidized. Chlorine and oxygen are reduced, so their oxidation numbers show a decrease from their initial values. Note that the sum of the oxidation numbers of H and Cl in HCl (+1 and -1) is zero. Likewise, if we add the charges on S (+4) and two atoms of O [2 x (-2)], the total is zero. The reason is that the HCl and SO₂ molecules are neutral, so the charges must cancel.

The teacher explains the rules to assign oxidation numbers, and thereafter asks learners to assign oxidation numbers to all the elements in the following compounds and ion: (a) Li₂O, (b) HNO₃, (c) Cr₂O₃²⁻.

  1. In free elements (that is, in the uncombined state), each atom has an oxidation number of zero. Thus, each atom in H₂, Br₂, Na, Be, K, O₂, and P₄ has the same oxidation number zero.
  2. For ions composed of only one atom (that is, monatomic ions), the oxidation number is equal to the charge on the ion. Thus, the Li⁺ ion has an oxidation number of +1; Ba²⁺ ion, +2; Fe³⁺ ion, +3; I⁻ ion, -1; O²⁻ ion, -2; and so on. All alkali metals have an oxidation number of +1 and all alkaline earth metals have an oxidation number of +2 in their compounds. Aluminium has an oxidation number of +3 in all its compounds.
  3. The oxidation number of oxygen in most compounds (for example, MgO and H₂O) is -2, but in hydrogen peroxide (H₂O₂) and peroxide ion (O₂²⁻), it is -1.
  4. The oxidation number of hydrogen is +1, except when it is bonded to metals in binary compounds. In these cases (for example, LiH, NaH, CaH₂), its oxidation number is -1.
  5. Fluorine has an oxidation number of -1 in all its compounds. Other halogens (Cl, Br, and I) have negative oxidation numbers when they occur as halide ions in their compounds. When combined with oxygen, for example in oxoacids and oxoanions, they have positive oxidation numbers.
  6. In a neutral molecule, the sum of the oxidation numbers of all the atoms must be zero. In a polyatomic ion, the sum of oxidation numbers of all the elements in the ion must be equal to the net charge of the ion. For example, in the ammonium ion, NH₄⁺, the oxidation number of N is -3 and that of H is +1. Thus the sum of the oxidation numbers is -3 + 4(+1) = +1, which is equal to the net charge of the ion.
  7. Oxidation numbers do not have to be integers. For example, the oxidation number of O in the superoxide ion, O₂⁻, is -1/2.
Learners are expected to respond.
H₂(g) + Cl₂(g) → 2HCl(g)
S(s) + O₂(g) → SO₂(g)




H₂⁰(g) + Cl₂⁰(g) → 2H⁺¹Cl⁻¹(g)
S⁰(s) + O₂⁰(g) → S⁺⁴O₂⁻²(g)

























Remember, all alkali metals have an oxidation number of +1, and in most cases, hydrogen has an oxidation number of 11 and oxygen has an oxidation number of -2 in their compounds.

SOLUTION
  1. Lithium has an oxidation number of +1 (Li⁺¹) and oxygen’s oxidation number is -2 (O²⁻).
  2. This is the formula for nitric acid, which yields an H⁺ ion and a NO₃⁻ ion in solution. H has an oxidation number of +1. Thus the other group (the nitrate ion) must have a net oxidation number of -1. Oxygen has an oxidation number of -2, and if we use n to represent the oxidation number of nitrogen, then the nitrate ion can be written as
    [N⁽ⁿ⁾O₃⁽²⁻⁾]⁻
    n + 3(-2) = -1
    n = +5
  3. oxidation numbers in the dichromate ion Cr₂O₇²⁻ must be -2.
    the oxidation number of O is -2, so all that remains is to determine the oxidation number of Cr, which we call n. The dichromate ion can be written as.
    [Cr₂⁽ⁿ⁾O₇²⁻]²⁻
    2(n) + 7(-2) = -2
    n = +6






CHECK
In each case, does the sum of the oxidation numbers of all the atoms equal the net charge on the species?
An atom’s oxidation number, also called oxidation state, signifies the number of charges the atom would have in a molecule (or an ionic compound) if electrons were transferred completely.

