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,

Ni^{2 +} + 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