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Electrochemistry

Electrochemistry. Oxidation-Reduction (Redox) Reactions. TRANSFER OF ELECTRONS Oxidation: Electrons are lost (product side of equation), charge increases. The substance that is oxidized is the reducing agent.

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Electrochemistry

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  1. Electrochemistry

  2. Oxidation-Reduction (Redox) Reactions TRANSFER OF ELECTRONS Oxidation: Electrons are lost (product side of equation), charge increases. The substance that is oxidized is the reducing agent. Reduction: Electrons are gained (reactant side of equation), charge decreases. The substance that is reduced is the oxidizing agent.

  3. Oxidation-Reduction (Redox) Reactions 2 Fe2O3(s) + 3 C(s) 4 Fe(s) + 3 O2(g) 2 Fe2O3(s) 4 Fe(s) + 3 CO2(g) Rusting of iron: an oxidation of Fe Manufacture of iron: a reduction of Fe3+

  4. Oxidation-Reduction (Redox) Reactions Oxidation: The loss of one or more electrons by a substance, whether element, compound, or ion. Reduction: The gain of one or more electrons by a substance, whether element, compound, or ion.

  5. Oxidation-Reduction (Redox) Reactions Oxidation Number (State): A value that indicates whether an atom is neutral, electron-rich, or electron-poor. Rules for Assigning Oxidation Numbers An atom in its elemental state has an oxidation number of 0. Na H2 Br2 S Ne Oxidation number 0

  6. Oxidation-Reduction (Redox) Reactions An atom in a monatomic ion has an oxidation number identical to its charge. Na+ +1 Ca2+ +2 Al3+ +3 Cl– –1 O2– –2

  7. Oxidation-Reduction (Redox) Reactions 1– H O H H Ca O H H • An atom in a polyatomic ion or in a molecular compound usually has the same oxidation number it would have if it were a monatomic ion. • Hydrogen can be either +1 or –1. –1 –1 +2 –2 +1 • Oxygen usually has an oxidation number of –2. H O O H –2 –1 –1 +1 +1 +1 +1

  8. Oxidation-Reduction (Redox) Reactions Cl O Cl • Halogens usually have an oxidation number of -1. 3. H Cl –1 –2 +1 +1 +1

  9. Oxidation-Reduction (Redox) Reactions The sum of the oxidation numbers is 0 for a neutral compound and is equal to the net charge for a polyatomic ion. H2SO3 2(+1) + x + 3(–2) = 0 (net charge) x = +4 –2 +1 x Cr2O72– 2(x) + 7(-2) = –2 (net charge) x = +6 –2 x

  10. Identifying Redox Reactions Reducing Agent • Causes reduction • Loses one or more electrons • Undergoes oxidation • Oxidation number of atom increases Oxidizing Agent • Causes oxidation • Gains one or more electrons • Undergoes reduction • Oxidation number of atom decreases

  11. Identifying Redox Reactions Reducing Agent oxidation 0 +3 4 Fe(s) + 3 O2(g) 2 Fe2 O3 (s) 0 –2 Oxidizing Agent reduction

  12. Identifying Redox Reactions Reducing Agent oxidation 0 +4 2 Fe2O3 (s) + 3 C (s) 4 Fe(s) + 3 C O2 (g) +3 0 Oxidizing Agent reduction

  13. In the redox reaction indicated above, which is the oxidizing agent? a) MnO4– b) Fe2+ c) H+ d) H2O e) None of the above

  14. In the redox reaction indicated above, which is the oxidizing agent? a) MnO4– b) Fe2+ c) H+ d) H2O e) None of the above

  15. Worked Example 7.9Identifying Oxidizing and Reducing Agents Assign oxidation numbers to all atoms, tell in each case which substance is undergoing oxidation and which reduction, and identify the oxidizing and reducing agents. a. b. Strategy and Solution a. The elements Ca and H2 have oxidation numbers of 0; Ca2+ is +2 and H+ is +1. Ca is oxidized, because its oxidation number increases from 0 to +2, and H+ is reduced, because its oxidation number decreases from +1 to 0. The reducing agent is the substance that gives away electrons, thereby going to a higher oxidation number, and the oxidizing agent is the substance that accepts electrons, thereby going to a lower oxidation number. In the present case, calcium is the reducing agent and H+ is the oxidizing agent.

  16. Worked Example 7.9Identifying Oxidizing and Reducing Agents Continued b. Atoms of the neutral element Cl2 have an oxidation number of 0; the monatomic ions have oxidation numbers equal to their charge: Fe2+ is oxidized because its oxidation number increases from +2 to +3, and Cl2 is reduced because its oxidation number decreases from 0 to –1. Fe2+ is the reducing agent, and Cl2 is the oxidizing agent.

  17. The Activity Series of the Elements The elements that are higher up in the table are more likely to be oxidized. Thus, any element higher in the activity series will reduce the ion of any element lower in the activity series.