The numbers above the element symbols are oxidation numbers. In both of the reactions shown, there is no charge on the atoms in the reactant molecules. Thus, their oxidation number is zero. For the product molecules, however, it is assumed that complete electron transfer has taken place and that atoms have gained or lost electrons.

The oxidation numbers reflect the number of electrons “transferred.”
STEP 4
Balancing Redox Reaction
5 mins
The teacher asks learners to write a chemical reaction for the formation of magnesium oxide (MgO) from magnesium and oxygen, identifying the gain and loss of ions.


The teacher may assist the learners in writing the half-reaction.

Magnesium oxide (MgO) is an ionic compound made up of Mg²⁺ and O²⁻ ions. In this reaction, two Mg atoms give up or transfer four electrons to two O atoms (in O₂). For convenience, we can think of this process as two separate steps, one involving the loss of four electrons by the two Mg atoms and the other being the gain of four electrons by an O₂ molecule:
Learners are expected to respond.
2Mg(s) + O₂(g) → 2MgO(s)





2Mg → 2Mg²⁺ + 4e⁻
O₂ + 4e⁻ → 2O²⁻

2Mg + O₂ + 4e⁻ → 2Mg²⁺ + 2O²⁻ + 4e⁻

(cancel the electrons that appear on both sides of the equation)

2Mg + O₂ → 2Mg²⁺ + 2O²⁻
(the 2Mg²⁺ and 2O²⁻ ions combine to form MgO)

2Mg²⁺ + 2O²⁻ → 2MgO
Splitting chemical reactions into two parts, each of these steps is called a half-reaction, which explicitly shows the electrons involved in a redox reaction. The sum of the half-reactions gives the overall reaction.

Note: that in an oxidation half-reaction, electrons appear as the product; in a reduction half-reaction, electrons appear as the reactant.









balancing redox reactions
  1. in Aqueous Solutions
  2. in Acidic Solutions and
  3. in Basic Solutions
The teacher explain to learners that Redox reaction could be
  1. in Aqueous Solutions
  2. in Acidic Solutions and
  3. in Basic Solutions
Each of these uses the Half-Reaction Method for Balancing Oxidation-Reduction Reactions and has rules for balancing the Redox reactions.

The teacher explains the rules to learners.
Rules for balancing redox reactions in aqueous solutions
  1. separate the reaction into two half-reactions: one involving oxidation reaction and the other involving reduction reaction.
  2. Find the overall oxidation-reduction (redox) reaction by summing up the two half-reactions


Rules for balancing redox reactions in acidic solutions
  1. Write separate equations for the oxidation and reduction half-reactions.
  2. For each half-reaction,
    1. balance all the elements except hydrogen and oxygen.
    2. then balance oxygen using H₂O
    3. then balance hydrogen using H⁺
    4. 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


Rules for balancing redox reactions in basic solution
  1. when balancing equations of oxidation-reduction reactions that take place in a basic solution, we go through the first steps of the procedure as if the reaction were occurring in an acidic solution.
  2. add the appropriate number of OH⁻ ions to each side of the equation, combining H⁺ and OH⁻ to form H₂O.
The teacher guide learners to balance the following redox reactions.
  1. Balance the following equation of the reaction between cerium(IV) ion and tin(II) ion.
    Ce⁴⁺(aq) + Sn²⁺(aq) Ce³⁺(aq) + Sn⁴⁺(aq)
  2. Balance the following equation of the reaction between
    MnO₄⁻ + NO₂⁻ → Mn²⁺ + NO₃⁻
  3. Balance the following equation of the reaction between
    I⁻ + MnO₄⁻ → I₂ + MnO₂

The learners follow the teacher's guidelines
SOLUTION
  1. Ce⁴⁺(aq) + Sn²⁺(aq) → Ce³⁺(aq) + Sn⁴⁺(aq)

