3.1 Oxidation and Reduction

Oxidation and reduction
■ Definition of redox reaction

	►	A process involves oxidation and reduction that occurs simultaneously.
	►	In any redox reaction, there is only one oxidation process and one reduction process.

■ Oxidation and reduction can be defined in terms of

	►	loss or gain of hydrogen
	►	loss or gain of oxygen
	►	transfer of electrons
	►	change in oxidation number

■ Definition of oxidising agent

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A substance that is reduced in a reaction.

■ Definition of reducing agent

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A substance that is oxidised in a reaction.

■ This video contains information on the redox reaction

■ This video contains information on the redox reaction.

■ Oxidation and reduction in terms of loss and gain of hydrogen

	►	Oxidation: the process of losing hydrogen
	►	Reduction: the process of gaining hydrogen
	►	Example: In the redox reaction between hydrogen sulfide and chlorine gas. Chlorine gas is reduced to hydrogen chloride. Chlorine gas acts as an oxidising agent. Hydrogen sulfide is oxidized to sulfur. Hydrogen sulfide acts as a reducing agent.

■ This video contains information on the oxidation and reduction (loss and gain of hydrogen)

■ Oxidation and reduction in terms of loss and gain of oxygen

	►	Oxidation: the process of gaining oxygen
	►	Reduction: the process of losing oxygen
	►	Example: in the redox reaction between tin(IV) oxide (SnO2) and carbon (C) Tin (IV) oxide is reduced to metallic tin. Tin (IV) oxide acts as an oxidising agent. Carbon is oxidized to carbon dioxide (CO2). Carbon acts as a reducing agent.

■ This video contains information on the oxidation and reduction(Loss and gain of hydrogen)

Oxidation and reduction in terms of electron transfer
■ Oxidation and reduction in terms of electron transfer

	►	Oxidation : the process of losing (releasing) electrons
	►	Reduction : the process of gaining (receiving) electrons
	►	The redox reaction can be defined as a process that involves electron transfer.
	►	The redox reaction can be defined as a process that involves electron transfer.
	►	Oxidising agent can be defined as an electron receiver.
	►	The following mnemonic method can be used to memorize the definition of oxidation and reduction in terms of electron transfer: ○ Mnemonic method : OIL (Oxidation Is Losing – electrons) RIG (Reduction Is Gaining - electrons)

■ This video contains information on the oxidation and reduction (transfer of electron)

■ Oxidation and reduction in terms of electron transfer

	►	Redox reaction between sodium and chlorine gas. Each sodium atom has lost an electron to form ions Na+ (in Na+Cl−) according to the following half ionic equation: Na → Na+ + e− (oxidation → loss of electrons) Each atom of chlorine has received an electron to form Cl− ion by ion half following equation: Cl2 + 2e− → 2Cl− (reduction→ electron received)
	►	Redox reaction between magnesium and oxygen gas. Each magnesium atom has lost two electrons to form ions Mg2+ (in Mg2+O2−) according to the following half ionic equation: Mg – 2e− → Mg2+ (oxidation → loss of electrons) Each oxygen atom has received an electron to form ions O2− ions by half following equation: O + 2e− → O2− (reduction → electron received)

Oxidation and reduction in terms of change in oxidation number
■ Oxidation number

	►	The charge that the atom of the element would have if complete transfer of electron take place.
	►	An atom of an element will have a positive oxidation number if it loses electron(s) easily.
	►	An atom of an element will have negative oxidation number if it obtains electron(s).
	►	The oxidation number is equal to the number of negative charge loses or obtains.
	►	Oxidation : the increase in oxidation number
	►	Reduction : the decrease in oxidation number
	►	Example: In the redox reaction between hydrogen sulfide and chlorine gas. Chlorine gas is reduced to hydrogen chloride. Chlorine gas acts as an oxidising agent. Hydrogen sulfide is oxidized to sulfur. Hydrogen sulfide acts as a reducing agent.

■ This video contains information on the oxidation and reduction (oxidation number)

Oxidation number
■ Assigning oxidation number of elements in a compound.

