Question

A steady current of 2 amperes was passed through two electrolytic cells x and y connected in series containing electrolytes feso4 and znso4 until 2.8g of fe deposited at the cathode of cell x. How long did the current flow. Calculate the mass of zn deposited at the cathode of cell y

Answer 1

Answer:

 

Explanation:

Given I = 2 A

Now weight of iron deposited at cathode of cell X = 2.8 g

n = 2

So weight of iron deposited W = Atomic mass x I x t / 96500 x n

    2.8 = 56 x 2 x t / 96500 x 2

So t = 2 x 96500 x 2.8 / 56 x 2

So t = 4825 secs

Now weight of zinc deposited W = 65.3 x 2 x 4825 / 96500 x 2

So W = 3.26 g