  18. The Activity Series of the Elements 2 Ag(s) + Cu2+(g) Cu(s) + 2 Ag+(g) 2 Ag+(aq) + Cu(s) Cu2+(aq) + 2 Ag(s) Which one of these reactions will occur?

  19. Silver cation will oxidize copper, becoming solid silver, and liberating copper (II) ions. Looking at the diagram to the left, which one of the four reactions below will occur? a) b) c) d) e)

  20. Silver cation will oxidize copper, becoming solid silver, and liberating copper (II) ions. Looking at the diagram to the left, which one of the four reactions below will occur? a) b) c) d) e)

  21. Worked Example 7.10Predicting the Products of a Redox Reaction Predict whether the following redox reactions will occur: a. b. Strategy Look at Table 7.5 to find the relative reactivities of the elements. Solution a. Zinc is above mercury in the activity series, so this reaction will occur. b. Copper is below hydrogen in the activity series, so this reaction will not occur.

  22. Balancing Redox Reactions by the Half-Reaction Method

  23. Balancing Redox Reactions by the Half-Reaction Method I–(aq) + Cr2O72–(aq) Cr3+(aq) + IO3–(aq) Balance the following net ionic equation in acidic solution:

  24. Balancing Redox Reactions by the Half-Reaction Method I–(aq) IO3–(aq) Cr2O72–(aq) Cr3+(aq) • Write the two unbalanced half-reactions.

  25. Balancing Redox Reactions by the Half-Reaction Method I–(aq) IO3–(aq) Cr2O72–(aq) 2 Cr3+(aq) • Balance both half-reactions for all atoms except O and H.

  26. Balancing Redox Reactions by the Half-Reaction Method 3 H2O(l) + I–(aq) IO3–(aq) + 6 H+(aq) 14 H+(aq)+ Cr2O72–(aq) 2 Cr3+(aq) + 7 H2O(l) • Balance each half-reaction for O by adding H2O, and then balance for H by adding H+.

  27. Balancing Redox Reactions by the Half-Reaction Method 3 H2O(l) + I–(aq) IO3–(aq) +6 H+(aq) + 6 e– 6 e- + 14 H+(aq)+ Cr2O72–(aq) 2 Cr3+(aq) + 7 H2O(l) • Balance each half-reaction for charge by adding electrons to the side with greater positive charge.

  28. Balancing Redox Reactions by the Half-Reaction Method IO3–(aq) + 6 H+(aq) + 6 e– 3 H2O(l) + I–(aq) 6 e– +14 H+(aq) + Cr2O72–(aq) 2 Cr3+(aq) + 7 H2O(l) • Multiply each half-reaction by a factor to make the electron count the same in both half-reactions. reduction: oxidation:

  29. Balancing Redox Reactions by the Half-Reaction Method IO3–(aq) + 6 H+(aq) + 6 e– 3 H2O(l) + I–(aq) 6 e– +14 H+(aq) + Cr2O72–(aq) 2 Cr3+(aq) + 7 H2O(l) • Add the two balanced half-reactions together and cancel species that appear on both sides of the equation. reduction: oxidation: 8 H+(aq) + I–(aq) + Cr2O72–(aq) IO3–(aq) + 2 Cr3+(aq) + 4 H2O(l)

  30. Example #1 1. MnO4-(aq) + Fe2+(aq) Mn2+(aq) + Fe3+(aq) 2. H+(aq) +Cr2O72-(aq) + C2H5OH(l) Cr3+(aq) + CO2(g) +H2O(l)

  31. Balancing Redox Reactions by the Half-Reaction Method MnO4–(aq) + Br–(aq) MnO2(s) + BrO3–(aq) Balance the following net ionic equation in basic solution:

  32. Balancing Redox Reactions by the Half-Reaction Method Br–(aq) BrO3–(aq) MnO4–(aq) MnO2(s) • Write the two unbalanced half-reactions.

  33. Balancing Redox Reactions by the Half-Reaction Method Br–(aq) BrO3–(aq) MnO4–(aq) MnO2(s) • Balance both half-reactions for all atoms except O and H.

  34. Balancing Redox Reactions by the Half-Reaction Method 3 H2O(l) + Br–(aq) BrO3–(aq) + 6 H+(aq) 4 H+(aq) + MnO4–(aq) MnO2(s) + 2 H2O(l) • Balance each half-reaction for O by adding H2O, and then balance for H by adding H+.

