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# Tang 02 balancing redox reactions 2

## on Jan 07, 2012

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## Tang 02 balancing redox reactions 2Presentation Transcript

• BALANCING REDOX REACTIONS
• An overall redox reaction can be balanced using half-reactions once the correct number of electrons have been accounted for.
BALANCING REDOX REACTIONS
• Cu 2+ (aq) + Al (s)  Cu (s) + Al 3+ (aq)
+2 0 0 +3 Cu 2+ (aq) + 2e -  Cu (s) Al (s)  Al 3+ (aq) + 3e - x 3 3 Cu 2+ (aq) + 6 e -  3 Cu (s) 2 Al (s)  2 Al 3+ (aq) + 6 e - x 2 3Cu 2+ (aq) + 2Al (s)  3Cu (s) + 2Al 3+ (aq) BALANCING REDOX REACTIONS
• Write a balanced equation to describe the reaction between zinc metal with aqueous lead (II) nitrate.
Zn (s) + Pb(NO 3 ) 2(aq)  Pb (s) + Zn(NO 3 ) 2(aq) 0 +2 -2 +5 0 +2 -2 +5 Pb 2+ (aq) + 2e -  Pb (s) Zn (s)  Zn 2+ (aq) + 2e - Redox reaction: Pb 2+ (aq) + Zn (s)  Pb (s) + Zn 2+ (aq) BALANCING REDOX REACTIONS Example #1
• Example #2
• Write the balanced redox equation for the reaction of MnO 4 - in an acidic solution with H 2 SO 3 resulting with the products of Mn 2+ and SO 4 2- .
Reduction: MnO 4 - (aq) + 8H + (aq) + 5e -  Mn 2+ (aq) + 4H 2 O (l) Oxidation: H 2 SO 3(aq) + H 2 O (l)  SO 4 2- (aq) + 4H + (aq) + 2e - You can use half-reactions from the Half Reactions Table x 2 x 5 2 MnO 4 - (aq) + 16 H + (aq) + 10 e -  2 Mn 2+ (aq) + 8 H 2 O (l) 5 H 2 SO 3(aq) + 5 H 2 O (l)  5 SO 4 2- (aq) + 20 H + (aq) + 10 e - 2MnO 4 - (aq) + 5H 2 SO 3(aq)  2Mn 2+ (aq) + 3H 2 O (l) + 5SO 4 2- (aq) + 4H + (aq) BALANCING REDOX REACTIONS
• Sometimes, the half-reaction you need is not part of the given tables. Therefore, you need to come up with the reaction yourself.
BALANCING REDOX REACTIONS
• Example #3
• Write the balanced equation for the half-reaction where NO is reduced to N 2 O in an acidic solution.
NO (g)  N 2 O (g) 1. Write down the reactant and product and balance atoms other than oxygen and hydrogen. 2. Add H 2 O molecules to balance the oxygen atoms. 3. Add H+ ions to balance the hydrogen atoms. 4. Add electrons to balance the charge on both sides of the equation. Step 1: 2 NO (g)  N 2 O (g) Step 2: 2NO (g)  N 2 O (g) + H 2 O (l) Step 3: 2NO (g) + 2H + (aq)  N 2 O (g) + H 2 O (l) Step 4: 2NO (g) + 2H + (aq) + 2e -  N 2 O (g) + H 2 O (l) BALANCING REDOX REACTIONS
• Example #4
• Write the balanced equation for the oxidation half-reaction where Cl 2 is oxidized to ClO 3 - in a basic solution.
• Write down the reactant and product and balance atoms other than oxygen and hydrogen.
• 2. Add H 2 O molecules to balance the oxygen atoms.
• 3. Add H+ ions to balance the hydrogen atoms.
• 4. Add OH - ions equal in number to the H + ions to both sides of the reaction.
• 5. Combine H + and OH - ions on the same side of the equation to form H 2 O molecules.
• 6. Add electrons to balance the charge.
Cl 2(g)  ClO 3 - (aq) Step 1: Cl 2(g)  2 ClO 3 - (aq) Step 2: Cl 2(g) + 6H 2 O (l)  2ClO 3 - (aq) Step 3: Cl 2(g) + 6H 2 O (l)  2ClO 3 - (aq) + 12H + (aq) Step 4: Cl 2(g) + 6H 2 O (l)  2ClO 3 - (aq) + 12H + (aq) + 12OH - (aq) 12OH - (aq) + Step 5: 12OH - (aq) + Cl 2(g) + 6H 2 O (l)  2ClO 3 - (aq) + 12H 2 O (l) (-12) (-2) Step 6: 12OH - (aq) + Cl 2(g) + 6H 2 O (l)  2ClO 3 - (aq) + 12H 2 O (l) + 10e - 12OH - (aq) + Cl 2(g)  2ClO 3 - (aq) + 6H 2 O (l) + 10e - BALANCING REDOX REACTIONS
