1 / 12

Balancing Redox Equations

Balancing Redox Equations. Reduction/Oxidation = Redox. Iron (Fe) rusts (is oxidized) when it reacts with oxygen (O 2 ) in the air. 2Fe + 3O 2  Fe 2 O 3 [iron(III) oxide]. A rusted (oxidized) bolt (rust is Fe 2 O 3 ). A new iron bolt (Fe). Redox Equations.

sharne
Télécharger la présentation

Balancing Redox Equations

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Balancing Redox Equations Reduction/Oxidation = Redox Iron (Fe) rusts (is oxidized) when it reacts with oxygen (O2) in the air. 2Fe + 3O2 Fe2O3 [iron(III) oxide] A rusted (oxidized) bolt (rust is Fe2O3) A new iron bolt (Fe)

  2. Redox Equations Rusting is a an example of a redox reaction. 2Fe + 3O2 Fe2O3 [iron(III) oxide] In redox reactions, one chemical is oxidized (e.g., Fe is oxidized) and another chemical is reduced (e.g., O2 is reduced). Oxidation and reduction always occur together. One cannot occur without the other. In this reaction, when iron is oxidized, it gains oxygen. This definition of oxidation (‘gain of oxygen’), while correct in this case, is inadequate, because many redox reactions do not involve oxygen. A better definition follows, but do not disregard the example of rusting. It is something we all have observed and will help you understand and remember the broader definitions.

  3. Redox Equations +3 -2 0 0 lose 3 e’s, LEO gain 2 e’s, GER What are the oxidation numbers of iron and oxygen in the rust equation? Definitions to Memorize: Loss of Electrons = Oxidation Gain of Electrons = Reduction To do redox chemistry, you must be able to determine which atoms have lost e’s (have been oxidized) and which atoms have gained e’s (have been reduced). Thus, iron has been oxidized by oxygen. Oxygen is an ‘oxidizing agent’. and oxygen has been reduced by iron. Iron is a ‘reducing agent’. GER LEO the Lion says 2Fe + 3O2 Fe2O3 [iron(III) oxide]

  4. Balancing Redox Equations by the Half-Reaction Method • The following rules describe the most versatile method for balancing redox equations ... • Break the eqn. into two ½-reactions, one for oxidation (oxid.) and one for reduction (red.) • Balance all atoms in each ½-reaction other than H and O • Balance O and H as follows ... • first balance O by adding H2O, • then balance H’s by adding H+ • finally, balance the charges by adding sufficient electrons • e-'s are always added to the right side of the oxid. ½-reaction because oxid. = loss of e-'s • e-'s are always added to the left side of the red. ½-reaction because red. = gain of e-'s • Multiply each ½-reaction by a coefficient so that the same number of e-'s are transferred in each ½-reaction. The number of e’s gained must equal the number of e’s lost. • Add the ½-reactions and cancel common terms (e.g., H2O and/or H+) from each side. • If the reaction is carried out in neutral or basic solution, add enough OH- to neutralize all H+ (combine them to make H2O) and be sure to add the same quantity of OH- to both sides to keep mass and charge balanced. • Cancel common terms (H2O and/or OH-) from each side and check for mass and charge balance

  5. Bromate and sulfite react as shown : BrO3- + SO3-2Br2+ SO4-2 +6 +5 +4 0 BrO3-, the oxidizer is reduced SO3-2, the reducer is oxidized gain 5 e’s, GER Oxidation ½-reaction: 1. 3a) balance O w. H2O 3b) balance H w. H+ 3c) balance charge w e’s 4. Reduction ½-reaction: balance Br 3a) balance O w. H2O 3b) balance H w. H+ 3c) balance charge w. e’s SO3-2 SO4-2 H2O + SO3-2  SO4-2 H2O + SO3-2  SO4-2+ 2H+ H2O + SO3-2  SO4-2+ 2H+ + 2e- (H2O + SO3-2  SO4-2+ 2H+ + 2e-) × 5 BrO3- Br2 2BrO3- Br2 2BrO3- Br2 + 6H2O 12H+ + 2BrO3-  Br2 + 6H2O 10e- + 12H+ + 2BrO3-  Br2 + 6H2O lose 2 e’s, LEO O can only be balanced w. H2O H can only be balanced w. H+ 4. Multiply both ½-reactions by the appropriate coefficient so that both reactions involve the same number of e’s. Find the lowest common multiple (LCM) of the e’s transferred (10 e’s is the LCM of 2e’s and 10e’s). Combine the left sides of the ½-reactions and combine the right sides of the ½-reactions and reduce them. 5H2O + 5SO3-2 +10e- + 12H+ + 2BrO3-  5SO4-2 + 10H+ + Br2 + 6H2O + 10e- 5SO3-2+2H+ + 2BrO3-  5SO4-2+ Br2 + H2O

