Question

 

 

Calculate e.m.f. and ∆G for the following cell:

Mg (s) |Mg²⁺ (0.001M) || Cu²⁺ (0.0001M) | Cu (s)

`"Given :" E_((Mg^(2+)"/"Mg))^0=−2.37 V, E_((Cu^(2+)"/"Cu))^0=+0.34 V.`

 

 

 

Answer 1

 

 

For the given cell representation, the cell reaction will be

Mg(s) + Cu²⁺(0.0001 M) → Mg²⁺(0.001 M) + Cu(s)

The standard emf of the cell will be given by

 

         = 0.34 - (-2.37)

         = 2.71 V

The Nernst equation for the cell reaction at 25 ºC will be

`E_(Cell)=E_(Cell)^@-(0.059/n)"log"([Mg^(2+)])/([Cu^(2+)])`

        `=2.71-0.059/2"log"0.001/0.0001`

        `=2.71-0.02955(log10)`

        `=2.71-0.02955(1)`

        `=2.68045V~~2.68V`

 We know

∆G = −nFE_cell

     =−2×96500×2.68

     =−517240 J mol⁻¹

     =−517.24 kJ mol⁻¹