• For the most part, the complete redox reactions you will need to balance will require you to develop your own half-reactions.
• The method used is the ion-electron method or the half-reaction method .
BALANCING REDOX REACTIONS
• It is important to know whether the reaction occurs in an acidic or basic environment, as different final products will be produced.
• This information will always be provided.
BALANCING REDOX REACTIONS
• Example #5
• Balance the following redox reaction which occurs in an acidic solution:
• Cr 2 O 7 2- + Fe 2+  Cr 3+ + Fe 3+
• Which half-reaction is reduction? Which is oxidation?
Cr 2 O 7 2-  Cr 3+ Fe 2+  Fe 3+ Separate into two skeleton half equations Cr 2 O 7 2-  Cr 3+ Cr 2 O 7 2-  2 Cr 3+ Cr 2 O 7 2-  2Cr 3+ + 7 H 2 O Cr 2 O 7 2- + 14 H +  2Cr 3+ + 7H 2 O Cr 2 O 7 2- + 14H + + 6e -  2Cr 3+ + 7H 2 O Reduction Fe 2+  Fe 3+ Fe 2+  Fe 3+ + e - Oxidation BALANCING REDOX REACTIONS
• Example #5
• Balance the following redox reaction which occurs in an acidic solution:
• Cr 2 O 7 2- + Fe 2+  Cr 3+ + Fe 3+
• Which half-reaction is reduction? Which is oxidation?
Cr 2 O 7 2-  Cr 3+ Fe 2+  Fe 3+ Reduction: Cr 2 O 7 2- + 14H + + 6e -  2Cr 3+ + 7H 2 O Oxidation: Fe 2+  Fe 3+ + e - x 6 6 Fe 2+  6 Fe 3+ + 6 e - 6Fe 2+ + Cr 2 O 7 2- + 14H +  2Cr 3+ + 7H 2 O + 6Fe 3+ BALANCING REDOX REACTIONS
• Example #5:
• Balance the following redox reaction which occurs in an acidic solution:
• Cr 2 O 7 2- + Fe 2+  Cr 3+ + Fe 3+
• Which half-reaction is reduction? Which is oxidation?
Cr 2 O 7 2- + Fe 2+  Cr 3+ + Fe 3+ Step 1: Cr 2 O 7 2- + Fe 2+  2 Cr 3+ + Fe 3+ Step 2: Cr 2 O 7 2- + Fe 2+  2Cr 3+ + Fe 3+ + 7H 2 O Step 3: Cr 2 O 7 2- + Fe 2+ + 14H +  2Cr 3+ + Fe 3+ + 7H 2 O (-2) + (+2) + (+14) = (+14) (+6) + (+3) = (+9) Step 4: Cr 2 O 7 2- + Fe 2+ + 14H + + 5e -  2Cr 3+ + Fe 3+ + 7H 2 O This is not a half reaction BALANCING REDOX REACTIONS
• Example #6
• Balance the following redox reaction which occurs in a basic solution:
• SO 3 2- + MnO 4 -  SO 4 2- + MnO 2
• Which half-reaction is reduction? Which is oxidation?
Separate into two skeleton half equations SO 3 2-  SO 4 2- MnO 4 -  MnO 2 SO 3 2- + H 2 O  SO 4 2- SO 3 2- + H 2 O  SO 4 2- + 2H + 2OH - + SO 3 2- + H 2 O  SO 4 2- + 2H + + 2OH - 2OH - + SO 3 2- + H 2 O  SO 4 2- + 2H 2 O 2OH - + SO 3 2- + H 2 O  SO 4 2- + 2H 2 O + 2e - MnO 4 -  MnO 2 MnO 4 -  MnO 2 + 2H 2 O 4H + + MnO 4 -  MnO 2 + 2H 2 O 4H + + MnO 4 - + 4OH -  MnO 2 + 2H 2 O + 4OH - 4H 2 O + MnO 4 -  MnO 2 + 2H 2 O + 4OH - 4H 2 O + MnO 4 - + 3e -  MnO 2 + 2H 2 O + 4OH - BALANCING REDOX REACTIONS
• Example #6
• Balance the following redox reaction which occurs in a basic solution:
• SO 3 2- + MnO 4 -  SO 4 2- + MnO 2
• Which half-reaction is reduction? Which is oxidation?
Oxidation: 2OH - + SO 3 2- + H 2 O  SO 4 2- + 2H 2 O + 2e - Reduction: 4H 2 O + MnO 4 - + 3e -  MnO 2 + 2H 2 O + 4OH - x 3 x 2 6OH - + 3SO 3 2- + 3H 2 O  3SO 4 2- + 6H 2 O + 6e - 8H 2 O + 2MnO 4 - + 6e -  2MnO 2 + 4H 2 O + 8OH - 3SO 3 2- + H 2 O + 2MnO 4 -  3SO 4 2- + 2MnO 2 + 2OH - BALANCING REDOX REACTIONS