Q -Four faradays of electricity were passed through AgNO3(1), CdS04(), AlCl3(e) and PbCl() kept in four
vessels using inert electrodes. The ratio of moles of Ag, Cd, Al and Pb deposited will be
1) 12:4:6:3
(2) 1:2:3:4
3) 12:6:4:3
(4) 4:3:2:1
Answers
ANSWER :
12 : 6 : 4 : 3
Explanation:
Ag+ +(aq) + e -> Ag(s)
1 mol 1F 1 mol
4F 4 mol
Cd2+ (aq) + 2e -> 2Cd(s)
2 mol 2F 1 mol
4F 4/2 mol =2 mol
Al3+ (aq) +3e -> 3Al(s)
3 mol 3F 1 mol
4F 4/3 mol
Pb4+ (aq) + 4e ->Pb(s)
4 mol 4F 1 mol
Therefore, ratio 4 : 2 : 4/3 : 1
= 4×3 : 2×3 : 4 : 1×3
= 12 : 6 : 4 : 3
The ratio would be 12:6:4:3.
1) Four Faradays of electricity would be equivalent to 4 moles of electron.
We can calculate the number of electrons it takes to convert one ion of the given metal into neutral atom and then find the moles that can be converted by four moles of electrons:
2) Ag+ + e- ---> Ag, one electron is needed per ion.
Cd2+ + 2e- -----> Cd, two electrons
Al3+ + 3e- -----> Al, three electrons
Pb4+ + 4e- ------> Pb, Four electrons
3) Hence the moles that can be prepared by four moles of electrons would be :
Ag: 4 moles
Cd: 2 moles
Al : 4/3 moles
Pb : 1 moles
4) The ratio would be 4:2:4/3:1 or 12:6:4:3