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Redox Reactions Oxidation - Reduction reactions Terms Oxidation loss of electrons electrons are a product Na --> Na + + e - Reduction gain of electrons.

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Presentation on theme: "Redox Reactions Oxidation - Reduction reactions Terms Oxidation loss of electrons electrons are a product Na --> Na + + e - Reduction gain of electrons."— Presentation transcript:

1 Redox Reactions Oxidation - Reduction reactions Terms Oxidation loss of electrons electrons are a product Na --> Na + + e - Reduction gain of electrons electrons are a reactant Br 2 + 2e - --> 2Br -

2 Oxidizing Agent Reactant that causes the oxidation (this reactant will be reduced) Reducing Agent Reactant that causes the reduction (this reactant will be oxidized) Oxidation Number charge on an atom can be explicit or implicit oxidation number of elements = 0 Na + oxidation number = +1 Br - oxidation number = -1

3 What is the oxidation number of iron in ferric chloride? A)+1 B)+2 C)+3 D)-1 E)-2

4 What is the oxidation number of iron in ferric chloride? Ferric chloride = ? Chloride has a charge of? So iron has to be? FeCl 3 +3

5 What is the oxidation number on S in sulfur dioxide? A)+2 B)+4 C)+6 D)-2 E)-4

6 What is the oxidation number on S in sulfur dioxide? sulfur dioxide = ? most of the time, oxygen in combination is the oxide oxygen. So each oxygen has an oxidation number of ? the molecule has to have an overall charge of 0 S + 2(-2) = 0 S = +4 SO 2 -2

7 What is the oxidation number of Cr in CrO 4 -2 ? A)+4 B)+6 C)+8 D)-4 E)-2

8 What is the oxidation number of Cr in CrO 4 -2 ? CrO 4 -2 is called? ion has a charge of -2 Cr + 4(O) = -2 Cr + 4(-2) = -2 Cr = +6 Chromate ion

9 Na + Br 2 --> NaBr 2 2 2 Na -- > 2 Na + + 2e - Oxidation half reaction Br 2 + 2e - -- > 2 Br - Reduction half reaction 2 Na + Br 2 -- > 2 NaBr

10 Balancing Redox Reaction Ion Electron Method Split the reaction into half reactions and balance each half reaction 1.Balance all elements other than H and O 2.Balance O by adding H 2 O 3.Balance H by adding H +. If the reaction is in acid go to 4, if in base go to 3A. 3A. If the reaction is in base, add OH - equal to the number of H + to both sides of the equation. Combine the H + and OH - that are on the same side to make H 2 O 4.Balance the charge by adding e - 5.Multiply the 2 half reactions so that the number of electrons is the same in each half reaction 6.Add the half reactions back together and eliminate redundancies

11 Cr 2 O 7 -2 + SO 3 -2 --> Cr +3 + SO 4 -2 Cr 2 O 7 -2 --> Cr +3 SO 3 -2 --> SO 4 -2 acid 2 +7H 2 O + 14H + +12 +6 + 6e - H 2 O ++ 2H + - 2 0 + 2e - 2 +7H 2 O + 14H + + 6e - Cr 2 O 7 -2 --> Cr +3 SO 3 -2 --> SO 4 -2 H 2 O + + 6H + + 6e - 3( 333 8 4 Cr 2 O 7 -2 +8H + +3SO 3 -2 -->2Cr +3 + 4H 2 O + 3SO 4 -2

12 MnO 4 - + C 2 O 4 -2 --> MnO 2 + CO 2 base MnO 4 - --> MnO 2 2MnO 4 - +4H 2 O+3C 2 O 4 -2 -->2MnO 2 + 8OH - + 6CO 2 C 2 O 4 -2 --> + CO 2 + 2 H 2 O + 4H + + 4OH - 4H 2 O -4 3e - + 2 0 2e - -2 2( 3( 2MnO 4 - --> 2MnO 2 + 4 H 2 O + 8H 2 O + 8OH - 6e - + 3C 2 O 4 -2 --> + CO 2 6 6e - 4

13 Br 2 + OH - --> BrO 3 - + Br - + H 2 O disproportionation Reaction where the same reactant is both oxidized and reduced Br 2 --> BrO 3 - Br 2 --> Br - 2 + 6H 2 O -2 - 12 +10e - 2 + 2e - +12H + + 12OH - 12H 2 O - 2 0 5( Br 2 --> BrO 3 - 2 + 6H 2 O+10e - + 12H 2 O + 12OH - 5 Br 2 + 10e - --> 10 Br - Br 2 + 5 Br 2 + 12OH - --> 2 BrO 3 - + 6 H 2 O + 10 Br - 6

14 Redox titration a known concentration of oxidizing agent is used to find an unknown concentration of reducing agent (or vice versa)

15 37.5 mL of 0.658 M KMnO 4 solution were required to completely react with 22.0 mL of a basic K 2 C 2 O 4 soluiton. What is the concentration of the oxalate solution? A)0.748 M B)1.12 M C)1.68 M D)3.89 M E)0.579 M 2MnO 4 - +4H 2 O+3C 2 O 4 -2 -->2MnO 2 + 8OH - + 6CO 2

16 37.5 mL of o 658 M KMnO 4 solution were required to completely react with 22.0 mL of a basic K 2 C 2 O 4 soluiton. What is the concentration of the oxalate solution? 1.Balance the redox reaction. K + is spectator. 4H 2 O + 2MnO 4 - + 3C 2 O 4 -2 --> 2MnO 2 + 6CO 2 + 8OH - 2. Concentration = M = moles C 2 O 4 -2 / L. So, need to find moles C 2 O 4 -2. Use the moles of MnO 4 -2 and the mole ratio from the balanced reaction.

17 37.5 mL of 0. 658 M KMnO 4 solution were required to completely react with 22.0 mL of a basic K 2 C 2 O 4 soluiton. What is the concentration of the oxalate solution? 4H 2 O + 2MnO 4 - + 3C 2 O 4 -2 --> 2MnO 2 + 6CO 2 + 8OH - Moles = ? Moles = VM

18 Activity Series oxidation series that shows the relative reactivity of metals

19 Li K Ba Ca Na Mg Al Mn Zn Cr Fe Cd Co Ni Sn Pb H 2 Cu Hg Ag Au Reactions are oxidation reactions. --> Li + --> Ca +2 --> Al +3 --> Zn +2 --> Cd +2 Metals react with ions that lie below them. Li + Ca +2 --> Li + + Ca Al + Cd +2 --> Al +3 + Cd Al + Ca +2 --> ? NR Reactions occur top left with bottom right

20 Li K Ba Ca Na Mg Al Mn Zn Cr Fe Cd Co Ni Sn Pb H 2 Cu Hg Ag Au --> Li + --> Ca +2 --> Al +3 --> Zn +2 --> Cd +2 Which of the following combination will react? A)Li + and Ca B)Ca +2 and Mg C)Ca and K + D)Ni and H + E)Ni and Sn


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