Question

calculate the time to deposit 1.17g of Ni at cathode when a current of 5A is passed through the solution Ni(NO3)2. (molar mass of Ni=58.5g/mol , 1F=96500C/mol)

Answer 1

mass of nickel , m = 1.17g

atomic mass of nickel , M = 58.5 g/mol

so, number of mole of Ni = 1.17/58.5 = 1/50 = 0.02

so, number of electron passed = 2 × 0.2 = 0.4 [ as nickel is diatomic ]

we know, 1mole of electrons is equivalent to 96,500C Chagres.

means, 96500C charges need to deposit 1 mole of electrons,

so, amount of charge required to deposit 0.4 mole of electron = 0.4 × 96500 = 3860C

now, current is given , I = 5A

so, time required to deposit 1.17g of nickel = charge/current = 3860/5

= 772 sec or 12 min 52 second.