State Faraday's first law of electrolysis . How much charge in terms of Faraday is required for the reduction of `1mol of `Cu^(2+)` to Cu.
(b) Calculate emf of the following cell at `298 K : Mg(s) | Mg ^(2+)(0.1 M)||Cu^(2+) (0.01)|Cu(s)`
[Given `E_(cell)^(@)=+ 2.71V, 1 F = 96500C mol^(-1)]`

Explanation

(a) Faraday's first law of electrolysis states that ''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 electrolytic solution or mell''
The reduction of one mol of CU2 + to Cu can be represented as : `Cu^(2+) + 2e to 2Cu`
Since, in the reaction there are two moles of electrons involved, so the amount of charge required is 2F.
(b) The cell reaction can be represented as `: Mg(s) + Cu^(2+) (aq) to Mg^(2+) (aq) + Cu(s)`
Given, `E_(cell)^(@)=+ 2.71V, T = 298 K` According to the Narsnt equation :
`E_(cell)^(@)=-(0.0591)/(2) "log"(0.1)/(0.01)`
` = 2.71 - (0.059)/(2) log 10 = 2 .68 V. `