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

Given -


Current = 5A


Time - 20 minutes


Mass of Ni deposited = ?

Quantity of electricity passed = 5 A × (20 × 60 sec)


= 6000 C ⇒ Equation 1


The electrode reaction is written as,


Ni2 + + 2e → Ni


Thus, the quantity of electricity required = 2F


= 2×96487 C


= 192974 C


∵ 192974 C of electricity deposits 1 mole of Ni, which is 58.7 g ⇒ Equation 2


Thus, equating equations 1 and 2, we get,


192974 C of electricity deposits = 58.7 g


6000 C of electricity will deposit =


= 1.825g of Ni


The mass of Ni deposited at the cathode is 1.825g of Ni


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