1. 9.2 Redox reactions

2. Deduce the number of electrons transferred and produce half-equations.
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. Cl2 + KI  I2 + KCl This is the unbalanced skeleton equation.
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 to Cl-
I-1 is oxidised to I2
K+ does not change = spectator
RED OX

6. 2I-1  I02+ 2e- CHARGE IS CONSERVED
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
REDuction Half-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 balance hydrogens
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-)
Cr2O e-  2Cr3+
Add H2O to balance oxygens
Cr2O e-  2Cr3+ + 7H2O
Add H+ to balance hydrogens
Cr2O H+ + 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: Cr2O H+ + 6e-  2Cr3+ + 7H2O
Multiply each to balance electrons transferred
[C2H6O + H2O C2H4O2 + 4e- + 4H+] x 3
[Cr2O H+ + 6e-  2Cr3+ + 7H2O] x 2
3C2H6O + 3H2O 3C2H4O2 + 12e- + 12H+
2Cr2O H+ + 12e-  4Cr H2O
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3C2H6O + 3H2O + 2Cr2O H+  3C2H4O2 + 12H+ + 4Cr H2O
3C2H6O + 2Cr2O H+  3C2H4O2 + 4Cr H2O
Don’t forget to check net charge