	►	The oxidation number of an atom in the elemental state is zero. ○ Example: The oxidation number of Mg, Zn, Cu, H2, O2, Cl2, Br2 is zero.
	►	The oxidation number of a monatomic ion is equal to its charge. ○ Example: In the compound sodium chloride, NaCl, the sodium has an oxidation number of +1 and the chlorine is -1.
	►	The algebraic sum of the oxidation numbers in the formula of a compound is zero. ○ Example: The oxidation numbers in sodium chloride, NaCl, is zero.
	►	The oxidation number of hydrogen in a compound is +1, except when hydrogen forms compounds called hydride with active metals, and then it is -1. ○ Example: The oxidation number of H is +1 in H2O, but is -1 in sodium hydride, NaH.
	►	The oxidation number of oxygen in a compound is -2, except in peroxides. ○ Example: In H2O the oxygen is -2, in H2O2 it is -1.
	►	The algebraic sum of the oxidation numbers in the formula for a polyatomic ion is equal to the charge on that ion. ○ Example: In the sulfate ion, SO42−, the oxidation numbers of the sulfur and the oxygen add up to -2. The oxygen are -2 each, and the sulfur is +6.

■ This video shows trends in oxidation numbers across the periodic table.

■ Oxidation number for monoatoms ions. (The oxidation number is equal to the number of charges it bring.)

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Group Oxidation number Group 1 ions: Li+, Na+, K+, Rb+ +1 Group 2 ions: Be2+, Mg2+, Ca2+, Ba2+ +2 Group13 ions: B3+, Al3+, Ga3+ +3 Group16 ions: O2−, S2− −2 Group17 ions: F−, Cl−, Br− −1 Group18 elements: He, Ne, Ar, Kr, Xe 0

✍ Worked-example 3.1(a) Calculate the oxidation number of chlorine in each of the following ion. Ion Oxidation number of halogen ClO− X + (-2) = -1 X = +1 ClO−3 X + 3(-2) = -1 X = +5 ClO−4 X + 4(-2) = -1 X = +7

✍ Worked-example 3.1(b) Calculate the oxidation number of manganese in each of the following compound/ion. Ion Oxidation number of halogen MnSO4 X + 6 + 4(-2) = 0 X = +2 MnO2 X + 2(-2) = 0 X = +4 MnO2−4 X + 4(-2) = -2 X = +6 MnO−4 X + 4(-2) = -1 X = +7

Naming compounds using the IUPAC nomenclature
■ Transition elements have more than one oxidation number.

	►	Copper Oxidation number Types of compound IUPAC name Formula +1 Oxide Copper (I) oxide Cu2O +2 Oxide Copper (II) oxide CuO
	►	Iron Oxidation number Types of compound IUPAC name Formula +2 Chloride Iron (II) chloride FeCl2 +3 Chloride Iron (III) chloride FeCl3
	►	Lead Oxidation number Types of compound IUPAC name Formula +2 Oxide Lead (II) oxide PbO +4 Oxide Lead (IV) oxide PbO2

■ Oxidation for special chromium and iron compounds

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Compound name Oxidation number IUPAC name Potassium chromate, K2CrO4 Cr = +6 Potassium chromate (VI) Potassium dichromate, K2Cr2O7 Cr = +6 Potassium dichromate Potassium hexacyanoferrite, K4Fe(CN)6 Fe = +2 Potassium hexacyanoferrate (II) Potassium hexacyanoferrate, K3Fe(CN)6 Fe = +3 Potassium hexacyanoferrate (III)

■ Oxidation number for some common non-metals.

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Common Oxidation number IUPAC name Sulphuric acid, H2SO4 S = +6 Sulphuric (VI) acid Sulphurous acid,H2SO3 S = +4 Sulphuric (IV) acid Nitric acid, HNO3 N = +5 Nitric (V) acid Nitrous acid, HNO2 N = +3 Nitric (III) acid Potassium hypochlorite, KClO Cl = +1 Potassium chlorate (I) Potassium chlorate, KClO3 Cl = +5 Potassium chlorate (V) Potassium iodate, KIO3 I = +5 Potassium iodate (V) Sodium nitrate, NaNO3 N = +5 Sodium nitrate (V) Sodium nitrate, NaNO2 N = +3 Sodium nitrate (III)

Half reaction in redox reaction
■ In the redox reaction

	►	Half of the reaction is an oxidising reaction and the other half of the reaction is a reducing reaction.
	►	The equation for this half reaction is called a half equation (ion-electron equations).

