Question

Amount of copper deposit on the cathode of an electrolytic cell containing copper sulphate solution by the passage of 2 amperes for 30 minut -( at mass of Cu- 63.5)e​

Answer 1

Answer:  The amount of copper deposited is 1.18 grams.

Explanation:

$Q=I\times t$

where Q= quantity of electricity in coloumbs

I = current in amperes = 2A

t= time in seconds =$30\times 60=1800s$

$Q=2A\times 1800s=3600C$

$CuSO_4\rightarrow Cu^{2+}+SO_4^{2-}$

$Cu^{2+}+2e^{-1}\rightarrow Cu$

$96500\times 2=193000Coloumb$ of electricity deposits 1 mole of  Cu

193000 C of electricity deposits 63.5 g of Cu.

3600 C of electricity deposits =$\frac{63.5}{193000}\times 3600=1.18g$ of Cu

Thus the amount of copper deposited is 1.18 grams