i) How many hours does it take to reduce 3 mol of Fe⁺⁺⁺to Fe⁺⁺with 2.0 A current? ( F= 96500 C)
ii) State faraday laws of electrolysis.

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Solution

1) The electrode reaction will be :
Fe³⁺+ e^-→ Fe²⁺
The reduction of 1 mol of Fe³⁺ to Fe²⁺requires 1 F of electricity
1 F = 96500 C mol^-1
For reduction of above reaction , for 3 moles we require 3 F .ie. 3 $\times$ 96500 C of electricity.
Given current (I) = 2 A
Time (t) = $\frac{Charge (Q)}{Current (I)}$ = $\frac{3×96500}{2}$ = 144750 s (40.2 hours)
2) Faraday's law of electrolysis :
a) First law : The amount of chemical reaction which occurs at any electrode during electrolysis by a current is proportional to the quantity of electricity passed through the electrolyte (solution or melt).
b) Second law : The amounts of different substances liberated by the same quantity of electricity passing through the electrolytic solution are proportional to their chemical equivalent weights ( Atomic weights of metal $\div$ Number of electrons required to reduce the cation).

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