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Electrochemistry

Electrochemistry. Chapters 18-19. Defining Oxidation and Reduction. Oxidation – old definition – reacts with oxygen - current definition – loss of electrons Reduction - old definition – formation of metals from its compounds - current definition – gain of electrons.

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Electrochemistry

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  1. Electrochemistry Chapters 18-19

  2. Defining Oxidation and Reduction • Oxidation – old definition – reacts with oxygen - current definition – loss of electrons • Reduction - old definition – formation of metals from its compounds - current definition – gain of electrons Zn (s) + Cu2+(aq) → Cu (s) + Zn2+(aq) As the reaction proceeds, the blue color of the Cu2+ solution fades and Cu is deposited on the metal strip. The Zn strip is in direct contact with the Cu2+ solution and the exchange of electrons between the reactants can occur.

  3. Equation: Zn (s) + CuSO4(aq) → Cu (s) + ZnSO4(aq) Ionic Equation: Zn (s) + Cu2+(aq) + SO42-(aq)→ Cu (s) + Zn2+(aq) + SO42-(aq) Net Ionic Equation: Zn (s) + Cu2+(aq) → Cu (s) + Zn2+(aq) By looking at the charges – there is obviously a change... Zinc atoms lose electrons  undergoes oxidation Copper ions gain electrons  undergoes reduction

  4. When both oxidization and reduction occurs in a reaction, it is known as an Oxidation-Reduction Reaction, or a Redox Reaction. LEO the lion goes GER Loss of Electrons is Oxidation Gain of Electrons is Reduction OIL RIG Oxidation Is Loss Reduction Is Gain

  5. In a redox reaction, both reactants have specific names: • Oxidizing Agent – the reactant that oxidizes another reactant. • The oxidizing agent is reduced. • Reducing Agent – the reactant that reduces the other reactant. • The reducing agent is oxidized. • Practice Problems #1-4 Page 715

  6. Half Reactions • To monitor the transfer of electrons in a redox reaction, you can represent the oxidation and reduction separately. • A half-reaction is a balanced equation that shows the number of electrons involved in either oxidation or reduction. • It requires two half-reactions to represent a redox reaction. One that shows oxidation; one that shows reduction.

  7. Redox Reaction: Zn (s) + Cu2+(aq) → Cu (s) + Zn2+(aq) Zn (s) is oxidized Cu2+(aq) is reduced. Oxidation Half-Reaction: Zn (s)→ Zn2+(aq) + 2 e- Reduction Half Reaction: Cu2+(aq) + 2e- → Cu (s) Note: Each side balances.... We use lowest whole numbers.

  8. There is one special type of redox reaction. A disproportionation reaction is a redox reaction where one element undergoes both oxidation and reduction. Ex. 2 Cu+(aq) → Cu (s) + Cu2+ The two half reactions are: Oxidation: Cu+(aq) → Cu2+ + 1 e- Reduction: Cu+(aq) + 1 e- → Cu (s) Practice Problems #5-8 Page 716

  9. Oxidation Numbers • Many redox reactions involve reactants and products with covalent bonds – • Including elements that exist as covalent compounds, such as O2 • covalent compounds, such as H2O • Including polyatomic ions that are not spectator ions, such as MnO4-, NO3-, etc. • For reactions involving covalent compounds, we need to use oxidation numbers to determine whether or not a compound has gained or lost electrons. Oxidation Number – an actual or hypothetical charge that is assigned by a set of rules.

  10. Not all oxidation numbers are integers – you can still use them. Charges are 2-, 3-, 2+ Oxidation Numbers are -2, -3, +4 Practice Problems #9-12 Page 726

  11. Identifying Redox Reactions • A Redox reaction must contain an oxidation and a reduction. • If it does not contain both, it is not a redox reaction. When looking at any reaction, 1. Determine the oxidation numbers of all elements. 2. Identify any elements with an increase in oxidation number and a decrease in oxidation number. 3. If you have one of each – REDOX!!! Sample Problem Page 727 Practice Problems #13-16 Page 728

  12. Balancing Half Reactions • As redox reactions may not take place under neutral situations, one must consider the H+ and the OH- ions. • As acidic and basic solutions are different, there are specific steps in balancing both acidic and basic conditions.

  13. Balancing Half-Reactions in Acidic Solution • Write an unbalanced half-reaction for the given reactants and products. • Balance any atoms except Oxygen and Hydrogen. • Balance any Oxygen atoms by adding water molecules. • Balance and Hydrogen atoms by adding hydrogen atoms. • Balance the charges by adding electrons.

  14. Ex. Write a balanced half-reaction that shows the reduction of permanganate, MnO4-, to manganese (II) ions in acidic solution. Step One: MnO4- → Mn2+ Step Two: 1 MnO4- → 1 Mn2+ Step Three: 1 MnO4- → 1 Mn2+ + 4 H2O Step Four: 1 MnO4- + 8 H+ → 1 Mn2+ + 4 H2O Step Five 7+ → 2+ So add 5e- to balance Therefore: 1 MnO4- + 8 H+ + 5e- → 1 Mn2+ + 4 H2O Practice Problems 17-20 on Page 732

  15. Balancing Half-Reactions For Basic Solutions • Write an unbalanced half-reaction for the given reactants and products. • Balance any atoms except Oxygen and Hydrogen. • Balance any Oxygen atoms by adding water molecules. • Balance and Hydrogen atoms by adding hydrogen atoms. • Adjust for basic conditions... Add the same number of hydroxide ions to both sides of the equation. • Combine and H+ ions with OH- ions to make water. • Remove any water molecules present on both sides of the equation. • Balance the charges by adding electrons.

  16. Write a balanced half-reaction that shows the oxidation of thiosulfate ions, S2O32-, to SO32-, in basic solution. Step One: S2O32- → SO32- Step Two: S2O32- → 2 SO32- Step Three: S2O32- + 3 H2O → 2 SO32- Step Four: S2O32- + 3 H2O → 2 SO32- + 6 H+ Step Five: S2O32- + 3 H2O + 6 OH- → 2 SO32- + 6 H+ + 6 OH- Step Six: 6 H2O Step Seven: S2O32- + 6 OH- → 2 SO32- + 3 H2O Step Eight: 8- → 4- So add 4e- Therefore: S2O32- + 6 OH- → 2 SO32- + 3 H2O + 4e- Practice Problems #21-24 Page 734

  17. Half-Reaction Method For Balancing RedoxRxns • Write an unbalanced net ionic equation – if it is not already given. • Split the unbalanced net ionic equation into its two half-reactions. • Balance the oxidation half separately. • Balance the reduction half separately. • Determine the Least Common Multiple (LCM) for the electrons from both half-reactions. • Multiply the half-reactions by the appropriate numbers to eliminate the electrons from both sides. • Add the two half-reactions. • Remove any identical species from both sides of the equations. • Add any spectator ions and any states if necessary. Sample Problem on Page 737

  18. Problems • Balance the following redox reactions: • Cr (s) + Sn4+(aq) → Cr3+(aq) + Sn2+(aq) Acidic Soln • Al (s) + H+(aq) → Al3+(aq) + H2(g) Acidic Soln • Zn (s) + Ag+(aq) → Zn2+(aq) + Ag (s) Acidic Soln • Br2(l) + SO32-(aq) → Br-(aq) + SO42-(aq) Basic Soln • IO3-(aq) + H2S(g)→ I2(g) + SO32-(aq) Basic Soln • NO2 (g) + ClO-(aq) → NO3-(aq) + Cl-(aq) Basic Soln • Do Practice Problems on Page 738-739 #s 25 - 28

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