Question

The cell potential of a redox reaction occurring in an electrochemical cell under any set of temperature and concentration conditions can be determined from the standard cell potential of the cell using the Nernst equation E = E° − RT nF ln Q where E is the cell potential of the cell, E° is the standard cell potential of the cell, R is the gas constant, T is the temperature in kelvin, n is the moles of electrons transferred in the reaction, and Q is the reaction quotient. Use this relationship to answer the problem below.

For the following oxidation-reduction reaction Cu2+(aq) + Ni(s) → Cu(s) + Ni2+(aq) the standard cell potential is 0.60 V. What is the actual cell potential of the cell if the temperature is 314 K, the initial Cu2+ concentration is 0.00107 M, and the initial Ni2+ concentration is 0.0183 M? (Note that the reaction involves the transfer of 2 moles of electrons, and the reaction quotient is 17.1.)

_______________ V

Answer 1

Cu²⁺(aq) + Ni(s) → Cu(s) + Ni²⁺(aq)

E= E^o- (RT/nF) ln Q

            E= actual cell potential
            E^o = standard cell potential = 0.60V
            R = constant = 8.31 J/mole K
            T = absolute temperature = 314 K
            F = Faraday's constant = 96,485 C/mole e^-
            n = number of moles of electrons transferred in the balanced equation for the reaction occurring
                  in the cell =2

           Q = quotient for the reaction = 17.1

           ln Q = 2.303 log Q

E= E^o- (RT/nF) ln Q = 0.60 - (8.31 x 314)/ (2 x 96485) x 2.303 log 17.1

                              = 0.6 - 0.03 x log 17.1 = 0.6 - 0.03 x 1.233 = 0.6- 0.037 = 0.563