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.
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
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.
0
Cl
2
+
+1
-1
K I

0
I
2
+
+1
-1
K Cl
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.
RED
0
Cl
2
+
+1
-1
K I

OX
 Cl2 is reduced to Cl I-1 is oxidised to I2
 K+ does not change = spectator
0
I
2
+
+1
-1
K Cl
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
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
 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
 CH3CH2OH + Cr2O72-  CH3COOH + Cr3+
+ Cr2O72-  C2H4O2 + Cr3+
 Assign oxidation numbers (H+ and O2- stay constant here)
 (C2+)2H6O + (Cr6+)2O72-  (C0)2H4O2 + Cr3+
 Rewritten: C2H6O
 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
(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.
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 balance hydrogens
 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
 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