Lesson Notes By Weeks and Term v3 - Senior Secondary 2
Oxidation-Reduction (Redox) Reactions
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Oxidation-Reduction (Redox) Reactions
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Subject: Chemistry
Class: Senior Secondary 2
Term: 1st Term
Week: 4
Theme: Chemistry And Industry
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Define oxidation as:addition of oxygen;removal of hydrogen;process of electron loss;process of in crease of oxidation number of as ubstance Define reduction as;the reverse of any of the.above processes Calculate oxidationnumbers of elements using as et of arbitrary rules viz:(a) oxidation number of freeelements = 0(b) oxidation number of oxygen in any compound is -2, except in peroxides whereit is - 1(c) oxidation number of His + 1, except in hydrideswhere it is - 1(d) oxidation number of aneutral molecule or compound is zero e.g. H2SO4 =0 etc.
Example: For SO42−, (1 × ON of S) + (4 × ON of O) = -2. 2.
3. Calculation of Oxidation Numbers (Worked Examples)
Example 1: Calculate the oxidation number of Sulphur (S) in H2SO
4. Hydrogen (H) = +1 Oxygen (O) = -2 The compound is neutral, so the sum of ON = 0. (2 × +1) + (1 × ON of S) + (4 × -2) = 0 +2 + ON of S - 8 = 0 ON of S - 6 = 0 ON of S = +6 Therefore, the oxidation number of Sulphur in H2SO4 is +
6. Example 2: Calculate the oxidation number of Manganese (Mn) in KMnO
4. Potassium (K) (Group 1) = +1 Oxygen (O) = -2 The compound is neutral, so the sum of ON = 0. (1 × +1) + (1 × ON of Mn) + (4 × -2) = 0 +1 + ON of Mn - 8 = 0 ON of Mn - 7 = 0 ON of Mn = +7 Therefore, the oxidation number of Manganese in KMnO4 is +
7. Example 3: Calculate the oxidation number of Chromium (Cr) in the dichromate ion, Cr2O72−. Oxygen (O) = -2 The ion has a charge of -2, so the sum of ON = -2. (2 × ON of Cr) + (7 × -2) = -2 2(ON of Cr) - 14 = -2 2(ON of Cr) = +14 - 2 2(ON of Cr) = +12 ON of Cr = +12 / 2 ON of Cr = +6 Therefore, the oxidation number of Chromium in Cr2O72− is +
6. Example 4: Calculate the oxidation number of Sulphur (S) in sodium thiosulphate, Na2S2O
3. Sodium (Na) (Group 1) = +1 Oxygen (O) = -2 The compound is neutral, so the sum of ON = 0. (2 × +1) + (2 × ON of S) + (3 × -2) = 0 +2 + 2(ON of S) - 6 = 0 2(ON of S) - 4 = 0 2(ON of S) = +4 ON of S = +2 Therefore, the average oxidation number of Sulphur in Na2S2O3 is +2. (
Note: In such compounds, Sulphur atoms might have different oxidation states, but the calculated value is the average). 2.
4. Naming Inorganic Compounds Using Oxidation Numbers (Stock Notation) The Stock notation uses Roman numerals in parentheses after the metal's name to indicate its oxidation number. This is crucial for metals that can form ions with different charges (transition metals, post-transition metals).
Iron(II) chloride: FeCl2 (Fe has ON +2)
Iron(III) chloride: FeCl3 (Fe has ON +3)
Copper(I) oxide: Cu2O (Cu has ON +1)
Copper(II) oxide: CuO (Cu has ON +2)
Sulphur(IV) oxide: SO2 (S has ON +4)
Sulphur(VI) oxide: SO3 (S has ON +6)
Nitrogen(II) oxide: NO (N has ON +2)
Nitrogen(IV) oxide: NO2 (N has ON +4)
Manganese(VII) oxide: Mn2O7 (Mn has ON +7)
Naming Hydrated Compounds: For hydrated compounds, the number of water molecules is indicated using Greek prefixes (mono-, di-, tri-, tetra-, penta-, hexa-, hepta-, octa-, nona-, deca-) followed by "hydrate". Copper(II) sulphate-5-water or Copper(II) sulphate pentahydrate: CuSO4·5H2O 2.
5. Oxidizing and Reducing Agents Oxidizing Agent (Oxidant): Causes oxidation in another substance. Itself undergoes reduction. Gains electrons. Its oxidation number decreases.
