(may be broke/outdated!)
Copper can be electroplated at the cathode of an electrolysis cell by the following half-reaction.
Cu2+ + e- —> Cu(s)
How much time would it take for 229 mg of copper to be plated at a current of 8.8 A?
can someone tell explain how to do this problem?
how did u get the answer?
thanks
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2 Responses
Convert mg –> g to find the number of moles. Once you have done that multiply by (2 mol e/1 mol Cu) x (1C/96,5000 e) x (1 C/1 sec x 1 amp) x 8.8 amp
This will give you the answer in seconds.
1st convert the 229mg to grams.That’s .229grams. multiply that by the molar mass of Cu, which is about 63.546g/mol. You get .00360mol of Cu. now you multiply that by the mol e- ratio for this equation. That is 2 mol of e- per 1 mol of Cu. This leaves you with .007207 mol e-. We then use Faraday’s constant, 96500coulombs. We multiply the .007207mol of e- by 96500Coulombs/1mole of e- to give us 695.5119 Coulombs.Then you use the formula Amps*time=Coulombs. You manipulate it to be Time=Coulombs/Amps. So, 695.51C/8.8Amps=79.035 seconds.
Note: time must be in seconds.