✍ Worked-example 3.1(c) A redox reaction occurs when magnesium, Mg, burns in oxygen gas, O2. 2Mg(s) + O2(g) → 2MgO(s) Solution: Each magnesium atom has lost two electrons to form magnesium ions, Mg2+. Half ionic equation: Mg → Mg2+ + 2e− (oxidation → loss of electrons) Each oxygen atom has received an electron to form oxygen ions O2−. Half ionic equation: O2 + 4e− → O2− (Reduction → electron received)

Redox reaction
■ Redox reaction

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Redox reaction is a reaction involving oxidation and reduction processes that occur simultaneously. ○ The combustion of metal (magnesium) in oxygen/halogen. ○ Redox reaction involving the change of iron(II) ions to ion(III) ions(Fe2+ → Fe3+). ○ Redox reaction involving the change of iron(III) ions to ion(II) ions(Fe3+ → Fe2+). ○ The displacement of metal from its salt solution. ○ The displacement of halogens from their halide solution by other halogens. ○ Transfer of electrons at a distance.

■ Systematic way of redox reaction analysing

	►	Write two half-equations. State the oxidising agent and reducing agent based on the two half-equations.
	►	Write overall equation.
	►	State the observation recorded.
	►	State confirmatory test for the product of the reaction(optional).

■ Non-redox reaction

	►	The reaction that do not involve change in the oxidation number.
	►	Example: ○ Neutralisation reaction between acid and alkali or base ○ Reaction between carbonate and acid ○ Precipitation reaction

Redox reaction involving the combustion of metal in oxygen/halogen
■ Combustion of metal in oxygen

	►	Metals like magnesium, zinc burn in oxygen to form metal oxides when heated strongly.
	►	The combustion of metal and oxygen is a redox reaction.

■ Summary of a redox reaction between magnesium and chlorine gas

	►	Half ionic equation: ○ Mg → Mg2+ + 2e− (Oxidation) → Reducing agent: Magnesium metal ○ Cl2 + 2e− → 2Cl− (Reduction) → Oxidising agent: chlorine gas
	►	Overall equation: ○ Mg(s) + Cl2(g) → MgCl2(s)
	►	Observation: ○ The magnesium ribbon burned with shinny white flame producing a white powder.

■ This video contains information on the redox reaction (Combustion of metal in oxygen & halogen)

Laboratory Activity 3.1.1 : The combustion of magnesium in oxygen

Redox reaction involving the change of iron (II) to iron (III) ions
■ Iron(II) ions, Fe²⁺ and iron(III) ions, Fe³⁺

	►	Ferum (iron) form two types of positive ions, which is iron (II) ions and iron (III) ions in the respective compounds.
	►	Colour of the iron (II) ions, Fe2+ solution is pale green, colour of iron (III) ions, Fe3+ is yellowish brown solution.
	►	Iron (II) ions and iron (III) ion in aqueous solution can be identified by chemical tests qualitative analysis. Confirmatory test Observation Addition of sodium hydroxide solution into Iron (II) ions, Fe2+ Green precipitate Addition of sodium hydroxide solution into Iron (III) ions, Fe3+ Brown precipitate

■ Changing iron(II) ions, Fe²⁺ to iron(III) ions, Fe³⁺

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Oxidising agents can be used in the changing of iron(II) ions, Fe2+ to iron(III) ions, Fe3+ .

■ This video contains information on the redox reaction(Change Fe²⁺ to Fe³⁺)

■ Other oxidising agents that can replace bromine water to convert Fe²⁺ to Fe³⁺ ions are as follows:

	►	Acidified potassium dichromate (VI)
	►	Hydrogen peroxide
	►	Concentrated nitric acid
	►	Acidified potassium manganate (VII)
	►	Chlorine water

Laboratory Activity 3.1.2 : Change of iron (II) ions to iron (III) ions

Redox reaction involving the change of iron (III) ions to iron (II) ions
■ Changing iron(III) ions, Fe³⁺ to iron(II) ions, Fe²⁺

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Reducing agents can be used in the changing of iron(III) ions, Fe3+ to iron(II) ions, Fe2+ .

■ This video contains information on the redox reaction (change Fe³⁺ to Fe³⁺)

■ Summary of a redox reaction between zinc and iron(III) nitrate

	►	Half ionic equation: ○ Zn(s) → ZN2+(aq) + 2e− (oxidation → reducing agent : zinc) ○ 2Fe3+(aq) + 2e− → 2Fe2+(aq) (reduction → oxidising agent: iron (III) ions, Fe3+)
	►	Overall equation: ○ Zn(s) + 2Fe3+(aq) → Zn2+(aq) + Fe2+(aq)
	►	Zinc powder dissolves in the solution and the brown solution turn to green.
	►	A green precipitate is produced when the sodium hydroxide solution is added. This confirm the presence of iron (II). ○ Fe2+(aq) + 2OH−(aq) → Fe(OH)2 (greenish precipitate)

