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Unit 28 Electrochemistry (Chapter 8)

Unit 28 Electrochemistry (Chapter 8). Electrochemistry as Applied to Batteries (8.3) Corrosion (8.4) Important Oxidizing Agents (8.6 – 8.7) Important Reducing Agents (8.8 – 8.9). Electrochemistry (8.3).

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Unit 28 Electrochemistry (Chapter 8)

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  1. Unit 28Electrochemistry (Chapter 8) Electrochemistry as Applied to Batteries (8.3) Corrosion (8.4) Important Oxidizing Agents (8.6 – 8.7) Important Reducing Agents (8.8 – 8.9)

  2. Electrochemistry (8.3) Electricity is due to the motion of electrons. Since oxidation-reduction reactions involve an exchange of electrons, such reactions can be used to generate electricity through applications such as batteries. Image from http://www.vistatutor.com

  3. Electrochemistry Cont. (8.3) In the reaction from the previous page, thecopper goes from being in the elemental form todissolving in solution as Cu2+ ions which causethe blue color in solution. The Ag+ ions previously dissolved in solutionbecome elemental silver and form the “hangings” seen in the second beaker. In equation form: Cu (s) → Cu2+ (aq) + 2 e- Ag+ (aq) + e- → Ag (s) An important aspect in understanding electrochemistry is to understand how these two equations may be combined to form the overall reaction.

  4. Electrochemistry Cont. (8.3) Each of the reactions below is called a half-reaction. One represents an oxidation and the other a reduction. Cu (s) → Cu2+ (aq) + 2 e- Ag+ (aq) + e- → Ag (s) Since the electrons donated by the copper are the ones accepted by the silver, the number of electrons being accepted and donated must match. In order for the electrons to balance, each half-reaction is multiplied by an integer as necessary to ensure that the number of electrons donated matches those accepted. In this example, the Cu equation involves two electrons and the Ag equation involves only one so multiplying the Ag equation by 2 will give two electrons accepted to go along with the two electrons donated by copper (continued on next page).

  5. Electrochemistry Cont. (8.3) Multiplying the first equation by “1” and the second by “2” gives: 1 × (Cu (s) → Cu2+ (aq) + 2 e- ) 2 × (Ag+ (aq) + e- → Ag (s) ) These simplify to: Cu (s) → Cu2+ (aq) + 2 e- 2 Ag+ (aq) + 2 e- → 2 Ag (s) Adding these two equations gives: Cu (s) + 2 Ag+ (aq) → Cu2+ (aq) + 2 Ag (s)

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