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Published byGwen Lloyd Modified over 9 years ago
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Redox Reactions. Oxidation Reduction
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Oxidation and Reduction Oxidation: Gain of oxygen Loss of electrons Reduction: Loss of oxygen Gain of electrons Increase in oxidation number Decrease in oxidation number
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4 Experiments: 1.Burning magnesium 2.Copper in silver nitrate solution 3.Chlorine solution and potassium iodide solution 4.Exploding hydrogen Word equation Balanced symbol equation
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2Mg(s) + O 2 (g) 2MgO(s) Oxidised – gains oxygen Must be a redox! Mg Mg 2+ O O 2- CHARGE OF A - = GAINED ELECTRONS CHARGE OF A += LOST ELECTRONS Put the e - in. +2e - Oxidised – loss of e - Reduced – gain of e -
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Cu(s) + 2AgNO 3 (aq) Cu(NO 3 ) 2 (aq) + 2Ag(s) Ag + Ag Cu Cu 2+ Complete the half- equations +e - +2e - Oxidised? Reduced? Oxidised – loss of e - Reduced – gain of e -
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H 2 (g) + ½ O 2 (g) H 2 O(g) Covalent! No H + or OH - Need a new definition. oxidation reduction Reducing agent Oxidising agent
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agents An oxidising agent is a substance that brings about oxidation(itself reduced) example- hydrogen peroxide for bleaching hair A reducing agent is a substance that brings about reduction.(itself oxidised)aa sulphur dioxide used to bleach straw
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In terms of oxidation number Oxidation: Gain of oxygen Loss of electrons Reduction: Loss of oxygen Gain of electrons Increase in oxidation number Decrease in oxidation number
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Oxidation Numbers- the seven rules The oxidation number of an atom in an uncombined element is zero. E.g. Mg in Mg, O in O 2. The oxidation number of an ion of an element is the same as its charge. O.N Br-= -1 O.N Mg in Mg 2+ = +2 ALKALI METALS= +1 ALKALINE EARTH METALS=+2 HALOGENS= -1
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Oxidation Numbers The oxidation numbers of atoms in a compound add up to zero. F O-2 H+1 Cl Oxidation state of C in CO 2 ? x – 4 = 0 x = +4 Put the +!
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Oxidation Numbers The oxidation numbers of atoms in a compound add up to zero. F O-2 H+1 Cl Oxidation state of Mg in MgCl 2 ? +2
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Oxidation Numbers The oxidation numbers of atoms in a compound add up to zero. F O-2 H+1 Cl Oxidation state of N in NH 3 ? -3
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Oxidation Numbers The oxidation numbers of atoms in an ion add up to the charge on the ion. F O-2 H+1 Cl Oxidation state of S in SO 4 2- ? x – 8 = -2 x = +6
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Oxidation Numbers The oxidation numbers of atoms in an ion add up to the charge on the ion. F O-2 H+1 Cl Oxidation state of S in S 2- ? -2
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Oxidation Numbers The oxidation numbers of atoms in an ion add up to the charge on the ion. F O-2 H+1 Cl Oxidation state of N in NH 4 + ? -3
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OXYGEN HYDROGEN Oxygen has a charge Oxygen has a charge of –2 EXCEPT in peroxides where the charge is –1 In the compound OF 2 Where it has a value of +2. This is because F has a greater electronegative number than oxygen Hyrogen has a charge of +1 except in Metal hydrides where it has an O.N of –1 Metal Hydrides are ionic compounds NaH (+1)(-1)
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Halogens- assign charge of –1 unless bonded to more electronegative element Cl 2 O (+1)2(-2) Cl= +1 when writing formulas the most electronegative is placed second. Halogen
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H 2 (g) + ½ O 2 (g) H 2 O(g) Covalent! No H + or OH - Need a new definition.
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Oxidation: Gain of oxygen Loss of electrons Reduction: Loss of oxygen Gain of electrons Increase in oxidation number Decrease in oxidation number
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H 2 (g) + ½ O 2 (g) H 2 O(g) Covalent! No H + or OH - Need a new definition. +1 0 -20 O H
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H 2 (g) + ½ O 2 (g) H 2 O(g) +1 0 -20 O H Oxidised? H – increase in oxidation number Reduced? O – decrease in oxidation number
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Balancing Redox Reactions Using Oxidation Numbers balance the following equation. Solution: 1 assign oxidation number 2. Note element that changes oxidation number. 3.show the number of electrons lost and gained.
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4.Work out ratio of oxidising agent to reducing agent. 5.Balance remaining items by inspection method.
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Worked Example MnO 4 ֿ+Fe 2+ +H + Mn 2+ +Fe 3+ +H 2 O (+7)4(-2) (+2)(+1) (+2) (+3) 2 (+1)(-2) ֿֿֿֿֿֿֿֿֿֿֿֿֿ MnO 4 ֿ+Fe 2+ +H + Mn 2+ +Fe 3+ +H 2 O (+7)4(-2) (+2)(+1) (+2) (+3) 2 (+1)(-2) GAINS 5 ELECTRONS LOSES 1 ELECTRON
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Balance remaining items 1MnO 4 ֿ:5Fe 2+ 1MnO 4 ֿ+5Fe 2+ +H+ 1Mn 2+ +5Fe3+ +H2O 1MnO 4ֿ+5Fe2+ +8H+ 1Mn2+ +5Fe3+ +4H2O
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Well done!
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