■ Other reducing agents that can replace bromine water to convert Fe³⁺ to Fe²⁺ ions are as follows:

	►	Hydrogen sulphide gas
	►	Stanum (II) chloride solution
	►	Sodium sulphite / potassium sulphite
	►	Sulphur dioxide gas

Laboratory Activity 3.1.3 : Change of iron (III) ions to iron (II) ions

Displacement reaction of metal from its salt solution as a redox reaction
■ Electrochemical series

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■ Displacement reaction of metal from its salt solution

	►	Higher metal in the electrochemical series have a higher tendency to form positive ions by losing electrons. ○ M(s) → Mn+(aq) + ne−
	►	Ions of metals which is located below the electrochemical series, the stronger is the oxidising ability of their ions.
	►	Metals which act as a strong reducing agents will have ions with weak oxidising agent, and vice versa.
	►	Thus, the metal at the top of the series is able to displace metals at the bottom of the series.

■ This video shows a more reactive metal will displace a less reactive metal from solution.

■ Displacement reaction of zinc from copper(II) sulphate solution

	►	Half ionic equation: ○ Zn(s) → Znn+(aq) + 2e− (oxidation → reducing agent: zinc) ○ Cun+(aq) + 2e− → Cu(s) (reduction→ oxidising agent: copper(II) ion / copper sulphate solution)
	►	Overall equation: ○ Zn(s) + Cun+(aq)→ Znn+(aq) + Cu(s)
	►	Observation: ○ Zinc piece dissolves and the blue copper(II) sulphate solution fades. ○ A brown solid is formed.
	►	Confirmatory test: ○ Blue precipitate dissolves in excess sodium hydroxide solution, NaOH(aq). This confirm the presence of copper ion in the solution.

Laboratory Activity 3.1.4 : Displacement of metals from its salt solution

Displacement reaction of halogen from their halide solution by other halogen as a redox reaction
■ Halogen

	►	Non-metallic elements in Group 17 of the Periodic Table.
	►	Highly electronegative due to a high tendency to accept electrons.

■ Displacement reaction of halogens

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	►	Tendency to accept electrons into their halide ion decreases from top to bottom in Group 17.
	►	A more reactive halogen (located at the top of the Group 17) can displace a less reactive halogen from its salts.

■ Colour of halogens

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The colour of halogens in their aqueous solution and the colour of aqueous solution with a little tetrachloromethane solution. Halogen Colour of aqueous solution Colour in halogen in the tetrachloromethane layer. (Confirmatory test for halogens) Cl2 Almost colorless to pale yellow Yellow Br2 Brown (orange) Yellow to brown I2 Yellow Purple

■ This video contains information on the displacement Reaction of halogens

■ Summary of a redox reaction between bromine water and potassium iodide solution as redox reaction

	►	Half ionic equation: ○ 2I−(aq) → I2(aq) + 2e− (oxidation → reducing agent: iodide ion) ○ Br2(aq) + 2e− → 2Br−(aq) (reduction→ oxidising agent: bromine solution)
	►	Overall equation: ○ Br2(aq) + 2I−(aq) → 2Br−(aq) + I2(aq)
	►	Observation ○ The brown coloured bromine colour water is decolourised. ○ The colourless potassium iodide solution changes to brown.

Transfer of electrons at a distance as a redox reaction
■ Transfer of electron at a distance

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When oxidising agent and reducing agent solutions is separated by an electrolyte in a U-Tube, redox reactions occur by transfer of electron using connecting wire.

■ U-tube redox cell

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	►	Electrolyte acts as a salt bridge to separate two solutions but allows ions to pass through to complete the circuit
	►	Electron transfer from reducing agent to oxidising agent through a connecting wire.
	►	Carbon/graphite electrode that is immersed in reducing agent act as negative terminal.
	►	Carbon/graphite electrode that is immersed in oxidising agent act as positive terminal.
	►	The deflection of the galvanometer needle shows the electron flowing/moving.

■ This video contains information on the transfer of electrons at a distance as a redox reaction.

