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Redox Reactions AP Chemistry Unit 3.

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Presentation on theme: "Redox Reactions AP Chemistry Unit 3."— Presentation transcript:

1 Redox Reactions AP Chemistry Unit 3

2 Titration: volumetric analysis to determine concentration; lab work!
Stoichiometric point = equivalence point Moles of acid (H+) = moles of base (OH-) [in acid-base titrations] How many mL of M sodium hydroxide solution are needed to neutralize 20.0 mL of M sulfuric acid solution? Write a balanced equation 2. Identify moles of one reactant. 3. Use mole ratios to predict moles of other reactant. 4. Solve for Molarity or mass.

3 2NaOH + H2SO4  2 H2O + Na2SO4 0.245 M = ___x____ 0.0200 L
moles moles moles = M x L = 16.1 mL **since we are finding molarity of entire reactant (not net ionic) use the molecular equation…usually common in titration problems

4 Redox reaction: transfer of electrons
Reduction: process of gaining electrons; becoming more negative LEO the lion Oxidation: process of losing electrons; becoming more positive goes GER

5 Oxidation and Reduction
Reactant reduced is the oxidizing agent. H+ oxidizes Zn by taking electrons from it. (e- acceptor) Reactant oxidized is the reducing agent. Zn reduces H+ by giving it electrons. (e- donor)

6 Oxidation Numbers In order to keep track of what loses electrons and what gains them, we assign oxidation numbers.

7 Oxidation numbers = charge of an atom Ca+2 F-1 Mn+7
-2 -2 x 4 = -8 K2SO4 CO3-2 ___ + 3(-2) = -2 Fe3O4 3(__) + 4(-2) = 0 Rules: Elements = 0 Hydrogen +1; except in metal hydrides (NaH), then it is Placement is the key! Oxygen = -2; except in peroxides (H2O2; O = -1) In a compound, oxidation numbers must balance out to net charge of zero. 2(+1) + ___ + (-8) = 0

8 Balancing simple redox reactions
Cu+2 + Al → Cu + Al+3 1. Cu+2 → Cu Al → Al+3 2. no changes e- + Cu+2 → Cu Al → Al+3 + 3e- 4. (2e- + Cu+2 → Cu)3 (Al → Al+3 + 3e-)2 5. 6e- + 3Cu+2 → 3Cu 2Al → 2Al+3 + 6e- 2Al + 3Cu+2 → 3Cu + 2Al+3 2(0) + 3(+2) = 3(0) + 2(+3) Half reaction method: Write ½ reactions. Balance atoms, if necessary (except H & O) Balance charges by adding electrons. Multiply (if needed) to cancel electrons. Check work: charges & atoms Balancing simple redox reactions

9 Half-Reaction Method MnO4− + C2O42-  Mn2+ + CO2
First, we assign oxidation numbers. MnO4− + C2O42-  Mn2+ + CO2 +7 +3 +4 +2 Since the manganese goes from +7 to +2, it is reduced;and changes color. Since the carbon goes from +3 to +4, it is oxidized.

10 Balancing redox reactions in acidic solutions
MnO4−(aq) + C2O42−(aq)  Mn2+(aq) + CO2(aq) 1. C2O42−  CO2 MnO4−  Mn2+ 2. C2O42−  2 CO2 3. MnO4−  Mn H2O 4. 8 H+ + MnO4−  Mn H2O 5. C2O42−  2 CO2 + 2 e− 5 e− + 8 H+ + MnO4−  Mn H2O 6. 5 C2O42−  10 CO e− 10 e− + 16 H+ + 2 MnO4−  2 Mn H2O 16 H+ + 2 MnO4− + 5 C2O42−  2 Mn H2O + 10 CO2 Write ½ reactions Balance all atoms except H & O, if needed. Balance oxygen atoms by adding H2O to one side since aqueous. Balance hydrogen atoms by adding H+ Balance electric charge by adding electrons Make electrons equal, to cancel Check atoms & charge. Balancing redox reactions in acidic solutions

11 Balancing redox reactions in basic solutions
Zn + VO3-1  VO Zn+2 Zn  Zn+2 VO3-1  VO+2 VO3-1  VO+2 + 2H2O 4 H+ + VO3-1  VO+2 + 2H2O 2 (e- + 4 H+ + VO3-1  VO+2 + 2H2O) Zn  Zn e- 6. 2e- + 8 H+ + 2VO3-1 2VO+2 +4H2O Zn + 2 VO H+  2VO+2 + 4H2O + Zn+2 OH OH- 8 H2O Zn + 2 VO H2O  2VO+2 + 4H2O + Zn OH- Zn + 2 VO H2O  2VO+2 + Zn OH- Write ½ reactions Balance all atoms except H & O, if needed. Balance oxygen atoms by adding H2O to one side since aqueous. Balance hydrogen atoms by adding H+ Balance electric charge by adding electrons Make electrons equal, to cancel Check atoms & charge. Now add OH- to both sides to eliminate H+ Balancing redox reactions in basic solutions

12 Half-Reaction Method Consider the reaction between MnO4− and C2O42− :
MnO4−(aq) + C2O42−(aq)  Mn2+(aq) + CO2(aq) **This is an example of a redox titration…next lab!


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