What mass of copper will be deposited by the liberation of Cu²⁺ when 0.1F of electricity flows through an aqueous solution of a copper (ll) salt? [Cu = 64]

DesmondEmmanuel

5 years ago

Cu + 2es ---- Cu^2+
2F------->64g
0.1F------> x
:. x= (64x0.1)/2
x=3.2g

Alisha101

2 years ago

The formula m=ArIt/vf
It/Q=0.1*96500=9650
m=64*9660/2*96500
m=3.2g

Kevin_Aro

3 years ago

The number of moles of electrons (n) in 0.1 F of electricity is given by the formula n = Q/F, where Q is the quantity of electricity in coulombs and F is Faraday’s constant (96485 C/mol). Plugging in the values gives us n = 0.1 F / 96485 C/mol = 1.036 x 10^-6 mol.

Since the copper (II) ion (Cu2+) has a charge of +2, it takes two moles of electrons to reduce one mole of Cu2+ to Cu. Therefore, the number of moles of Cu deposited is half the number of moles of electrons: 1.036 x 10^-6 mol / 2 = 5.18 x 10^-7 mol.

The molar mass of copper is given as 64 g/mol, so the mass of copper deposited is (5.18 x 10^-7 mol) * (64 g/mol) = 0.033 g, which is closest to E. 3.2g.