How many g of Ni (s) will accumulate on an electrode if a 9.5 A current is run t

Question

How many g of Ni (s)   will   accumulate   on   an   electrode   if   a   9.5   A   current   is   run  
through   an   electrolytic   cell   for   25   minutes?      

Sorry for the strange formatting of the question, i just copied and pasted it straight on to here. I can't seem to figure this out, even though it's supposedly easy. Like, there's no charge on the Nickel so how do you know how many electrons are transfered and stuff? A detailed explanation on how to get the answer would be very awesome! :) the answer is 4.34g, and i'm seriously lost on how to get that.

Explanation / Answer

Ni2+ + 2 e- => Ni

Charge = current x time in s

= 9.5 x 25 x 60 = 14250 C

Moles of electrons = charge/Faraday constant

= 14250/96485 = 0.148 mol

Moles of Ni = 1/2 x moles of electrons

= 1/2 x 0.148 = 0.074 mol

Mass of Ni = moles x molar mass of Ni

= 0.074 x 58.7

= 4.34 g

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