if you wanted to plate out solid Na metal in an electrolytic cell (anhydrous con

Question

if you wanted to plate out solid Na metal in an electrolytic cell (anhydrous conditions6, how much Na would be played out if you applied a current of 8.00 Amperes for 584 sec? if you wanted to plate out solid Na metal in an electrolytic cell (anhydrous conditions6, how much Na would be played out if you applied a current of 8.00 Amperes for 584 sec? if you wanted to plate out solid Na metal in an electrolytic cell (anhydrous conditions6, how much Na would be played out if you applied a current of 8.00 Amperes for 584 sec?

Explanation / Answer

The sodium will be plate out as

Na+   + e --> Na(s)

So for every Faraday of electricity used up will plate out 1 moles of sodium

The charge used up = current X time = 8X 584 = 4672 coulombs

96500 coloumbs = 1 Faraday
4672 coulombs / 96500 coulombs/faraday= 0.048 faradays

So from 1 Faraday 1 mole of sodium

So from 0.048 Faraday = 1 X 0.048 moles = 0.048 moles of sodium

1 mole of sodium = 23 grams

so 0.048 moles = 23 X 0.048 = 1.104 grams

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