j4yman Posted May 25, 2006 Posted May 25, 2006 Consider the following reduction potentials: Cu2+ + 2e- --> Cu E° = +0.34 V Pb2+ + 2e- --> Pb E° = -0.13 For a galvanic cell employing the Cu, Cu2+, Pb, Pb2+ couples, calculate the maximum amount of work (in kJ) that would accompany the reaction of one mole of lead under standard conditions. Hmmmmm help plz
Tartaglia Posted May 28, 2006 Posted May 28, 2006 V = High - low = 0.34-(-.13) = 0.47 V Faradays used = 2 as Pb is oxidised to dication E = V*Q = 0.47 * 2 *96500 J
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