Nernst Equation

1. Nernst equation
    (a) Most redox reactions are not performed under standard conditions. The cell potential (Ecell) for such a reaction is related to its standard cell potential (E^ocell) by Nernst equation.
    (b) Consider a redox reaction,
                aA + bB → cC + dD
   
   
    
   
    Example:

   

    Calculate the cell potential for the following cell at 25^oC.  
    Zn (s) | Zn²⁺ (aq, 0.050 M) || Cu²⁺ (aq, 5.0 M) | Cu (s)
    [E^oZn²⁺/Zn = -0.76 V; E^oCu²⁺/Cu = +0.34 V]

    Remember the rule in writing cell notation (ABC rule), it shows that zinc half-cell is the anode and copper half-cell is the cathode.   
    Overall cell reaction equation: Zn (s) + Cu²⁺ (aq) → Zn²⁺ (aq) + Cu (s)
    Since there are 2 mole of electrons transferred, n =2.

    E^ocell = E^ocathode - E^oanode
                = +0.34 - (-0.76)
            = +1.10 V

   
   
   
   
    (c) For galvanic cell involving gas, the pressure of the gas (in atm) is included in the reaction quotient, Q.
    Example:       













    For the following cell reaction,
    Zn (s) + 2 H⁺ (aq) → Zn²⁺ (aq) + H2 (g)
    Determine the concentration of H⁺ ion if the cell potential at 25^oC is +0.45 V, [Zn²⁺] = 1.0 M and PH2 = 1.0 atm. [E^oH⁺/H2 = +0.00 V; E^oZn²⁺/Zn = -0.76 V]

    From the overall cell reaction equation, it shows that zinc half-cell is the anode (since it undergoes oxidation: an increase in oxidation number) and hydrogen half-cell is the cathode (since it undergoes reduction: a decrease in oxidation number).
    E^ocell = E^ocathode - E^oanode
            = +0.00 - (-0.76)
            = +0.76 V
    Since there are 2 mole of electrons transferred, n =2.

     
     
     



    (d) At equilibrium, there is no net transfer of electrons. Thus, Ecell = 0 and Q = K, where K is the equilibrium constant.
   
   
   
    E^ocell = (0.0592/n) log K
    Example:
    Find the value of the equilibrium constant (K) for the following reaction using standard reduction potential at 25^oC.
    Cu (s) + 2 Ag⁺ (aq) → Cu²⁺ (aq) + 2 Ag (s)
    [E^oCu²⁺/Cu = +0.34 V; E^oAg⁺/Ag = +0.80 V]

    From the overall cell reaction equation, it shows that copper half-cell is the anode (since it undergoes oxidation: an increase in oxidation number) and silver half-cell is the cathode (since it undergoes reduction: a decrease in oxidation number).
    E^ocell = E^ocathode - E^oanode
            = +0.80 - (+0.34)
            = +0.46 V

    Ecell = E^ocell - (0.0592/2) log K
        0 = 0.46 - (0.0296) log K
    (0.0296) log K = 0.46
                 log K = 15.54
                      K = 3.47 x 10¹⁵