  6. Permanganate reacts w. hydrogen peroxide: MnO4- + H2O2MnO2+ O2 +7 -1 +4 0 Oxidation ½-reaction: 1. 3b) balance H w. H+ 3c) balance charge w e’s 6. neutralize H+ w. OH- Reduction ½-reaction: 3a) balance O w. H2O 3b) balance H w. H+ 3c) balance charge w. e’s 6. neutralize H+ w. OH- 7. cancel H2O gain 3 e’s, GER H2O2O2 H2O2 O2 + 2H+ H2O2 O2 + 2H++ 2e- 2OH- + H2O2 O2 + 2H2O+ 2e- (2OH-+ H2O2 O2 + 2H2O + 2e-) × 3 MnO4- MnO2 MnO4-  MnO2+ 2H2O 4H+ + MnO4-  MnO2 + 2H2O 3e- + 4H+ + MnO4-  MnO2 + 2H2O 3e-+ 4H2O + MnO4-  MnO2 + 2H2O + 4OH- 3e-+ 2H2O + MnO4-  MnO2 + 4OH- (3e-+ 2H2O + MnO4-  MnO2 + 4OH-) × 2 lose 1 e’s, LEO e’s are always on the right side of an oxidation ½-reaction MnO4-, the oxidizer is reduced H2O2, the reducer is oxidized e’s are always on the left side of a reduction ½-reaction Multiply both ½-reactions by the appropriate coefficient so that both reactions involve the same number of e’s. Find the lowest common multiple (LCM) of the e’s transferred (6 e’s is the LCM of 2e’s and 3e’s) 6OH- + 3H2O2+6e- + 4H2O + 2MnO4-  2MnO2 + 8OH- + 3O2 + 6H2O + 6e- 3H2O2+ 2MnO4- 2MnO2 + 2OH- + 3O2 + 2H2O

  7. Dichromate and methane react as shown : Cr2O7-2 + C2H6 CO2+ Cr+3 +3 +6 +4 -3 lose 7 e’s, LEO gain 3 e’s, GER Oxidation ½-reaction: 1. 2. balance C’s 3a) balance O w. H2O 3b) balance H w. H+ 3c) balance charge w e’s 6) neutralize H+ w. OH- 7) cancel H2O Reduction ½-reaction: 1. 2. balance Cr 3a) balance O w. H2O 3b) balance H w. H+ 3c) balance charge w. e’s 6. neutralize H+ w. OH- 7. cancel H2O C2H6CO2 C2H6 2CO2 4H2O + C2H6 2CO2 4H2O + C2H6 2CO2 + 14H+ 4H2O + C2H6 2CO2 + 14H+ + 14e- 4H2O + C2H6 + 14OH- 2CO2 + 14H2O + 14e- C2H6+ 14OH- 2CO2 + 10H2O + 14e- Cr2O7-2 Cr+3 Cr2O7-2 2Cr+3 Cr2O7-2 2Cr+3 + 7H2O 14H+ + Cr2O7-2 2Cr+3 + 7H2O 6e- + 14H+ + Cr2O7-2 2Cr+3 + 7H2O 6e- + 14H2O + Cr2O7-2 2Cr+3 + 7H2O + 14OH- 6e- + 7H2O + Cr2O7-2 2Cr+3 + 14OH- Cr2O7-2, the oxidizer is reduced C2H6, the reducer is oxidized