    Ce⁴⁺(aq) + e⁻ → Ce³⁺(aq)       (reduction half-reaction)
    Sn²⁺(aq) → Sn⁴⁺(aq) + 2e⁻      (oxidation half-reaction)

    2Ce⁴⁺(aq) + 2e⁻ → 2Ce³⁺(aq)
    Sn²⁺(aq) → Sn⁴⁺(aq) + 2e⁻

    2Ce⁴⁺(aq) + Sn²⁺(aq) → 2Ce³⁺(aq) + Sn⁴⁺(aq)      (overall redox reaction)


  2. MnO₄⁻ + NO₂⁻ → Mn²⁺ + NO₃⁻

    MnO₄⁻ → Mn²⁺       (reduction half-reaction)
    NO₂⁻ → NO₃⁻       (oxidation half-reaction)

    MnO₄⁻ → Mn²⁺ + 4H₂O
    NO₂⁻ + H₂O → NO₃⁻

    MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O
    NO₂⁻ + H₂O → NO₃⁻ + 2H⁺

    MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
    NO₂⁻ + H₂O → NO₃⁻ + 2H⁺ + 2e⁻

    2x[MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O]
    5x[NO₂⁻ + H₂O → NO₃⁻ + 2H⁺ + 2e⁻]

    2MnO₄⁻ + 16H⁺ + 10e⁻ → 2Mn²⁺ + 8H₂O
    5NO₂⁻ + 5H₂O → 5NO₃⁻ + 10H⁺ + 10e⁻

    2MnO₄⁻ + 6H⁺ + 5NO₂⁻ → 2Mn²⁺ + 3H₂O + 5NO₃⁻      (overall redox reaction)


  3. I⁻ + MnO₄⁻ → I₂ + MnO₂

    MnO₄⁻ → MnO₂
    I⁻ → I₂

    MnO₄⁻ → MnO₂
    2I⁻ → I₂

    MnO₄⁻ → MnO₂ + 2H₂O
    2I⁻ → I₂

    MnO₄⁻ + 4H⁺ + 4OH⁻ → MnO₂ + 2H₂O + 4OH⁻
    2I⁻ → I₂

    MnO₄⁻ + 4H₂O → MnO₂ + 2H₂O + 4OH⁻
    2I⁻ → I₂

    MnO₄⁻ + 2H₂O → MnO₂ + 4OH⁻
    2I⁻ → I₂

    MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻
    2I⁻ → I₂ + 2e⁻

    2x[MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻]
    3x[2I⁻ → I₂ + 2e⁻]

    2MnO₄⁻ + 4H₂O + 6e⁻ → 2MnO₂ + 8OH⁻
    6I⁻ → 3I₂ + 6e⁻

    2MnO₄⁻ + 6I + 4H₂O → 2MnO₂ + 3I₂ + 8OH⁻      (overall redox reaction)

We provide educational resources/materials, curriculum guide, syllabus, scheme of work, lesson note & plan, waec, jamb, O-level & advance level GCE lessons/tutorial classes, on various topics, subjects, career, disciplines & department etc. for all the Class of Learners

