5.4 Balancing half-reactions 8 HSO 3 S 2 O H + + 2HSO H 2 O. + MnO 4 - Mn 2+ + Answer: 4H + on the left will cancel these 4H +

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1 Notes on Unit 5 OxidationReduction 5.4 Balancing halfreactions Some halfreactions are on the table, but not all. You may be asked to answer these Acid Soln. E.g.) S 2 O 8 2 HSO 3 (acid soln.) (1) Balance Major Atoms (other than O & H S in this case) S 2 O 2 8 HSO 3 (2) Balance O atoms, by adding H 2 O (to the side with less O s) S 2 O 2 8 2HSO 3 + (3) Balance H atoms by adding H + (to the side with less H s) S 2 O HSO 3 + 2H 2 O (4) Balance charge by adding e s (to the more + side) S 2 O H + + 2HSO 3 + 2H 2 O TIC = (2) TIC = (2) Always doublecheck these 4 Step Summary: Major atoms O H e E.g. 2 MnO 4 Mn (acid soln) + MnO 4 Mn + In basic solution Do the first steps of the balancing just like an acid E.g.) MnO 2 MnO 4 (basic solution) Major (Mn already balanced) Oxygen + MnO 2 MnO 4 Hydrogen 2H 2 O + MnO 2 MnO 4 + Charge 2H 2 O + MnO 2 MnO 4 + 4H + + In basic solution: write the reaction H + + OH H 2 O or H 2 O H + + OH In whichever way is needed to cancel out the H + s Add to the halfrx E.g.) 2H 2 O + MnO 2 MnO 4 + 4H + + 3e Answer: 4H + on the left will cancel these 4H + Reduce H 2 O s from both sides Unit 5 Notes 1

2 Notes on Unit 5 OxidationReduction E.g. 4: Pb HPbO 2 (basic soln) Pb HPbO 2 Hebden p # Balancing Overall Redox Reactions Using the Halfreaction Method (1) Break up Rx into 2 halfrx s. (5.1) (2) Balance each one (in acidic or basic as given) (5.4) (3) Multiply each half rx by whatever is needed to cancel out e s (Note: balanced halfrx have e s (on left reduction on right oxidation) Balanced redox don t have e s) (4) Add the 2 halfrx s canceling e s and anything else (usually H 2 O s, H + s or OH s) in order to simplify. Example: U 4+ + MnO 4 Mn + UO 2 (acidic) Balance each ½ rx U 4+ UO 2 (Major (U) balanced already) Oxygen U 4+ + UO 2 Hydrogen U H 2 O UO 2 + Charge U H 2 O UO 2 + 4H + + MnO 4 Mn (Major (Mn) balanced already) MnO 4 Mn + MnO 4 + Mn + 4H 2 O MnO 4 + 8H + + Mn + 4H 2 O (U H 2 O UO 2 4H + + 2e ) 5 (MnO 4 + 8H + + 5e Mn + 4H 2 O) 2 Multiply by 5 to get 10e Multiply by 2 to get 10e 5U H 2 O + 2MnO H e 5UO H + + 2Mn + 8H 2 O + 10e To simplify: Take away e from both sides Take away H + s from both sides Take away H 2 O s from both sides 5U H 2 O + 2MnO 4 5UO 2 + 4H + + 2Mn Quick check by finding TIC s on both sides TIC = +18 TIC = +18 Unit 5 Notes 2

3 Notes on Unit 5 OxidationReduction Try this one: SO 2 + IO 3 SO I 2 (basic solution) Eg) Br 2 Br + BrO 3 (basic) (found in some hot tubs) Some redox equations have just one reactant Use this as the reactant in both halfrx s. These are called selfoxidationreduction or Disproportionation reactions. Half rx s are: Br 2 Br Br 2 BrO 3 Answer: Hebden p Exercise # 24 Unit 5 Notes 3

4 Notes on Unit 5 OxidationReduction 5.6 Balancing Redox Equations Using Oxidation Numbers In a redox equation the overall change in oxidation numbers ( ON) must be zero. Therefore the overall increase must equal the overall decrease. Eg. 1) Balance the following equation: U 4+ + MnO 4 Mn + UO 2 (in acidic solution) 1) Break into half reactions and assign oxidation numbers (ON) U 4+ UO 2 MnO 4 Mn 2) We need to look at the species with a ON (not oxygen) and then cross multiply to get total ON to cancel out. U 4+ UO ON U = (+2) x = 10 MnO 4 Mn ON Mn = (5) x = 10 3) Multiply half reactions by chosen value and add the half reactions. 5 x (U 4+ UO 2 ) 2 x (MnO 4 Mn ) 4) Balance O and H as in 5.4 (H 2 O & H + ) 5 U MnO 4 2 Mn + 5 UO 2 Check charges: E.g 2) Balance the following equation in basic solution Zn + As 2 O 3 AsH 3 + Zn Hebden # 25 Unit 5 Notes 4

