A solution of Ni(NO3)2 is electrolysed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode?

Q.15: A solution of Ni(NO3)2 is electrolysed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode?

Solution:

Given that ,

Current, I = 5A

Time, t = 20 × 60 = 1200 s (1 min = 60 second)

Charge = I × t

            = 5 × 1200

  = 6000 Coulombs

Charge on Ni

No₃  has –1 charge always so that

Ni +2(NO₃) =0

Ni + 2(–1)  = 0

Ni             = +2

Charge required to deposited 1 mol of Ni = nF

                                              = 2 × 96487 Coulombs

                                              = 192974 Coulombs

1 mol of Ni = 58.7 g (use periodic table to get this value)

192974 Coulombs will generate = 58.7 g of Ni

6000 Coulombs will generate     = 58.7 g × 6000 Coulombs /192974 Coulombs

                                               = 1.825 g

Hence, 1.825 g of nickel will be deposited at the cathode.