9 2 redox reactions
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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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9 2 redox reactions

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:

  • 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.


9 2 redox reactions

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


9 2 redox reactions

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


9 2 redox reactions

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


9 2 redox reactions

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


9 2 2 deduce redox equations using half equations

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.


Balancing redox in acid

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


Oxidation of ethanol using acidified dichromate

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+


Oxidation half reaction in acid

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.


Reduction half reactions in acid

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.


Combine half reactions

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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