EVALUATION
3 mins
The teacher asks questions to the learners to evaluate the lesson.
  1. Define oxidation and reduction.
  2. Suggest at least two chemical equations where oxidation and reduction reactions occurred.
  3. In the following unbalanced equations, indicate the reactant oxidized, the reactant reduced, the oxidizing agent, and the reducing agent. Indicate the products that contain the elements that were oxidized or reduced.
    (i) Al + HCl → AlCl₃ + H₂
    (ii) CH₄ + O₂ → CO₂ + H₂O
    (iii) MnO₂ + HCl → MnCl₂ + Cl₂ + H₂O
    (iv) K₂Cr₂O₇ + SnCl₂ + HCl → CrCl₃ + SnCl₄ + KCl + H₂O
  4. Working using oxidation numbers only, determine whether the species under-lined has undergone oxidation or reduction
    1. C(s) + H₂O(g) → CO(g) + H₂(g)
    2. 2Na(s) + H₂(g) 2NaH(s)
    3. 2FeCl₂(aq) + Cl₂(g) → 2FeCl₃(aq)
  5. What is the oxidation number of chromium in each of the following compounds or ions?
    1. Cr
    2. CrO₃
    3. CrCl₃
    4. Cr₂O₃
    5. Cr₂O₇²⁻
    6. CrO₄²⁻
  6. Balance the following equations for reactions
    1. MnO₄⁻(aq) + SO₂(g) + H₂O(l) → Mn²⁺(aq) + SO₄²⁻(aq) + H⁺(aq)
    2. Cu(s) + NO₃⁻(aq) → Cu²⁺(aq) + H₂O + NO(g)
    3. Bi₂O₃(s) + NO₃⁻(aq) + OH⁻(aq) → BiO₃⁻(aq) + NO₂⁻(aq) + H₂O(l)
Learners respond to the teacher's questions.
  1. oxidation reaction refers to the half-reaction that involves the loss of electrons, the combination of elements with oxygen.
    A reduction reaction is a half-reaction that involves the gain of electrons.
    1. 2Mg(s) + O₂(g) → 2MgO(s)
    2. Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s)
    3. H₂(g) + Cl₂(g) → 2HCl(g)
    4. S(s) + O₂(g) → SO₂(g)
  2. HINT: we wish to identify the species that contain atoms of an element changing oxidation state. At first, it may be necessary to calculate the oxidation state of every atom in the equation until you can recognize the changes by inspection.

    1. Al⁰ + H⁺¹Cl⁻¹ → Al⁺³Cl₃⁻¹ + H₂⁰
      Reactant Al is oxidized
      Reactant Al is the reducing agent
      The product is AlCl₃

      Reactant HCL is reduced
      Reactant HCL is the oxidizing agent
      The product is H₂
    2. C⁻⁴H₄⁺¹ + O₂⁰ → C⁺⁴O⁻²₂ + H₂⁺¹O⁻²
      Reactant CH₄ is oxidized
      Reactant CH₄ is the reducing agent
      The product is CO₂

      Reactant O₂ is reduced
      Reactant O₂ is the oxidizing agent
      The product is CO₂, H₂O
    3. Mn⁺⁴O₂⁻² + H⁺¹Cl⁻¹ → Mn⁺²Cl₂⁻¹ + Cl₂⁰ + H₂⁺O⁻²
      Reactant HCl is oxidized
      Reactant HCl is the reducing agent
      The product is Cl₂

      Reactant MnO₂ is reduced
      Reactant MnO₂ is the oxidizing agent
      The product is MnCl₂
    4. K⁺¹₂Cr₂⁺⁶O₇⁻² + Sn⁺²Cl₂⁻¹ + H⁺¹Cl⁻¹ → Cr⁺³Cl₃⁻¹ + Sn⁺⁴Cl₄¹⁻ + K⁺¹Cl⁻¹ + H₂⁺¹O⁻²
      Reactant SnCl₂ is oxidized
      Reactant SnCl₂ is the reducing agent
      The product is SnCl₄