  8. Cr2O7-2 + C2H6 CO2+ Cr+3 continued: +3 +6 -3 +4 lose 7 e’s, LEO gain 3 e’s, GER Oxidation ½-reaction: Reduction ½-reaction: C2H6 + 14OH- 2CO2 + 10H2O + 14e- 7H2O + Cr2O7-2 + 6e- 2Cr+3 + 14OH- Multiply both ½-reactions by the appropriate coefficient so that both reactions involve the same number of e’s. Find the lowest common multiple (LCM) of the e’s transferred ( 42 is the LCM’s of 14e’s and 6e’s) (C2H6 + 14OH- 2CO2 + 10H2O + 14e-) × 3 (7H2O + Cr2O7-2 + 6e- 2Cr+3 + 14OH-) × 7 Combine the two ½-reactions: 3C2H6+ 42OH- 49H2O + 7Cr2O7-2+ 42e- 14Cr+3+ 98OH- + 6CO2+ 30H2O + 42e- 3C2H6 + 19H2O + 7Cr2O7-2 14Cr+3 + 56OH- + 6CO2

  9. Cr2O7-2 + C2H6 CO2+ Cr+3 Oxidation Number Method for Balancing Redox Equations +6 +5 +4 0 BrO3- + SO3-2Br2+ SO4-2 each Br atom in BrO3- gains 5 e’s. each S atom in SO3-2 loses 2 e’s, 10e’s is the LCM 5 e’s × 2BrO3- = 10e’s gained 2 e’s × 5SO3-2 = 10e’s lost 2BrO3- + 5SO3-2 Br2 + 5SO4-2 gain 5 e’s, GER lose 2 e’s, LEO each Mnatom in MnO4-gains 3 e’s. each Oatom in H2O2loses 1 e’s or each H2O2 lose 2 e’s, 6e’s is the LCM 3 e’s × 2MnO4- = 6e’s gained 2 e’s × 3H2O2 = 6e’s lost 2MnO4- + 3H2O2 2MnO2 + 3O2 +7 -1 +4 0 MnO4- + H2O2MnO2+ O2 gain 3 e’s, GER lose 1 e’s, LEO each Cr atom in Cr2O7-2 gains 3 e’s or each Cr2O7-2 gains 6 e’s each C atom in C2H6loses 7 e’s or each C2H6 lose 14 e’s, 42e’s is the LCM 6 e’s × 7Cr2O7-2 = 42e’s gained 14e’s × 3C2H6 = 42e’s lost 7Cr2O7-2+ 3C2H6 14Cr+3 + 6CO2 +3 +6 -3 +4 lose 7 e’s, LEO gain 3 e’s, GER

  10. A Difficult Redox Equation to Balance +7 -8/5 +4 +4 MnO4- + C5H8MnO2+ CO2 gain 3 e’s, GER lose 8/5 e’s, LEO Equations like the one above are not uncommon in organic chemistry, where carbon may have fractional oxidation numbers. Although this can be balanced by the oxidation number method (with a bit of mental arithmetic), most students will likely have more success using the Half-Reaction method. The balanced equation is shown below. Can you obtain this answer? 28MnO4- + 3C5H8+ 2H2O 28MnO2+ 15CO2 + 28 OH-

  11. Practice is the best teacher

  12. Activity Series of Metals and Oxidation Potential (V = emf) A positive oxidation potential means that a reaction is favorable (spontaneous) compared to hydrogen and will produce the emf shown (under standard conditions). These very reactive metals will be oxidized by H2O and liberate H2 gas. These moderately reactive metals are not oxidized by H2O, but are oxidized by dilute acids (H2SO4, HCl, etc.) and liberate H2 gas. A negative oxidation potential means that a reaction is unfavorable compared to hydrogen and will require that a emf greater than the value shown be applied to force the reaction to occur as written. The oxidation potential of hydrogen is set to zero and used as a reference for other redox reactions. Reversing an oxidation reaction yields a reduction reaction with the same magnitude of emf, but with opposite sign. These inert (‘noble’) metals are unreactive. They are not oxidized by dilute acids.

More Related