Solution:
SOLUTION
AgNO3 + e−?Ag+NO3−
1 mole of AgNO3 required 1 mole of e− or 1 Faraday of electricity.
Thus 3F is used for 3 mole of AgNO3
CuSO4 + 2e− ? Cu+SO42−
1 mole of CuSO4 required 2 Faraday of charge.
Thus 33 Faraday of charge will deposit / 32 mole of CuSO4
Al(MO3)3 + 3e− + 3e− ? Al + 3NO− 3Al+3NO3−
3F of charge is required for 1 mole of Al(NO3)3
NaCl + e− ? Na + Cl−
1F of charge required for 1 mole of NaCl .
Thus, 3F of charge is required for 3 mole of NaCl
The molar ratio will be: 6:3:2:6