Common examples: Oxygen, chlorine, concentrated sulphuric acid, nitric acid, potassium permanganate (KMnO4), potassium dichromate (K2Cr2O7).
Reducing Agent (Reductant): Causes reduction in another substance. Itself undergoes oxidation. Loses electrons. Its oxidation number increases.
Common examples: Hydrogen, carbon, carbon monoxide, hydrogen sulphide (H2S), sulphur dioxide (SO2), metals (Na, Mg, Zn), H2S, SO2, sodium thiosulphate (Na2S2O3).
Example: 2FeCl3 + H2S → 2FeCl2 + 2HCl + S Fe in FeCl3: ON = +3 Fe in FeCl2: ON = +2 (Decrease, so FeCl3 is reduced. FeCl3 is the oxidizing agent.)
S in H2S: ON = -2 S in S: ON = 0 (Increase, so H2S is oxidized. H2S is 2.
1. Defining Oxidation and Reduction Oxidation can be defined in four ways:
1. Addition of Oxygen: A substance gains oxygen atoms.
Example: 2Mg(s) + O2(g) → 2MgO(s) (Magnesium is oxidized)
2. Removal of Hydrogen: A substance loses hydrogen atoms.
Example: H2S(g) + Cl2(g) → S(s) + 2HCl(g) (Hydrogen sulphide is oxidized)
3. Process of Electron Loss (LEO - Loss of Electrons is Oxidation): A substance donates one or more electrons. This is the most fundamental definition.
Example: Na → Na+ + e− (Sodium is oxidized)
Example: Fe2+ → Fe3+ + e− (Iron(II) ion is oxidized)
4. Process of Increase in Oxidation Number: The oxidation number of an element in a substance increases.
Example: Mn in MnO2 (+4) → Mn in MnO4− (+7) (Manganese is oxidized) Reduction is the reverse of any of the above processes:
1. Removal of Oxygen: A substance loses oxygen atoms.
Example: CuO(s) + H2(g) → Cu(s) + H2O(g) (Copper(II) oxide is reduced)
2. Addition of Hydrogen: A substance gains hydrogen atoms.
Example: CH2=CH2 + H2 → CH3-CH3 (Ethene is reduced)
3. Process of Electron Gain (GER - Gain of Electrons is Reduction): A substance accepts one or more electrons.
Example: Cl + e− → Cl− (Chlorine is reduced)
Example: MnO4− → Mn2+ (Manganese is reduced)
4. Process of Decrease in Oxidation Number: The oxidation number of an element in a substance decreases.
Example: Mn in MnO4− (+7) → Mn in MnO2 (+4) (Manganese is reduced)
Redox Reaction: A reaction in which both oxidation and reduction occur simultaneously. One substance is oxidized while another is reduced. 2.
2. Rules for Assigning Oxidation Numbers Oxidation number (or oxidation state) is a hypothetical charge an atom would have if all bonds were 100% ionic.
1. Free elements: The oxidation number of an atom in its elemental form is zero.
Examples: Na, O2, Cl2, Fe, S8, P4 all have an oxidation number of 0.
2. Monatomic Ions: The oxidation number of a monatomic ion is equal to its charge.
Examples: Na+ = +1, Cl− = -1, Al3+ = +3, O2− = -2.
3. Oxygen: The oxidation number of oxygen in most compounds is -
2. Examples: H2O (-2), CO2 (-2), H2SO4 (-2).
Exceptions: In peroxides (e.g., H2O2, Na2O2), oxygen's oxidation number is -
1. In superoxides (e.g., KO2, RbO2), oxygen's oxidation number is -1⁄2. (Less common for SS2, but good to mention). When bonded to fluorine (e.g., OF2), oxygen's oxidation number is +2 (because fluorine is more electronegative).
4. Hydrogen: The oxidation number of hydrogen in most compounds is +
1. Examples: H2O (+1), HCl (+1), H2SO4 (+1).
Exception: In metal hydrides (e.g., NaH, CaH2), hydrogen's oxidation number is -1 (as metals are less electronegative than hydrogen in these cases).
5. Group 1 Metals: Always have an oxidation number of +1 in compounds (e.g., Li, Na, K).
6. Group 2 Metals: Always have an oxidation number of +2 in compounds (e.g., Be, Mg, Ca).
7. Fluorine: Always has an oxidation number of -1 in compounds. Other halogens (Cl, Br, I) usually have -1, but can have positive oxidation numbers when bonded to oxygen or a more electronegative halogen.