  35. Balancing Redox Reactions by the Half-Reaction Method • Balance each half-reaction for charge by adding electrons to the side with greater positive charge. 3 H2O(l) + Br–(aq) BrO3–(aq) + 6 H+(aq) + 6 e– 3 e– + 4 H+(aq) + MnO4–(aq) MnO2(s) + 2 H2O(l)

  36. Balancing Redox Reactions by the Half-Reaction Method • Multiply each half-reaction by a factor to make the electron count the same in both half-reactions. 3 H2O(l) + Br–(aq) BrO3–(aq) + 6 H+(aq) + 6 e– 2 3 e– + 4 H+(aq) + MnO4–(aq) MnO2(s) + 2 H2O(l)

  37. Balancing Redox Reactions by the Half-Reaction Method 3 H2O(l) + Br–(aq) BrO3–(aq) + 6 H+(aq) + 6 e– 6 e– + 8 H+(aq) + 2 MnO4–(aq) 2 MnO2(s) + 4 H2O(l) • Add the two balanced half-reactions together and cancel species that appear on both sides of the equation. 2 H+(aq) + 2 MnO4–(aq) + Br–(aq) 2 MnO2(s) + H2O(l) + BrO3–(aq)

  38. Balancing Redox Reactions by the Half-Reaction Method • Since the reaction occurs in a basic solution, “neutralize” the excess H+ by adding OH– and cancel any water (if possible). 2 H2O 2 OH–(aq) + 2 H+(aq) + 2 MnO4–(aq) + Br–(aq) 2 MnO2(s) + H2O(l) + BrO3–(aq) + 2 OH–(aq) H2O(l) + 2 MnO4–(aq) + Br–(aq) 2 MnO2(s) + BrO3–(aq) + 2 OH–(aq)

  39. Worked Example Balancing an Equation for a Reaction in Base Aqueous sodium hypochlorite (NaOCl; household bleach) is a strong oxidizing agent that reacts with chromite ion [Cr(OH)4–] in basic solution to yield chromate ion (CrO42–) and chloride ion. The net ionic equation is Balance the equation using the half-reaction method. Strategy Follow the steps outlined in Figure 7.4. Solution Steps 1 and 2. The unbalanced net ionic equation shows that chromium is oxidized (from +3 to +6) and chlorine is reduced (from +1 to –1). Thus, we can write the following half-reactions: Step 3. The half-reactions are already balanced for atoms other than O and H. Figure 7.4 Using the half-reaction method to balance redox equations for reactions in acidic solution.

  40. Worked Example Balancing an Equation for a Reaction in Base Continued Step 4. Balance both half-reactions for O by adding H2O to the sides with less O, and then balance both for H by adding H+ to the sides with less H: Step 5. Balance both half-reactions for charge by adding electrons to the sides with the greater positive charge: Next, multiply the half-reactions by factors that make the electron count in each the same. The oxidation half-reaction must be multiplied by 2, and the reduction half-reaction must be multiplied by 3 to give 6 e–in both: Step 6. Add the balanced half-reactions:

  41. Worked Example Balancing an Equation for a Reaction in Base Continued Now, cancel the species that appear on both sides of the equation: Finally, since we know that the reaction takes place in basic solution, we must add 2 OH– ions to both sides of the equation to neutralize the 2 H+ ions on the right giving 2 additional H2O. The final net ionic equation, balanced for both atoms and charge, is

  42. Example #2 1. CN-(aq) + MnO4-(aq) CNO-(aq) + MnO2 (s) 2. S2-(aq) + MnO4-(aq) S(s) + MnO2 (s)

  43. Galvanic Cells Electrochemistry: The area of chemistry concerned with the interconversion of chemical and electrical energy. In a Galvanic (Voltaic) Cell: A spontaneous chemical reaction generates an electric current. In an Electrolytic Cell: An electric current drives a nonspontaneous reaction.

  44. How to draw a Galvanic Cell The oxidation reaction occurs at the anode. The reduction reaction occurs at the cathode. You will be give the unbalanced net ionic reaction or a list of the substances present (line notation). From the information given you need to decide what half reactions occur in each beaker. Draw each beaker with its substances present. The electrons leave the anode and travel to the cathode. Remember to draw a salt bridge. If the substance reduced produces a gas you will have a platinum electrode and an inverted test tube to catch the gas. Use platinum as electrode for redox involving ions; otherwise, use the metal given.

  45. Example #3 Draw the galvanic cell for the reaction: 1. Cu(s) + Ag+(aq) Ag(s) + Cu2+(aq) 2. Zn(s) + 2H+(aq) Zn2+(aq) + H2(g) 3. Pt(s) |Fe2+(aq), Fe3+(aq) |Ag+(aq), Ag(s)|

  46. Cell Potentials and Free-Energy Changes for Cell Reactions Electromotive Force (emf): The force (or electrical potential) that pushes the negatively charged electrons away from the anode (– electrode) and pulls them toward the cathode (+ electrode). It is also called the cell potential (eo) or the cell voltage.

  47. Cell Potentials and Free-Energy Changes for Cell Reactions 1 J = 1 C × 1 V joule SI unit of energy volt SI unit of electric potential coulomb Electric charge 1 coulomb is the amount of charge transferred when a current of 1 ampere flows for 1 second.

  48. Cell Potentials and Free-Energy Changes for Cell Reactions faraday or Faraday’s constant the electric charge on 1 mol of electrons 96,500 C/mol e– ∆G = –nFe ∆G° = –nFeo or free-energy change cell potential number of moles of electrons transferred in the reaction

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