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Copper is deposited when a current of 3.00 A flows for 485 seconds through a cell composed of two Pt electrodes in a solution of Cu(NO3)2 in dilute HNO3. If 0.36 g of copper is obtained calculate the current efficiency for depositing copper.

Respuesta :

Answer:

Explanation:

Cu(NO3)2- Cu2+ + NO3 2-

Cu2+ + 2e- Cu(s)

63.5gCu= 2× 96500 (1F= 96500)

1g= 2×96500/63. 5

0. 36g Cu needs=>2×96500/63.5×0. 36

1094C

So, 1094C of charge needed to deposit 0. 36g of Cu at cathode

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