8. Neutral Molecules: The sum of the oxidation numbers of all atoms in a neutral molecule is zero.
Example: For H2SO4, (2 × ON of H) + (1 × ON of S) + (4 × ON of O) = 0.
9. Polyatomic Ions: The sum of the oxidation numbers of all atoms in a polyatomic ion is equal to the charge of the ion.
Example: For SO42−, (1 × ON of S) + (4 × ON of O) = -2. 2.
3. Calculation of Oxidation Numbers (Worked Examples)
Example 1: Calculate the oxidation number of Sulphur (S) in H2SO
4. Hydrogen (H) = +1 Oxygen (O) = -2 The compound is neutral, so the sum of ON = 0. (2 × +1) + (1 × ON of S) + (4 × -2) = 0 +2 + ON of S - 8 = 0 ON of S - 6 = 0 * ON right means 1H2O remaining on right): 2Cr(OH)3 + 3ClO− → 2CrO42− + 4H+ + 3Cl− + H2O. (This is the correctly balanced equation in acidic medium). Now, convert to basic medium: Add 4OH− to both sides (to neutralize 4H+): 2Cr(OH)3 + 3ClO− + 4OH− → 2CrO42− + 4H+ + 3Cl− + H2O + 4OH− Combine 4H+ and 4OH− to 4H2O on the right: 2Cr(OH)3 + 3ClO− + 4OH− → 2CrO42− + 4H2O + 3Cl− + H2O Combine H2O on the right: 2Cr(OH)3 + 3ClO− + 4OH− → 2CrO42− + 5H2O + 3Cl− Verify: Atoms: Left: Cr=2, O=(33 from Cr(OH)3 + 3 from ClO− + 4 from OH−) = 9+3+4 =
1
6. H=(32 from Cr(OH)3 + 4 from OH−) = 6+4=
1
0. Cl=
3. Right: Cr=2, O=(24 from CrO42− + 5 from H2O) = 8+5 =
1
3. H=(52 from H2O) =
1
0. Cl=
3. Self-correction again! My oxygen count is off. Let's re-verify from the "acidic medium" step: 2Cr(OH)3 + 3ClO− → 2CrO42− + 4H+ + 3Cl− + H2O Left side: Cr=2, O=(32)=6, H=(32)=6, Cl=3, O=
3. Total O=9, H=
6. Right side: Cr=2, O=(42)=8, H=4, Cl=3, O=1, H=
2. Total O=9, H=
6. So, the acidic equation is correct: 2Cr(OH)3 + 3ClO− → 2CrO42− + 4H+ + 3Cl− + H2O Now, convert to basic: Add 4OH− to both sides: 2Cr(OH)3 + 3ClO− + 4OH− → 2CrO42− + 4H+ + 3Cl− + H2O + 4OH− Combine 4H+ + 4OH− → 4H2O on the right: 2Cr(OH)3 + 3ClO− + 4OH− → 2CrO42− + 4H2O + 3Cl− + H2O Simplify H2O (1 H2O on right cancels, leaving 4H2O from the 4H2O term): 2Cr(OH)3 + 3ClO− + 4OH− → 2CrO42− + 3Cl− + 4H2O Final Verification in Basic Medium: Atoms: Left: Cr=2, O=(32 + 3 + 4) = 6+3+4 =
1
3. H=(32 + 4) = 6+4 =
1
0. Cl=
3. Right: Cr=2, O=(24 + 4) = 8+4 =
1
2. H=(42) =
8. Cl=
3. Still not balanced! This shows how tricky it can be. The issue might be in balancing Cr(OH)3 + H2O → CrO42− + 5H+ + 3e− Let's restart the Cr(OH)3 half-reaction for basic conditions from scratch. Alternative (More Direct) for Basic Medium Balancing:
1. Split: Cr(OH)3 → CrO42− and ClO− → Cl−
2. Balance non-O, non-H: (Already done)
3. Balance O: Cr(OH)3 + H2O → CrO42− (Added H2O, 4O on right, 3O from Cr(OH)3 + 1O from H2O = 4O on left. Correct) ClO− → Cl− + H2O
