Question

in a solution of cuso4 how much time will be required to precipitate 2g of cu by 0.5 ampere current​

Answer 1

Answer:

Explanation:

We know that m=eit/96500

m=2g

i=0.5amperes

Equivalent weight of copper=127/2=63.5

Now, t=2×96500/63.5×0.5

Solve it to get time required..

Hope u will understand....

Answer 2

The time required is 12157.48 seconds

Given,

Solution contains CuSO4

Current(I) = 0.5

Weight of Cu  precipitated = 2g

To Find,

Time required to precipitate 2g of Cu by 0.5 amperes current​ =?

Solution,

The reaction that takes place is,

$Cu^{2+} + 2e^{-} = Cu$

It means 2 moles  of electrons are required for 1 mole of Cu

The atomic mass of Cu = 63.5 g

To deposit 63.5g of Cu = 2 faraday is required

To deposit 1 g of Cu = 2 / 63.5 faraday of charge

To deposit 2 g of Cu = 2 x (2 / 63.5) faraday of charge

We know that 1 faraday = 96500 c

To deposit 2 g of Cu = 2 x (2 / 63.5) x 96500

To deposit 2 g of Cu = 6078.74 C

Charge required(Q) = 6078.74 C

From the formula of current, we get

I = Q / t

Q = I x t

6078.74 C = 0.5 A x t

t = 6078.74 / 0.5

t = 12157.48 seconds

Hence, the time required to precipitate 2g of Cu by 0.5 A current​ is 12157.48 seconds.

#SPJ6