Question

how many minutes t would take to reduce 0.20mol of Cu2+ to Cu using a current of 10A?

Answer 1

Answer:

It would take 64.33 minutes to reduce from 0.20 mol of Cu2+ to CU.

Explanation:

Current, I = 10 A

Since Cu2+ has 2 moles of electrons  

Cu2+ has 0.20 mol …. [given data]

Therefore, every mole of Cu will require = 2 * 0.20 = 0.40 mole of electrons

We know that 1 mole of electrons require 96500 C of charge, to flow.

Now, the charge required for the flow of 0.40 mol of electrons, is given by

Q = 96500 * 0.40 = 38600 C

Hence,  

Q(in Coulombs) = I (in Ampere) * t (seconds)

Or, 38600 = 10 * t

Or, t = 3860 seconds = 3860/60 = 64.33 minutes

Answer 2

Answer:

I hope it help you.......