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9.2 Redox reactions

9.2 Redox reactions. 9.2.1: Deduce simple oxidation and reduction half-equations given the species involved in a redox reaction. Step 1: Identify the oxidation states of the species on either side of the reaction. Step 2 :

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9.2 Redox reactions

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  1. 9.2 Redox reactions

  2. 9.2.1: Deduce simple oxidation and reduction half-equations given the species involved in a redox reaction • Step 1: • Identify the oxidation states of the species on either side of the reaction. • Step 2: • Identify the oxidation half reaction by identifying which reactant undergoes oxidation. • Identify the reduction half reaction by identifying which reactant undergoes reduction. • Step 3: • Deduce the number of electrons transferred and produce half-equations.

  3. Example: In the reaction between chlorine and potassium iodide solution the products are iodine and potassium chloride solution. • Cl2 + KI  I2 +KCl • This is the unbalanced skeleton equation. • Redox half-reactions can be used to balance complex reaction equations

  4. Example: In the reaction between chlorine and potassium iodide solution the products are iodine and potassium chloride solution. • Step 1: • Identify the oxidation states of the species on either side of the reaction. • Cl02 + K+1I-1 I02 + K+1Cl-1

  5. Example: In the reaction between chlorine and potassium iodide solution the products are iodine and potassium chloride solution. • Step 2: • Identify the OXidation half reaction by identifying which reactant undergoes oxidation. • Identify the REDuction half reaction by identifying which reactant undergoes oxidation. • Cl02 + K+1I-1 I02 + K+1Cl-1 • Cl2 is reduced toCl- • I-1 is oxidised to I2 • K+ does not change = spectator RED OX

  6. Example: In the reaction between chlorine and potassium iodide solution the products are iodine and potassium chloride solution. • Step 3: • Deduce the number of electrons transferred and produce half-equations. • Oxidation Half-Reaction(Oxidation is Loss of Electrons) • Electrons are removed from the reactant (appear as product) • Balance number of atoms • 2I-1 I02+ 2e-CHARGE IS CONSERVED • REDuctionHalf-Reaction(Reduction is Gain of Electrons) • Electrons are added to a reactant. • Cl02+ 2e-2Cl-1

  7. 9.2.2: Deduce redox equations using half-equations. • This means balance chemical equations using electrons in half-reactions • H+ and H2O should be used where necessary to balance half -equations in acid solution. • The balancing of equations for reactions in alkaline solution will not be assessed.

  8. Balancing Redox in Acid • Add these steps to balance oxygen and hydrogen atoms in redox half-reactions • Step 1: Balance oxygen by adding water (H2O) • Step 2: Balance the hydrogen by adding H+ ions

  9. Oxidation of Ethanol using Acidified Dichromate • CH3CH2OH + Cr2O72- CH3COOH + Cr3+ • Rewritten:C2H6O + Cr2O72- C2H4O2+ Cr3+ • Assign oxidation numbers (H+ and O2- stay constant here) • (C2+)2H6O + (Cr6+)2O72- (C0)2H4O2+ Cr3+ • Carbon in ethanol gets oxidised from 2+ to 0 • Chromium in dichromate gets reduced from 6+ to 3+

  10. Oxidation Half-Reaction in Acid • Oxidation Half-Reaction(2e- per C = 4e-) • C2H6OC2H4O2+ 4e- • Add H2O to balance oxygens • C2H6O+ H2OC2H4O2 + 4e- • Add H+to balancehydrogens • C2H6O+ H2OC2H4O2 + 4e- + 4H+ • net charge on each side is balanced • (0) + (0) = (0) + (-4) + (+4) • If charge is not balanced, then it is wrong.

  11. Reduction Half-Reactions in Acid • Reduction Half-Reaction(3e- per Cr = 6e-) • Cr2O72-+ 6e- 2Cr3+ • Add H2O to balance oxygens • Cr2O72- + 6e- 2Cr3+ + 7H2O • Add H+to balancehydrogens • Cr2O72-+ 14H+ + 6e-2Cr3+ + 7H2O • net charge on each side is balanced • (-2) + (+14) + (-6) = (+6)+ (0) • If charge is not balanced, then it is wrong.

  12. Combine Half-Reactions • Oxidation:C2H6O+ H2OC2H4O2 + 4e- + 4H+ • Reduction: Cr2O72- + 14H+ + 6e- 2Cr3+ + 7H2O • Multiply each to balance electrons transferred • [C2H6O+ H2OC2H4O2 + 4e- + 4H+] x 3 • [Cr2O72- + 14H+ + 6e- 2Cr3+ + 7H2O] x 2 • 3C2H6O+3H2O3C2H4O2+12e-+12H+ • 2Cr2O72- + 28H+ + 12e- 4Cr3+ +14H2O _______________________________________________________ • 3C2H6O+ 3H2O + 2Cr2O72- + 28H+3C2H4O2 + 12H++ 4Cr3+ + 14H2O • 3C2H6O+ 2Cr2O72- + 16H+ 3C2H4O2 + 4Cr3+ + 11H2O • Don’t forget to check net charge

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