5 Notes on Unit 5 OxidationReduction 5.4 Balancing halfreactions Some halfreactions are on the table, but not all. You may be asked to answer these Acid Soln. E.g.) S 2 O 8 2 HSO 3 (acid soln.) (5) Balance Major Atoms (other than O & H S in this case) S 2 O HSO 3 (6) Balance O atoms, by adding H 2 O (to the side with less O s) S 2 O 2 8 2HSO 3 + 2H 2 O (7) Balance H atoms by adding H + (to the side with less H s) S 2 O H + 2HSO 3 + 2H 2 O (8) Balance charge by adding e s (to the more + side) S 2 O H e 2HSO 3 + 2H 2 O TIC = (2) TIC = (2) Always doublecheck these 4 Step Summary: Major atoms O H e E.g. 2 MnO 4 Mn (acid soln) 5 e + 8H + + MnO 4 Mn + 4H 2 O In basic solution Do the first steps of the balancing just like an acid E.g.) MnO 2 MnO 4 (basic solution) Major (Mn already balanced) Oxygen 2H 2 O + MnO 2 MnO 4 Hydrogen 2H 2 O + MnO 2 MnO 4 + 4H + Charge 2H 2 O + MnO 2 MnO 4 + 4H + + 3e In basic solution: write the reaction H + + OH H 2 O or H 2 O H + + OH In whichever way is needed to cancel out the H + s Add to the halfrx E.g.) 2H 2 O + MnO 2 MnO 4 + 4H + + 3e 4H + + 4OH 4H 2 O Answer: MnO 2 + 4OH MnO 4 + 2H 2 O + 3e 4H + on the left will cancel these 4H + Reduce H 2 O s from both sides Unit 5 Notes 5

6 Notes on Unit 5 OxidationReduction E.g. 4: Pb HPbO 2 (basic soln) 2H 2 O + Pb HPbO 2 + 3H + 3H + +3OH 3H 2 O Pb + 3OH HPbO 2 + H 2 O Hebden p # Balancing Overall Redox Reactions Using the Halfreaction Method (5) Break up Rx into 2 halfrx s. (5.1) (6) Balance each one (in acidic or basic as given) (5.4) (7) Multiply each half rx by whatever is needed to cancel out e s (Note: balanced halfrx have e s (on left reduction on right oxidation) Balanced redox don t have e s) (8) Add the 2 halfrx s canceling e s and anything else (usually H 2 O s, H + s or OH s) in order to simplify. Example: U 4+ + MnO 4 Mn + UO 2 (acidic) Balance each ½ rx U 4+ UO 2 (Major (U) balanced already) Oxygen U H 2 O UO 2 Hydrogen U H 2 O UO 2 + 4H + Charge U H 2 O UO 2 + 4H + + 2e MnO 4 Mn (Major (Mn) balanced already) MnO 4 Mn + 4H 2 O MnO 4 + 8H + Mn + 4H 2 O MnO 4 + 8H + + 5e Mn + 4H 2 O (U H 2 O UO 2 4H + + 2e ) 5 (MnO 4 + 8H + + 5e Mn + 4H 2 O) 2 Multiply by 5 to get 10e Multiply by 2 to get 10e 5U H 2 O + 2MnO H e 5UO H + + 2Mn + 8H 2 O + 10e To simplify: Take away 10 e from both sides Take away 16 H + s from both sides Take away 8 H 2 O s from both sides 5U H 2 O + 2MnO 4 5UO 2 + 4H + + 2Mn Quick check by finding TIC s on both sides TIC = +18 TIC = +18 Unit 5 Notes 6

7 Notes on Unit 5 OxidationReduction Try this one: SO 2 + IO 3 SO I 2 (basic solution) Eg) Br 2 Br + BrO 3 (basic) (found in some hot tubs) Some redox equations have just one reactant Use this as the reactant in both halfrx s. These are called selfoxidationreduction or Disproportionation reactions. Half rx s are: Br 2 Br Br 2 BrO 3 Answer: Hebden p Exercise # 24 Unit 5 Notes 7

8 Notes on Unit 5 OxidationReduction 5.6 Balancing Redox Equations Using Oxidation Numbers In a redox equation the overall change in oxidation numbers ( ON) must be zero. Therefore the overall increase must equal the overall decrease. Eg. 1) Balance the following equation: U 4+ + MnO 4 Mn + UO 2 (in acidic solution) 1) Break into half reactions and assign oxidation numbers (ON) U 4+ UO 2 MnO 4 Mn ) We need to look at the species with a ON (not oxygen) and then cross multiply to get total ON to cancel out. U 4+ UO ON U = +2 (+2) x 5 = 10 MnO 4 Mn ON Mn = 5 (5) x 2 = 10 3) Multiply half reactions by chosen value and add the half reactions. 5 x (U 4+ UO 2 ) 2 x (MnO 4 5 U MnO 4 2 Mn + 5 UO 2 Mn ) 5) Balance O and H as in 5.4 (H 2 O & H + ) 2 H 2 O + 5 U MnO 4 2 Mn + 5 UO 2 + 4H + Check charges: = E.g 2) Balance the following equation in basic solution Zn + As 2 O 3 AsH 3 + Zn Hebden # 25 Unit 5 Notes 8

Using data to make your own simple Redox table

Using data to make your own simple Redox table Example problem: 1) Four metals A, B, C, & D were tested with separate solutions of A 2+, B 2+, C 2+ & D 2+. Some of the results are summarized in the following