The Non-Standard Half-Cell
Up until now all the Cell Potentials (Eo) were measured at SATP and all the concentrations of the electrolytic solutions have been 1 M.  If the concentrations of the electrolytes change for any reason we must use the Nernst equation to determine the new voltage.  In the standard cell the voltage depended upon the electronegativity difference of the metals involved because the ions were keep equal to each other.  When the concentration of the ions is different the voltage is slightly difference.
 
The Nernst Equation:     For any reaction  a A  +  b B  <====>  c C  + d D

Ecell = Eo - (0.059 / n)  *  log ([C]c[D]d / [A]a[B]b)

Ecell = the Voltage potential of the non-standard cell.
Eo = the voltage potential of the cell at standard conditions
n = the # of electrons exchanged
[C]c = concentrations of the ions, not the metals
 

eg. #1: Calculate the cell voltage of:        Zn |  Zn+2  (0.001 M) || Ag+1 (0.1 M) | Ag

First find the Eo value for the standard cell.

Eo = +0.80 V    Ag+1(aq)  + e-1  --->  Ag(s)  reduction
Eo = -0.76 V    Zn+2(aq)  + 2e-1  ---> Zn(s)  oxidation
 

Eo = +0.80 V                  Ag+1(aq)  + e-1  --->  Ag(s)     reduction
Eo = +0.76 V                  Zn(s)  ---> Zn+2(aq)  + 2e-1     oxidation
_________________________________
Eo = +1.56 V        2 Ag+1(aq) + Zn(s) ---- > 2Ag(s)  + Zn+2(aq)
 
The Zn+2 is the product ion and Ag+1 is the reactant ion therefore the Nernst equation looks like:
 
Ecell = Eo - (0.059 / n)  *  log ([C]c[D]d / [A]a[B]b)
Ecell = Eo - (0.059 / n)  *  log ([Zn+2]1 / [Ag+1]2)
Ecell = +1.56 V - (0.059 / 2)  *  log ([0.001]1 / [0.1]2)
        = +1.56 V -0.03 * log (0.001/ 0.01)
        = +1.56 V -0.03 * log(0.1)
        = +1.56 V -0.03*(-1)
        = +1.56 V + 0.03
        = +1.59 V
 
Ex #2:     Zno  |Zn+2 (0.10 M) || Cu+2 (0.001 M) | Cuo

             Cu+2  +  2e-1  --->  Cuo     Eo = +0.34 V
             Zn+2  +  2e-1  --->  Zno      Eo = -0.76 V

Therefore
             Cu+2  +  2e-1  --->  Cuo     Eo = +0.34 V
              Zno  ---->   Zn+2  +  2e-1   Eo = +0.76 V
             ________________________________
             Zno  +  Cu+2  --->  Zn+2  +  Cuo     Eo = +1.10 V
 

Ecell = Eo - (0.059 / n)  *  log ([C]c[D]d / [A]a[B]b)
         = 1.10 V  -  (0.059/2) log [Zn+2] / [Cu+2]
         = 1.10 V  -  (0.059/2) log [0.1] / [0.01]
         = 1.10 V  -  0.03 log (1)
         = 1.07 V
 
Ex #3:     Coo |  Co+2 (0.1 M) || Ni+2 (0.001 M) | Nio

    Ni+2  +  2e-1   ---->  Nio      Eo = -0.25 V
    Co+2  +  2e-1  ---->  Coo     Eo = -0.28 V

Therefore
    Ni+2  +  2e-1   ---->  Nio      Eo = -0.25 V
    Coo   ----> Co+2  +  2e-1     Eo = +0.28 V
  _______________________________
    Ni+2  +  Coo    ---->  Nio  +  Co+2     Eo = +0.03 V
 

Ecell = Eo - (0.059 / n)  *  log ([C]c[D]d / [A]a[B]b)
         = 0.03 V  -  (0.059/2) log [Co+2] / [Ni+2]
         = 0.03 V  -  (0.059/2) log [0.1] / [0.001]
         = 0.03 V  -  0.03 log (2)
         = -0.029 V
 
In which direction does the reaction of      Ni+2  +  Coo    ---->  Nio  +  Co+2   with the above concentrations proceed spontaneously?  ( To the left because the voltage is negative)
 
At standard conditions the voltage is +0.03 V.  Which direction should it proceed?
(To the right)