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Published byChristopher Dickerson Modified over 9 years ago
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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 2+ + 2e – Reduction:Ag + + 1e – → Ag Half reactions are often shown as aqueous net ionic equations – spectator IONS not included. +2 +1 0 0
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Balancing redox reactions 2
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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.
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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.
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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 +3 +4 -2 +6 -2 +6 -2 Oxidation:SO 3 2- → SO 4 2- + 2e – Reduction:Cr 2 O 7 2- + 3e – → Cr 3+ Not ionic (no spectators) –keep together!
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2: Balance all elements except H and O. 3: Balance oxygen atoms by using H 2 O. Oxidation:SO 3 2- → SO 4 2- + 2e – Reduction:Cr 2 O 7 2- + 6e – → Cr 3+ 2 Oxidation:SO 3 2- + H 2 O → SO 4 2- + 2e – Reduction:Cr 2 O 7 2- + 6e – → 2 Cr 3+ + 7 H 2 O
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4: Balance hydrogen atoms using H + ions. 5: Balance the number of electrons lost and gained. Oxidation: SO 3 2- + H 2 O → SO 4 2- + 2e – Reduction: Cr 2 O 7 2- + 6e – → 2 Cr 3+ + 7 H 2 O + 2 H + 14 H + + Oxidation: 3 x ( SO 3 2- + H 2 O → SO 4 2- + 2e – + 2 H + ) Reduction: 14 H + + Cr 2 O 7 2- + 6e – → 2 Cr 3+ + 7 H 2 O 3 SO 3 2- + 3 H 2 O → 3 SO 4 2- + 6e – + 6 H +
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6: Add the two half-reactions. Oxidation: 3 SO 3 2- + 3 H 2 O → 3 SO 4 2- + 6e – + 6 H + Reduction: 14 H + + Cr 2 O 7 2- + 6e – → 2 Cr 3+ + 7 H 2 O 8 H + + Cr 2 O 7 2- + 3 SO 3 2- → 2 Cr 3+ + 3 SO 4 2- + 4 H 2 O
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MnO 4 – + I – → MnO 2 + I 2 8 H + + 2 MnO 4 – + 6 I – → 2 MnO 2 + 3 I 2 + 4 H 2 O Balance the following reaction in a acidic solution. +7 0 +4 Oxidation: I 1- → I 2 + 2e – Reduction: MnO 4 1- + 3e – → MnO 2 2 + 2 H 2 O 4 H + + 3 x () 2x ( ) Oxidation: 6 I 1- → 3 I 2 + 6e – Reduction: 8 H + + 2 MnO 4 1- + 6e – → 2 MnO 2 + 4 H 2 O
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Balancing Redox Reactions in Basic Solutions For basic solutions add hydroxide ions. MnO 4 – + C 2 O 4 2– → CO 2 + MnO 2 +4 +3 +7 +4 Oxidation: C 2 O 4 2- → CO 2 + 2e – Reduction: MnO 4 1- + 3e – → MnO 2 2 + 2 H 2 O4 H + + *5a: Add the same number of OH - as H + to BOTH sides of the equation.
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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 4 1- + 3e – → MnO 2 + 2 H 2 O + 4 OH - 4 OH - + 4 H 2 O 2 Oxidation C 2 O 4 2- → 2 CO 2 + 2e – Reduction: 2 H 2 O + MnO 4 1- + 3e – → MnO 2 + 4 OH - 3 x () 2x ( )
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4 H 2 O + 2 MnO 4 – + 3 C 2 O 4 2– → 2 MnO 2 + 6 CO 2 + 8 OH – Oxidation 3 C 2 O 4 2- → 6 CO 2 + 6e – Reduction: 4 H 2 O + 2 MnO 4 1- + 6e – → 2 MnO 2 + 8 OH -
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N 2 O + ClO – → NO 2 – + Cl – Balance the following reaction in a basic solution. +3+1 Oxidation: N 2 O → NO 2 - + 4e – Reduction: ClO - + 2e – → Cl - 2 3 H 2 O + 2 H + + + 6 H + + 1 H 2 O 5a: Add same number of OH - to BOTH side. 5b:Cancel any waters you can to simplify half reactions.
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2 OH – + 2 ClO – + N 2 O → 2 Cl – + 2 NO 2 – + H 2 O Oxidation: N 2 O → NO 2 - + 4e – Reduction: ClO - + 2e – → Cl - 23 H 2 O + 2 H + + + 6 H + + 1 H 2 O + 6 OH - 2 OH - + 2 H 2 O 3 + 2 OH - 6 OH - + 6 H 2 O 1 Reduction: ClO - + 2e – → Cl - 1 H 2 O + + 2 OH - Oxidation: N 2 O → NO 2 - + 4e – 2 + 3 H 2 O 6 OH - + 2x ( ) 2 H 2 O + 2 ClO - + 4e – → 2 Cl - + 4 OH - 12
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