j4yman Posted May 25, 2006 Share 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 Link to comment Share on other sites More sharing options...
Tartaglia Posted May 28, 2006 Share 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 Link to comment Share on other sites More sharing options...
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