4. Balance H: Cr(OH)3 + H2O + 5OH− → CrO42− (Balance H with H2O and OH−. 5H on left from Cr(OH)3 + H2
O. Add 5OH− to left. Needs 5H2O on right side to balance
H. Let's try another approach for basic media: add H2O to balance O, then OH− to balance H, then electrons to balance charge). Let's use the standard "acid first, then basic" conversion as it's typically taught. The previous acidic balance was confirmed correct. 2Cr(OH)3 + 3ClO− → 2CrO42− + 4H+ + 3Cl− + H2O (Acidic balanced equation) Add 4OH− to both sides: 2Cr(OH)3 + 3ClO− + 4OH− → 2CrO42− + 4H+ + 3Cl− + H2O + 4OH− Convert 4H+ + 4OH− to 4H2O: 2Cr(OH)3 + 3ClO− + 4OH− → 2CrO42− + 4H2O + 3Cl− + H2O Combine H2O on right: 2Cr(OH)3 + 3ClO− + 4OH− → 2CrO42− + 5H2O + 3Cl− Final verification of the basic equation: Left side: Cr: 2 O: (2 3) + 3 + 4 = 6 + 3 + 4 = 13 H: (2 3) + 4 = 6 + 4 = 10 Cl: 3 Charge: (3 -1) + (4 -1) = -3 - 4 = -7 Right side: Cr: 2 O: (2 4) + 5 = 8 + 5 = 13 H: (5 2) = 10 Cl: 3 Charge: (2 -2) + (3 -1) = -4 - 3 = -7 The equation is now balanced in basic medium! The earlier counting errors were due to miscalculation during verification. This example demonstrates the meticulous nature required for balancing redox equations. --- Oxidation: Cr(OH)3 → CrO42− Cr(OH)3 + H2O → CrO42− (Balance O) Cr(OH)3 + H2O → CrO42− + 5H+ (Balance H: 3H on left from Cr(OH)3, 2H from H2O, total 5
H. Needs 5H+ on right) Cr(OH)3 + H2O → CrO42− + 5H+ + 3e− (Balance Charge: -2 + 5 = +3 on right. +0 on left, so 3e− added to right)
Reduction: ClO− → Cl− ClO− → Cl− + H2O (Balance O) ClO− + 2H+ → Cl− + H2O (Balance H) ClO− + 2H+ + 2e− → Cl− + H2O (Balance Charge: -1 + 2 = +1 on left. -1 on right, so 2e− added to left)
Equalize Electrons: Multiply oxidation by 2, reduction by 3 (LCM of 3 and 2 is 6). 2Cr(OH)3 + 2H2O → 2CrO42− + 10H+ + 6e− 3ClO− + 6H+ + 6e− → 3Cl− + 3H2O Combine and Simplify (Acidic): 2Cr(OH)3 + 2H2O + 3ClO− + 6H+ + 6e− → 2CrO42− + 10H+ + 6e− + 3Cl− + 3H2O Cancel 6e− and simplify H+ and H2O: 2Cr(OH)3 + 3ClO− → 2CrO42− + 4H+ + 3Cl− + H2O
2. Neutralize H+ (Add OH−): There are 4H+ on the right. Add 4OH− to both sides. 2Cr(OH)3 + 3ClO− + 4OH− → 2CrO42− + 4H+ + 3Cl− + H2O + 4OH−
3. Combine H+ and OH−: On the right side, 4H+ + 4OH− combine to form 4H2O. 2Cr(OH)3 + 3ClO− + 4OH− → 2CrO42− + 4H2O + 3Cl− + H2O
4. Cancel H2O: 1H2O on the left, 5H2O on the right. Cancel 1 H2O from both sides. 2Cr(OH)3 + 3ClO− + 4OH− → 2CrO42− + 3H2O + 3Cl−
5. Verify: Atoms: Left (2Cr, 6O, 6H, 3Cl, 4O, 4H) = (2Cr, 10O, 10H, 3Cl). Right (2Cr, 8O, 3Cl, 6H, 3O, 0H) = (2Cr, 11O, 6H, 3Cl).
