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A metal undergoes electrolytic reduction according to ne- + Mn+ ===> M. What current (in Ampères) must be provided to deposit the metal at a rate of 0.583 mol/hr if n = 2?

2007-11-19 01:31:33 · 1 answers · asked by kamer37078 1 in Science & Mathematics Chemistry

1 answers

1 ampere = 1 coulomb/second

96,500 coulombs = 1 faraday = electricity needed to reduce 1 eq wt of substrate

0.583mol/1hr x 2eqwt/1mol x 96,500coul/1eqwt x 1hr/60min x 1min/60sec = 31.2coul/sec = 31.2 amperes

2007-11-19 01:46:48 · answer #1 · answered by steve_geo1 7 · 1 0

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