Oxidation and reduction half-reactions allow us to establish mole ratios for ions, elements, and electrons. Two moles of electrons are required to reduce.

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Presentation transcript:

Oxidation and reduction half-reactions allow us to establish mole ratios for ions, elements, and electrons. Two moles of electrons are required to reduce one mole of zinc ions, producing one mole of zinc metal. e.g. Zn2+(aq) + 2 e– → Zn(s) Q = charge (C) The unit of electrical charge is the coulomb (C). Q = It I = current (A) The unit of electrical current is the ampere (A). t = time (s) One coulomb (C) is the quantity of charge transferred by a current of one ampere (A) during a time of one second.

Faraday’s Law Michael Faraday (1791 – 1867) In an electrolytic cell, the mass of an element produced or consumed at an electrode is directly proportional to the time the cell operates at a constant current. The molar charge of electrons, or the Faraday constant, F: Michael Faraday (1791 – 1867) (data book pg. 3)

Electrical Units and Quantitative Problems Involving Electrolysis Unit of electrical charge = coulomb (C). One electron has a charge of 1.60 x 10 –19 C. One mole of electrons has a charge of 9.65 x 10 4 C. Molar charge = 9.65 x 10 4 (Faraday constant) I = current in amps (A) t = time (s) = # moles of electrons F = constant of 9.65 x 104 C/mol

Half-Cell Calculations Separate calculations are carried out for each electrode, although the same charge, and, therefore, the same amount of electrons passes through each electrode in a cell or a group of cells in series.

Homework: Read pgs. 652 – 656 pgs. 653, 654 Practice #’s 1 – 7 pg. 657 Section 14.4 Questions #’s 1 – 11