The Nernst Equation

As a voltaic cell is discharged, its emf falls until E = 0, at which point we say that the cell is dead.  Studies show that the emf depends on the concentrations of the reactants and products in the cell reaction.  Increasing the concentrations of the reactants will increase the cell emf, while increasing the concentrations of products will decrease the cell emf.  The emf generated under nonstandard conditions can be calculated by using an equation first derived by Walther Hermann Nernst (1864 - 1941).

The dependence of the cell emf on concentration can be obtained from the dependence of the free-energy change on concentration.  Recall that the free-energy change, ΔG, is related to the standard free-energy change, ΔGo, by the equation:

ΔG  = ΔGo + RT ln Q  

The quantity Q is the reaction quotient, which has the form of the equilibrium constant expression except the concentrations are those that exist before the system reaches equilibrium.  Substituting  ΔG  = -nFEcell into this equation gives:

-nFEcell = -nFEocell + RT ln Q  

Solving this equation for E gives the Nernst equation:

                 RT
Ecell = Eocell - ---- ln Q
                 nF

The equation is usually written in term of common logarithms:

                 2.303RT
Ecell = Eocell - -------- log Q
                   nF

At 298 K the quantity 2.303RT / F equals 0.0592 V-mol, so at 298 K we can write the following simplified form of the Nernst equation.

                 0.0592
Ecell = Eocell - -------- log Q
                   n

Example 1 

Calculate the emf of the following cell at 25°C;   Sn(s)|Sn2+(0.025 M)||Ag+(2.0 M)|Ag(s

E°cell = E°cathode - E°anode

E°cell = E°Ag - E°Sn = 0.80 V - (-0.14 V) = 0.94 V

                 0.0592                    0.0592      [0.025]
Ecell = Eocell - -------- log Q = 0.94 V - -------- log ------- = 1.01 V
                   n                          2         [2.0]2

Example 1 

The emf of the following cell at 25°C is 1.04 V, determine the value of [Sn2+];   Sn(s)|Sn2+(? M)||Ag+(1.00 M)|Ag(s

E°cell = E°cathode - E°anode

E°cell = E°Ag - E°Sn = 0.80 V - (-0.14 V) = 0.94 V

                          0.0592                    0.0592      [Sn2+]
Ecell = 1.04 V = Eocell - -------- log Q = 0.94 V - -------- log ------- 
                            n                          2        [1.00]2

 

              0.10           
log [Sn2+] = -------- = -3.38 
             -0.0296         

[Sn2+] = antilog(-3.38) = 4.2 x 10-4 M        (antilog is inverse log on your calculator)

 

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