Question

Refer to the Attachment !​

Answer 1

Given:

• current passed = 12 A
• current efficiency = 60%
• time taken = 60 min = 3600 sec

Solution :

As per the formula ,

$\boxed{ \frac{w}{e.wt} = \eta\frac{q}{F}} ...(1)$

here ,

• w = mass of nickel
• e.wt = equivalent weight
• q = charge
• F = Faraday constant

we know that ,

$e.wt = \frac{mol.wt}{n \: factor} ..(2)$

Ni ----> Ni ^2+. + 2e-

no of electron exchanged = n factor = 2

$q = it...(3)$

here ,

• i = current
• t = time

Substituting (2) and (3) in eq (1)

$\frac{w \times n \: factor}{mol.wt} = \eta\frac{it}{F}$

$w = \frac{ \eta \times i \times t \times mol.wt}{F \times n \: factor}$

$w = \frac{60 \times 12 \times 3600\times 59}{96500\times 2 \times 100 }$

$w = \frac{ 60 \times 12 \times 18\times 59}{96500}$

$\boxed{w = 7.923g}$