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Redox Reactions

Redox Reactions. Oxidation Number. The oxidation state of an element in an elemental state is zero. (O 2 , Fe, He) The oxidation state of an element in a monoatomic ion is the charge of that ion. (oxidation number of iron in Fe +3 is +3)

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Redox Reactions

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  1. Redox Reactions

  2. Oxidation Number • The oxidation state of an element in an elemental state is zero. • (O2, Fe, He) • The oxidation state of an element in a monoatomic ion is the charge of that ion. • (oxidation number of iron in Fe+3 is +3) • The oxidation number of Group 1 elements in all their compounds is +1. • The oxidation number of Group 2 elements in all their compounds is +2.

  3. Oxidation Number • The oxidation number for fluorine in a compound is -1. • The oxidation number of H is almost always +1. • The oxidation of O is ordinarily -2. • The sum of oxidation numbers in a molecule is zero. • H2SO4 • In a polyatomic ion, the sum of the oxidation numbers is the charge of the ion. • Cr2O7-2

  4. Definitions Oxidation: An increase in oxidation number Reduction: A decrease in oxidation number Reducing agent: Whatever is oxidized Oxidizing agent: Whatever is reduced Zn(s)+ 2H+(aq)Zn+2(aq) + H2(g) Zn is oxidized (ox # 0+2) Reducing agent H+ is reduced (ox # +10) Oxidizing agent

  5. Balancing Half-Equations Balance the atoms of the atoms of the element being oxidized or reduced. Balance the oxidation number by adding electrons. Balance charge by adding H+ in acidic solution, OH- in basic solution. Balance hydrogen by adding H2O molecules. Check to make sure oxygen is balanced.

  6. Balancing equations in Acidic Solution

  7. MnO4-(aq)Mn+2(aq) MnO4-Mn+2 MnO4- + 5e-Mn+2 MnO4- + 5e- + 8H+Mn+2 MnO4- + 5e- + 8H+ Mn+2 + 4H2O

  8. Balancing equations in Basic solution

  9. Cr(OH)3(s)  CrO4-2(aq) Cr(OH)3 CrO4-2 Cr(OH)3 CrO4-2 + 3 e- Cr(OH)3 + 5OH-CrO4-2 + 3e- Cr(OH)3 + 5OH-CrO4-2 + 3e- + 4H2O

  10. Balancing Redox Equations

  11. Split the equation into two half-equations. Balance one of the half equations with respect to atoms and charge. Balance the other half equation. Combine the two half equations in such a way as to eliminate electrons.

  12. Balance the following redox equation in acidic solution. Fe+3(aq) + MnO4-(aq)Fe+2(aq) + Mn+2(aq) Oxidation: Fe+3Fe+2 Reduction: MnO4-Mn+2 Balanced half equations Fe+2 -Fe+3+ e- MnO4- + 5e-+ 8H+ Mn+2 + 4H2O To eliminate e-, multiply the oxidation half-equation by 5 and add to the reduction half equation. 5Fe+2 + MnO4- + 5e- + 8H+5Fe+3 + Mn+2 + 4H2O + 5e- 5Fe+2 + MnO4- + 8H+5Fe+3 + Mn+2 + 4H2O

  13. Balance the following redox equation in basic solution. Cl2(g) + Cr(OH)3(s)Cl-(aq)+ CrO4-2(aq) Cl2+ 2e- 2Cl- Cr(OH)3CrO4-2 + 3e- + 4H2O Multiply reduction equation by 3 and the oxidation equation by 2. 3Cl2+ 6e- + 2Cr(OH)3 6Cl- + 2CrO4-2 + 6e- + 8H2O 3Cl2+ 2Cr(OH)3 6Cl- + 2CrO4-2 + 8H2O

  14. Stoichiometry

  15. What volume of of 0.684 M KMnO4 solution is required to react completely with 27.50 mL of 0.250 M Fe(NO3)2? 5Fe+2 + MnO4- + 8H+5Fe+3 + Mn+2 + 4H2O Moles Fe+2 = 0.02750 L x0.250 moles Fe(NO3)2x1 mole Fe+2 = 1 L 1 mole Fe(NO3)2 6.88 x 10-3moles Fe+2 Moles MnO4- = 6.88 x 10-3moles Fe+2 x1 mole MnO4- = 0.00138 moles MnO4- 5 moles Fe+2 Moles KMnO4 = moles MnO4- = 0.00138 moles KMnO4 V = moles/M 0.00138 moles KMnO4/0.684 M KMnO4 = 2.02 x 10-2L = 2.02 mL

  16. Redox Reaction Problems • Page 97 • 49-56 • 64, 66, 68 • 70, 72

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