0.1 Faraday of electricity was passed through a solution of copper (ll) sulphate. The maximum weight of copper deposited on the cathode would be [Cu = 64]

Kejieusual

1 year ago

1mole of Cu contains 64g
CuSO4 produced Cu2+and SO4^2-
1F is driven by 1e-(electron)
Cu losses two e-
i.e 2F decompose 64/2 Cu =32g
1F produced 32g of Cu
0.1F produced x
x=0.1 *32
x=3.2

bless. 🙏🙏🙏

civilianmf

1 year ago

Guys am I the only one who got 6.4g and I'm sure that I'm correct
Analyzing
According to the definition of Faraday
Faraday is the minimum quantity of electricity required to liberate one mole of electrons from a substance in it aqueous or molten state
In short
1F = 1 mole of electrons =1 mole of substance = 96500 coloumb
And from the question
64= RMM of copper
Then
1 mole of copper = 2mole of electrons=2F
x mole of copper= 0.1 mole of electrons=0.1F
Then
x= 0.1F
2=2F
x = 0.1 ×2/2
x= 0.1
x = number of moles of copper
From the formula
no of moles= mass /RMM
Hence
0.1 = mass /64
Mass = 6.4g

darkypearl

11 years ago

ans shud b 6.4

Kenhinde

4 years ago

0.1=64
2F=. X
Cross multiply
Ans:3.2g