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Electrochemistry

Electrochemistry. homework. Page 863 1-13, 17, 31-89 odd. Oxidation- Reduction Reactions. Redox or oxidation-reduction reactions are reactions that involve a transfer of electrons. O xidation i s the l oss of electrons. R eduction i s the g ain of electrons.

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Electrochemistry

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  1. Electrochemistry

  2. homework • Page 863 • 1-13, 17, 31-89 odd

  3. Oxidation- Reduction Reactions • Redox or oxidation-reduction reactions are reactions that involve a transfer of electrons. • Oxidation is the loss of electrons. • Reduction is the gain of electrons. • (think of the charge, OIL RIG) • So in the reaction • 4 K + O2→ 4 K+ + 2 O2- • Potassium get oxidized, oxygen get reduced

  4. Oxidation States • Oxidation state is the theoretical charge on all atoms if all bonds were ionic. • The sum of the oxidation states must be equal to the charge of the ion or molecule.

  5. Using oxidation states • In the reaction… • 2 Na +2 H2O →2 NaOH + H2 • 0 +1 -2 +1 -2 +1 0 • Note the changes • Sodium went from 0 to 1 • 2 of the hydrogen atoms went from +1 to 0 (the other two were unchanged)

  6. Identification of Redox Components. • Specify which of the following equations represents oxidation-reduction reactions, and indicate the oxidizing agent, the reducing agent, the species being oxidized, and the species being reduced. • CH4(g) + H2O(g) CO(g) + 3H2(g) • 2AgNO3(aq) + Cu(s) Cu(NO3)2(aq) + 2Ag(s) • H+(aq) + 2CrO42-(aq) Cr2O72-(aq) + H2O(l)

  7. Half reactions • Ce4+ + Sn2+→ Ce3+ + Sn4+ • Half reactions • Ce4+ + e-→ Ce3+ • Sn2+→ 2e- + Sn4+ • Electrons lost must equal electrons gained! • 2Ce4+ +2 e-→2 Ce3+ • Merge the two half reactions • 2 Ce4+ + Sn2+→ 2 Ce3+ + Sn4+

  8. Redox reactions in acidic solutions • It will be noted in the problem • Balance all elements except hydrogen and oxygen. • Balance oxygen by adding H2O (which is always prevalent in an acidic solution) • Balance hydrogen by adding H+ • Then balance the charge adding electrons and proceed normally.

  9. Example • In an acidic solution • Cr2O72- + Cl-→ Cr3+ + Cl2 • Half reactions • Cr2O72-→ Cr3+ • Cl-→ Cl2

  10. Reduction side • Cr2O72-→ Cr3+ • Cr2O72-→ 2 Cr3+ • Cr2O72-→ 2 Cr3+ + 7 H2O • Cr2O72- + 14 H+→ 2 Cr3+ + 7 H2O • Cr2O72- + 14 H++ 6 e-→2Cr3++7 H2O

  11. Oxidation side • Cl-→ Cl2 • 2 Cl-→ Cl2 • 2 Cl-→ Cl2 + 2 e- • I have to equal 6 e- so multiply by 3 • 6 Cl-→ 3 Cl2 + 6 e-

  12. Combine my half reactions • Cr2O72- + 14 H++ 6 e- → 2 Cr3+ + 7 H2O • 6 Cl-→ 3 Cl2+ 6 e- • And you get • Cr2O72-+14 H++6Cl-→2Cr3++3 Cl2+7H2O • The electrons cancel out .

  13. Example • In an acidic solution • MnO4- + H2O2→ Mn2+ + O2 • Half reactions • MnO4- → Mn2+ • H2O2→ O2

  14. Top Equation • MnO4- → Mn2+ • MnO4- → Mn2+ + 4 H2O • MnO4- + 8 H+→ Mn2+ + 4 H2O • MnO4- + 8 H+→ Mn2+ + 4 H2O • MnO4- + 8 H++ 5 e-→ Mn2+ + 4 H2O

  15. Bottom Equation • H2O2→ O2 • H2O2→ O2 + 2 H+ • H2O2→ O2 + 2 H+ + 2 e- • I need to equal 5 e- so… • That won’t work… • 2MnO4- + 16 H++ 10 e-→ 2 Mn2+ + 8 H2O • 5 H2O2→ 5 O2 + 10 H+ + 10 e-

  16. Add them together • 2MnO4- + 16 H++ 10 e-→ 2 Mn2+ + 8 H2O • 5 H2O2→ 5 O2 + 10 H+ + 10 e- • And you get • 2 MnO4- + 6 H++ 5 H2O2 → 2 Mn2+ + 5 O2 + 8 H2O • Notice the H+ canceled out as well.

  17. Balancing Redox Equations in a basic solution • Look for the words basic or alkaline • Follow all rules for an acidic solution. • After you have completed the acidic reaction add OH- to each side to neutralize any H+. • Combine OH- and H+ to make H2O. • Cancel out any extra waters from both sides of the equation.

  18. Example • We will use the same equation as before • In a basic solution • MnO4- + H2O2→ Mn2+ + O2 • 2 MnO4- + 6 H++ 5 H2O2 → 2 Mn2+ + 5 O2 + 8 H2O

  19. Basic solution • Since this is a basic solution we can’t have excess H+. • We will add OH- to each side to neutralize all H+ • 2 MnO4- + 6 H++ 5 H2O2 + 6OH- →2 Mn2+ +5 O2 +8 H2O + 6OH- • We added 6 OH- because there were 6H+

  20. Cont. • H+ + OH- → H2O • Combine the hydroxide and hydrogen on the reactant side to make water • 2 MnO4- + 6 H2O + 5 H2O2 → 2 Mn2++5 O2+ 8 H2O + 6OH- • Cancel out waters on both sides 2 MnO4- + 5 H2O2 →2 Mn2+ + 5 O2 +2 H2O +6OH-

  21. Another example • In a basic solution • MnO4 − + SO32-→MnO4 2− + SO42- • Half reactions • MnO4 − → MnO4 2− • SO32-→ SO42-

  22. Half reactions • MnO4 − → MnO4 2− • MnO4 - + e- → MnO4 2− • SO32-→ SO42- • H2O + SO32-→ SO42- • H2O + SO32-→ SO42- + 2 H+ • H2O + SO32-→ SO42- + 2 H+ +2e- • Double the top reaction

  23. 2 MnO4 - + 2 e- → 2 MnO4 2− • H2O + SO32-→ SO42- + 2 H+ +2e- • Combine them • 2 MnO4 - + H2O + SO32- → 2 MnO4 2− +SO42- + 2 H+ • Add OH- 2 MnO4 - + H2O + SO32- + 2 OH- → 2 MnO4 2−+SO42- +2 H++2 OH-

  24. 2 MnO4 - + H2O + SO32- + 2 OH- → 2 MnO4 2− +SO42- + 2 H2O • finishing • 2 MnO4 - + SO32- + 2 OH- → 2 MnO4 2− +SO42- + H2O

  25. Major Redox Points to Remember. • Any redox reaction can be treated as the sum of thereduction and oxidation half-reactions. • Mass (atoms) and charge are conserved in each half-reaction. • Electrons lost in one half-reaction are gained in the other. • Even though the half-reactions are treated separately,electron loss and electron gain occur simultaneously.

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