Question

In an aqueous solution of CuSO4 2 ampere electric current is passed for 10 minutes. Calculate the amount of copper deposited in electrode (EQ.wt of Cu = 31.75 gm)

Answer 1

Answer:

i am tried next timean aqueous solution of CuSO4 2 ampere electric current is passed for 10 minutes.

Answer 2

0.197 g of Copper gets deposited.

Explanation:

Electric current, I = 2 A

Time for which the current is passed, $t = 10 minutes = 10\times 60 = 600 seconds$

So, charge, $Q = I\times t = 2 \times 600 = 1200 C/mol$

Now, In the given question, the change taking place is

$Cu^{2+}(aq) + 2e \rightarrow Cu(s)$

Thus, for one Cu, there is 2 e change taking place

or For 1 mol = 2 F electricity  = 31.75 g of Cu getting deposited

or $2\times96500 C = 31.75 g$

Thus, in $1 C = \frac{31.75}{2\times96500}$

So, in $1200 C =\frac{31.75}{2\times96500}\times 600$ $= 0.197 g$

Thus, 0.197 g of Copper gets deposited.