How many Faraday of electricity are required to deposit 0.20 mole of nickel, if 0.10 Faraday of electricity deposited 2.98g of nickel during electrolysis of its aqueous solution?

(Ni = 58.7, 1F = 96 500 C mol^-)

Modupemii

4 years ago

n=m/mm
n=2.98/58.7
=0.05
Then
0.05m deposited 0.10faraday of electricity
0.05m----------0.10f
0.20m---------- X
X=0.10×0.20/0.05
=0.4f

olyezema

4 years ago

n=m/M
m=0.2×58.7=11.74g
96500x --------- 11.74g
96500×0.1 -------- 2.98g
x=0.39~ 0.4F

Kevin_Aro

3 years ago

First, we can calculate the amount of nickel deposited by 0.10 Faraday of electricity:
m = 2.98 g
n = m/M = 2.98 g / 58.7 g/mol = 0.0507 mol
1F = 96,500 C/mol

Now, we can use the ratio of Faraday to moles to calculate the amount of electricity needed to deposit 0.20 moles of nickel:
0.10 F deposits 0.0507 mol Ni
x F deposits 0.20 mol Ni
x = (0.20 mol Ni)(0.10 F / 0.0507 mol Ni) = 0.393 F

Therefore, 0.393 Faraday of electricity are required to deposit 0.20 mole of nickel.

Macprince00

3 years ago

which topic can i see this?

millionaire001

1 year ago

video lesson attached to correction