How many moles of electrons are required for the reduction of (i) 3 moles of `Zn^(2+) ` to Zn (ii) 1 mol of `Cr^(3+) ` to Cr?
How many Faradays of electricity will be required in each case ?

Answer

(i) 6 mol electrons and 6 Faradays
(ii) 3 mol electrons and 3 Faradays

Explanation

Given : for reduction of 3 mol `Zn^(3+) ` to Zn
Number of moles of electrons required = ?
Reduction half reaction .
`Zn^(2+) + 2e^(-) to Zn`
`:. 1 ` mole of `Zn^(2+) ` requires 2 moles of electrons
`:. `3 moles of `Zn^(2+) ` will requires `3xx2 = 6` moles of electrons
`:' 1` mole of electrons = 1 F
`:. 6` mol of electrons = 6 F
(ii) Given : Reduction of 1 mol of `Cr^(3+) ` to Cr
Reduction half reaction
`Cr^(3+) + 3e^(-) " to " Cr`

Hence 1 mole of `Cr^(3+)` will require 3 mole of electrons
`:' 1` mole of electrons `=1 F " " :. 3 ` Mole of electrons = 3 F `