      Reactant K₂Cr₂O₇ is reduced
      Reactant K₂Cr₂O₇ is the oxidizing agent
      The product is CrCl₃
Ask the learners questions to assess the achievement of the set objectives.
    In question No. 3
  • A N A LY S I S
    You will notice that any substance present as a free element in a reaction is involved in either the oxidation or the reduction process. Also, hydrogen and oxygen are generally not oxidized or reduced if they remain part of compounds. They are involved only if present as free elements in either the reactants or products. Further note that none of the analyses done for this problem requires a balanced reaction. It is sufficient simply to know the identities of the reactants and their products.
  • S Y N T H E S I S
    Consider the equation illustrating the decomposition of hydrogen peroxide:
    2H₂O₂(aq) → 2H₂O(l) + O₂(g)
    What has been oxidized and what has been reduced? In this case. H₂O₂ plays both roles. In one molecule of H₂O₂, the oxidation state of the oxygen has decreased from -1 to -2 to form H₂O, so it has been reduced. In the other molecule of H₂O₂, the oxygen has increased from -1 to 0, so it has been oxidized. One molecule of hydrogen peroxide is the oxidizing agent and one is the reducing agent. In such a case, we say that the H₂O₂ has been disproportionated.
CONCLUSION
2mins
The teacher guides the learners so that they come up with the concepts from the observations drawn. (Bridging the activities to the concept).The two processes (oxidation and reduction) complement each other. If there is oxidation occurs, then there must also be a reduction, giving us the basis for this classification of a chemical reaction. Reactions involving an exchange of electrons are known as oxidation-reduction, simply, redox reactions.Application: expansion of the concept
ASSIGNMENT
  1. What is the oxidation state of the following?
    1. Fe in FeO
    2. S in H₂SO₃
    3. N in N₂O₅
    4. As in AsO₄³⁻
  2. Classify the following redox reactions and indicate changes in the oxidation numbers of the elements:
    1. 2N₂O(g) → 2N₂(g) + O₂(g)
    2. 6Li(s) + N₂(g) → 2Li₃N(s)
    3. Ni(s) + Pb(NO₃)₂(aq) → Pb(s) + Ni(NO₃)₂(aq)
    4. 2NO₂(g) + H₂O(l) → HNO₂(aq) + HNO₃(aq)
Learners answer other questions.Improving their level of understanding of Redox's reaction.

Free Will Donation

We know times are tough right now, but if you could donate and support us, be rest assured that your great contributions are immensely appreciated and will be for the progress of our work to help us pay for the server cost, domain renewal, and other maintenance costs of the site. Nothing is too small; nothing is too little.

Account Details

BANK: UNITED BANK FOR AFRICA PLC

ACCOUNT NAME: OFAGBE GODSPOWER GEORGE

ACCOUNT NUMBER: 2250582550

SWIFT CODE: UNAFNGLA

ACCOUNT TYPE: SAVINGS

CURRENCY: DOLLAR (USD) ACCOUNT

ADDRESS: 1. M. Aruna Close, Ughelli, Delta State, Nigeria

PHONE: +234805 5084784, +234803 5586470



BANK: UNITED BANK FOR AFRICA Plc (UBA)

ACCOUNT NAME: OFAGBE GODSPOWER GEORGE

ACCOUNT NUMBER: 2042116266

SORT CODE: 033243371

ACCOUNT TYPE: SAVINGS

CURRENCY: NAIRA ACCOUNT

ADDRESS: 1. M. Aruna Close, Ughelli, Delta State, Nigeria

PHONE: +234805 5084784, +234803 5586470



Your active support gives strength to our Team and inspires to work. Each donated dollar is not only money for us, but it is also the confidence that you really need our project!
AseiClass is a non-profit project that exists at its founders' expense, it will be difficult to achieve our goals without your help.
Please consider making a donation.
Thank you.


AseiClass Team

Facts about Teachers

● ● ● Teachers Are Great No Controversy.

● ● ● Teachers are like candles, they burn themselves to light others.

● ● ● Teachers don't teach for the money.

● ● ● Every great mind was once taught by some brilliant teachers.

● ● ● Teachers are the second parents we have.

● ● ● If you can write your name, thank your teacher.

Teaching slogans

● ● ● Until the learner learns the teacher has not taught.

● ● ● I hear and forget, I see and remember, I do and know.

● ● ● The good teacher explains. The superior teacher demonstrates. The great teacher inspires.

We provide educational resources/materials, curriculum guide, syllabus, scheme of work, lesson note & plan, waec, jamb, O-level & advance level GCE lessons/tutorial classes, on various topics, subjects, career, disciplines & department etc. for all the Class of Learners