Half-reactions show the oxidation or reduction reaction separated. Cu (s) + 2 AgNO 3(aq) → Cu(NO 3 ) 2(aq) + 2 Ag (s) Oxidation:Cu → Cu e – Reduction:Ag + + 1e – → Ag Half reactions are often shown as aqueous net ionic equations – spectator IONS not included
Balancing redox reactions 2
Two methods for balancing redox reactions: 1.Oxidation number method 2.Half-reaction method. Balance redox equations using the oxidation number method. Balance redox equations in acidic and basic solutions using the half reaction method.
Redox reactions usually require an acidic or basic solution. Acid / base not oxidized or reduced in reaction. Usually converted to water. Half-reaction method is used for balancing redox reactions in the presence of acid or base.
Cr 2 O 7 2– (aq) + SO 3 2– (aq) → Cr 3+ (aq) + SO 4 2– (aq) 1: Assign ox.numbers and write half-reactions. Balancing Redox Reactions in Acidic Solutions Oxidation:SO 3 2- → SO e – Reduction:Cr 2 O e – → Cr 3+ Not ionic (no spectators) –keep together!
2: Balance all elements except H and O. 3: Balance oxygen atoms by using H 2 O. Oxidation:SO 3 2- → SO e – Reduction:Cr 2 O e – → Cr 3+ 2 Oxidation:SO H 2 O → SO e – Reduction:Cr 2 O e – → 2 Cr H 2 O
4: Balance hydrogen atoms using H + ions. 5: Balance the number of electrons lost and gained. Oxidation: SO H 2 O → SO e – Reduction: Cr 2 O e – → 2 Cr H 2 O + 2 H + 14 H + + Oxidation: 3 x ( SO H 2 O → SO e – + 2 H + ) Reduction: 14 H + + Cr 2 O e – → 2 Cr H 2 O 3 SO H 2 O → 3 SO e – + 6 H +
6: Add the two half-reactions. Oxidation: 3 SO H 2 O → 3 SO e – + 6 H + Reduction: 14 H + + Cr 2 O e – → 2 Cr H 2 O 8 H + + Cr 2 O SO 3 2- → 2 Cr SO H 2 O
MnO 4 – + I – → MnO 2 + I 2 8 H MnO 4 – + 6 I – → 2 MnO I H 2 O Balance the following reaction in a acidic solution Oxidation: I 1- → I 2 + 2e – Reduction: MnO e – → MnO H 2 O 4 H x () 2x ( ) Oxidation: 6 I 1- → 3 I 2 + 6e – Reduction: 8 H MnO e – → 2 MnO H 2 O
Balancing Redox Reactions in Basic Solutions For basic solutions add hydroxide ions. MnO 4 – + C 2 O 4 2– → CO 2 + MnO Oxidation: C 2 O 4 2- → CO 2 + 2e – Reduction: MnO e – → MnO H 2 O4 H + + *5a: Add the same number of OH - as H + to BOTH sides of the equation.
5b: Eliminate H + / OH - by forming water.. Cancel any waters you can to simplify each half reaction. Oxidation C 2 O 4 2- → 2 CO 2 + 2e – Reduction: 4 H + + MnO e – → MnO H 2 O + 4 OH - 4 OH H 2 O 2 Oxidation C 2 O 4 2- → 2 CO 2 + 2e – Reduction: 2 H 2 O + MnO e – → MnO OH - 3 x () 2x ( )
4 H 2 O + 2 MnO 4 – + 3 C 2 O 4 2– → 2 MnO CO OH – Oxidation 3 C 2 O 4 2- → 6 CO 2 + 6e – Reduction: 4 H 2 O + 2 MnO e – → 2 MnO OH -
N 2 O + ClO – → NO 2 – + Cl – Balance the following reaction in a basic solution Oxidation: N 2 O → NO e – Reduction: ClO - + 2e – → Cl H 2 O + 2 H H H 2 O 5a: Add same number of OH - to BOTH side. 5b:Cancel any waters you can to simplify half reactions.
2 OH – + 2 ClO – + N 2 O → 2 Cl – + 2 NO 2 – + H 2 O Oxidation: N 2 O → NO e – Reduction: ClO - + 2e – → Cl - 23 H 2 O + 2 H H H 2 O + 6 OH - 2 OH H 2 O OH - 6 OH H 2 O 1 Reduction: ClO - + 2e – → Cl - 1 H 2 O OH - Oxidation: N 2 O → NO e – H 2 O 6 OH - + 2x ( ) 2 H 2 O + 2 ClO - + 4e – → 2 Cl OH - 12