■ The reducing agent undergoes oxidation (loss of electron)

	►	Solution that can react as oxidising agent: Solution Half equation and changes colour of solution Acidified potassium manganate(VII), KMnO4 MnO4−(aq) + 8H+(aq) + 5e− → Mn2+(aq) + 4H2O(l) (purple to colourless) Acidified potassium dichromate(VI), K2Cr2O7 Cr2O72−(aq) + 14H+(aq) + 6e− → 2Cr3+(aq) + 7H2O(l) (orange to green) Chlorine, Cl2 Cl2(aq) + 2e− → Cl−(aq) (pale yellow to colourless) Bromine, Br2 Br2(aq) + 2e− → 2Br−(aq) (reddish brown to colourless) Iodine, I2 I2(aq) + 2e− → 2I−(aq) (brown to colourless)
	►	Solution Half equation and changes colour of solution Iron(II) sulphate, FeSO4 Fe2+(aq) → Fe3+(aq) + e− (green to brown) Stanum(II) chloride, SnCl2 Sn2+(aq) → Sn4+(aq) + 2e− (both of the ions is colourless) Potassium chloride, KCl 2Cl−(aq) → Cl2(aq) + 2e− (colourless to pale yellow) Potassium bromide, KBr 2Br−(aq) → Br2(aq) + 2e− (colourless to reddish brown) Potassium iodide, KI 2I−(aq) → I2(aq) + 2e− (colourless to brown)

■ Summary of a redox reaction between bromine solution and iron (II) sulphate solution as redox reaction

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	►	Half ionic equation: ○ Fe2+(aq) → Fe3+(aq) + e− (oxidation → reducing agent: iron (II) ions, Fe2+) ○ Br2(aq) + 2e− → 2Br−(aq) (reduction → oxidising agent: bromine water)
	►	Overall equation: ○ 2Fe2+(aq) + Br2(aq) → Fe2+(aq) + 2Br−(aq)
	►	Observation: ○ The brown coloured bromine colour water is decolourised. ○ The iron(II) sulphate solution turned from greenish colour to brownish colour.

■ Summary of a redox reaction between acidified potassium manganate (VII) solution and potassium iodide solution as redox reaction

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	►	Half ionic equation: ○ 2I−(aq) → I2(aq) + 2e− (oxidation → reducing agent: iodide ion) ○ MnO4−(aq) + 8H+(aq) + 5e− → Mn2+(aq) + 4H2O(l) (reduction → oxidising agent: manganate (VII) ion)
	►	Overall equation: ○ 10I−(aq) + 2MnO4−(aq) + 16H+(aq) → 5I2(aq) + 2Mn2+(aq) + 8H2O(l)
	►	Observation: ○ The purple coloured acidified potassium manganate (VII) is decolourised. ○ The colourless potassium iodide solution changes to brown.

■ Summary of a redox reaction between acidified potassium manganate (VII) solution and iron (II) sulphate solution as redox reaction

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	►	Half ionic equation: ○ Fe2+(aq) → Fe3+(aq) + e− (oxidation → reducing agent: iron (II) ions, Fe2+) ○ MnO4−(aq) + 8H+(aq) + 5e− → Mn2+(aq) + 4H2O(l) (reduction → oxidising agent: manganate (VII) ion)
	►	Overall equation: ○ 5Fe2+(aq) + MnO4−(aq) + 8H+(aq) → Fe3+(aq) + Mn2+(aq) + 4H2O(l)
	►	Observation: ○ The purple coloured acidified potassium manganate(VII) is decolourised. ○ The iron(II) sulphate solution turned from greenish colour to brownish colour.

✍ Worked-example 3.1(a) State whether the following displacement reaction of metals from salt solution can occur. Question Answer Magnesium and zinc chloride solution ✔ Aluminium and copper chloride solution ✔ Silver and zinc chloride solution ✘ Copper and magnesium chloride solution ✘ Lead and iron(II) sulphate solution ✘

✍ Worked-example 3.1(b) State whether the following displacement reaction of halogen from halide solution can occur. Question Answer Bromine water and zinc chloride solution ✘ Bromine water and potassium iodide solution ✔ Chlorine water and potassium iodide solution ✔ Iodide water and sodium bromide solution ✘ Iodide water and zinc chloride solution ✘

✍ Worked-example 3.1(c) State whether the following chemical reaction are redox reactions. Question Answer NaCl + AgNO3 → KNO3 + AgCl ✘ Pb(NO3)2 + 2NaCl → PbCl2 + 2NaNO3 ✘ Cl2 + 2KBr → 2KCl + Br2 ✔ Mg + ZnCl2 → MgCl2 + Zn ✔ CaCO3 + 2HCl → CaCl2 + H2O + CO2 ✘

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