140.
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?
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.
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