1. Electrochemistry Mr. Weldon

2. 1. Definition: Field that deals with chemical changes caused by electric current and the production of electricity by chemical reactions

3. 2. Divisions of electricity A. Conduction of current out of the reaction system 1) Current in and out via electrodes 2) electrodes: surface area on which oxidation and reduction occur 3) categories of electrodes i. Cathode: electrode where reduction occurs ii. Anode: electrode where oxidation occurs

4. B. Cell Define: Area of a contained electrochemical reacting system Types i. Electrolytic Cell: outside electrical source causes internal nonspontaneous chemical reaction. (Electrolysis) ii. Voltaic Cell: spontaneous chemical reaction produces electricity supply

5. C. Electrolysis for Chemical Production i. Electrolysis of Sodium Chloride - Solid NaCl + Heat  Melted NaCl  Positive Na ions and Negative Chlorine ions  Solid Sodium at negative electrode and Chlorine gas at positive electrode. images.tutorvista.com

6. ii. Aqueous Sodium Chloride Solid NaCl + Water  Sodium and Chlorine ions + Hydrogen and Hydroxide ions  Chlorine gas at anode and Hydrogen gas at cathode www.greener-industry.org.uk

7. D. Faraday’sLaw of Electrolysis The amount of oxidation and reduction in an electrolysis is proportional to the added electricity. – i. One faraday = the amount of electricity in one mole of electrons – ii. 1 coulomb (C) = 1 A x 1 s= The charge of 6.25 * 10^18 electrons. – iii. 1 A = 1 C/s – iiii. 1 faraday = 96, 485 C – The process is called coulometry

8. D.1 Converting Electricity to Metal Production. Ex. How much Cu will be produced by 2.50 A for 50.0 min in CuSO4? Solution – 1) current x time = C (50.0 min x 60 s/min x 2.50A) – 2)C/96 485 C/mole = moles of e- 7.50 x 10^3 C/96485 c/mole =.078 moles of e- 3) Moles of e-/ # of e- per atom of metal = moles of metal.078 moles of e-/ 2 e-/atoms of metal =.039 moles Cu 4) Moles of Cu x Atomic Mass = Mass of Cu.039 moles Cu x 63.5 g/mole = 2.48 g Cu

9. F. Voltaic Cells (Galvanic Cells) A redox reaction separated into parts and connected by electrical pathways. (both chemical and metallic)

10. G. Calculating Standard Voltage of a Cell Calculating the voltage of a cell is just like calculating Heat of reaction Standard Electrode Potential – Standard Hydrogen Electrode – H2  2H+ + 2e- – 2H+ + 2e-  H2 0 V -By comparison all other half reactions receive a Standard Potential when compared to the SHE. Tables of Standard Potentials are then created. (See pg. 810)

11. G.1 Calculating Standard Voltage of a Cell Ex. A cell uses CuSO4 and ZnSO4 The SO4 is a spectator ion and therefore ignored. The two half reactions are Zn + 2e-  Zn -.763V Cu + 2e-  Cu +.337 V Since the Cu has the greater pull on electrons we reverse the Zn Zn  Zn + 2e- +.763V Cu + 2e-  Cu +.337V Cu+2e + Zn  Zn+2 e + Cu +1.100V will not happen spontaneously.

12. H. Calculating Non-Standard Voltage- The Nernst Equation E = E – (2.303RT/nF) log Q Reaction : xOx + ne  y Red then Q = [Red]^y [Ox]^x F = 96 485 J/V*mol e-