Consider a cell composed of the following half cells:
  1. Mg(s) | Mg2+ (aq)  
  2.  Ag(s)  | Ag​+ (aq)
The emf of the cell is 2.96 V at [Mg2+] = 0.130 M and [Ag+] = 1.0 * 10-4 M. Calculate the standard emg of the cell ( R = 8.3 JK-1 mol-1  ,  F = 96500 C ) 
I am asking this question 2nd time soo plzz give me the answer this time.

Dear Student,


The standard EMF of the cell has to be calculated

Mg (s) / Mg2+ (aq)  Ag (s) / Ag+ (aq)


The reactions would be:

Mg (s) ---> Mg2+ (aq) + 2e- ------ (-2.37)

Ag+ (aq) + e- ----> Ag (s) -----(+0.8)


Mg (s) + 2Ag+ (aq)  Mg2+ (aq) + 2Ag (s)


E ocell  = 2.96 V


According to Nernst Equation:

Ecell  =  E ocell 0.05912 log kc

Because, 2 electrons are transferring, therefore, n = 2

Placing the values we get, 

E cell  = 2.96 - 0.05912 log [Mg2+] [Ag (s)][Ag+] [Mg (s)]

         =  2.96 - 0.05912 log (0.130 M (1 x 10-4) 2M)

        = 2.96 - 0.05912 log 0.1301 x 10-8

         = 2.96 - 0.05912 log (13000000) = 2.96 - 0.05912 (-7.11)

         =  2.96 + 0.210 = 3.17 V


[Another Approach:

Eocell = Eoright - Eoleft

Eo Mg2+/ Mg = -2.36 V and Eo Ag+/Ag = +0.80

So, Eocell = (-2.36 - 0.80) V = +3.17 V]



Regards

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