Answer:
Step-by-step 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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