Self-correction: Error in counting atoms or earlier balancing. Let's recheck step 1 for the reduction half-reaction after balancing in acidic medium. Re-doing the reduction half-reaction in acidic medium: ClO− → Cl− ClO− → Cl− + H2O (Balance O) ClO− + 2H+ → Cl− + H2O (Balance H)
Charge: ClO− (+2H+) = -1 + 2 = +
1. Cl− + H2O = -1 + 0 = -
1. Add 2e− to left: ClO− + 2H+ + 2e− → Cl− + H2O. (Charge: +1 - 2 = -1 on left, -1 on right). This is correct. Re-doing the oxidation half-reaction in acidic medium: Cr(OH)3 → CrO42− Cr(OH)3 + H2O → CrO42− (Balance O: 3O from Cr(OH)3 + 1O from H2O = 4
O. Correct). Cr(OH)3 + H2O → CrO42− + 5H+ (Balance H: 3H from Cr(OH)3 + 2H from H2O = 5H on left. Needs 5H+ on right). Correct.
Charge: Cr(OH)3 + H2O =
0. CrO42− + 5H+ = -2 + 5 = +
3. Needs 3e− on the right. Cr(OH)3 + H2O → CrO42− + 5H+ + 3e−. (Charge: 0 on left, +3 - 3 = 0 on right). This is correct.
Re-combining: Multiply oxidation by 2: 2Cr(OH)3 + 2H2O → 2CrO42− + 10H+ + 6e− Multiply reduction by 3: 3ClO− + 6H+ + 6e− → 3Cl− + 3H2O Add: 2Cr(OH)3 + 2H2O + 3ClO− + 6H+ + 6e− → 2CrO42− + 10H+ + 6e− + 3Cl− + 3H2O Cancel electrons: 2Cr(OH)3 + 2H2O + 3ClO− + 6H+ → 2CrO42− + 10H+ + 3Cl− + 3H2O Simplify H+ (6H+ on left, 10H+ on right means 4H+ remaining on right): 2Cr(OH)3 + 3ClO− + 2H2O → 2CrO42− + 4H+ + 3Cl− + 3H2O Simplify H2O (2H2O on left, 3H2O on right means 1H2O remaining on right): 2Cr(OH)3 + 3ClO− → 2CrO42− + 4H+ + 3Cl− + H2O. (This is the correctly balanced equation in acidic medium). Now, convert to basic medium: Add 4OH− to both sides (to neutralize 4H+): 2Cr(OH)3 + 3ClO− + 4OH− → 2CrO42− + 4H+ + 3Cl− + H2O + 4OH− Combine 4H+ and 4OH− to 4H2O on the right: 2Cr(OH)3 + 3ClO− + 4OH− → 2CrO42− + 4H2O + 3Cl− + H2O Combine H2O on the right: 2Cr(OH)3 + 3ClO− + 4OH− → 2CrO42− + 5H2O + 3Cl−
Corrosion Prevention and Material Science: Nigerian Context: Rusting of iron is a common sight on metal roofs, vehicles, and infrastructure like bridges in Nigeria. This is an oxidation reaction of iron. Teachers can discuss methods of corrosion prevention, such as painting, galvanizing (coating with zinc, which is more easily oxidized than iron and thus acts as a sacrificial anode), and cathodic protection (used for pipelines, common in oil-rich regions). Understanding redox helps students appreciate the science behind protecting valuable assets from environmental degradation.
Batteries and Energy Storage: Nigerian Context: Dry cell batteries (e.g., for torches, radios) and lead-acid batteries (for vehicles, inverters) are ubiquitous in Nigeria. All these devices rely on controlled redox reactions to generate electrical energy. The teacher can explain how the chemical reactions (oxidation at the anode, reduction at the cathode) produce a flow of electrons. This connects directly to the need for reliable power sources and the growing interest in renewable energy storage solutions in the country.
Water Treatment and Sanitation: Nigerian Context: Chlorination of water (using chlorine or hypochlorite, strong oxidizing agents) is a vital process for making water safe for drinking, especially in urban areas and some rural communities. These oxidizing agents kill bacteria and pathogens. Similarly, bleaching powders used in laundries and industries (e.g., textile factories in Kaduna or Kano) work by oxidizing coloured impurities. This highlights the importance of redox in public health and industrial hygiene.
Food Processing and Preservation: Nigerian Context: The spoilage of food, especially oils and fats (e.g., palm oil, groundnut oil), involves oxidation reactions (rancidity). Antioxidants, which are reducing agents, are added to processed foods to slow down these reactions and extend shelf life. This relates to food security and quality control in local food industries. ---