Question

What is the calculated value of the cell potential at 298K for an electrochemical cell with the following reaction, when the Cu2+ concentration is 1.23 M and the Cr3+ concentration is 5.82×10-4 M ? 3Cu2+(aq) + 2Cr(s)-->3Cu(s) + 2Cr3+(aq) Answer: V The cell reaction as written above is spontaneous for the concentrations given

Answer 1

3Cu2+(aq) + 2Cr(s)   -------------->   3Cu(s) + 2Cr3+(aq)

Oxidation : 2 Cr (s)   ------------> 2 Cr+3 + 6 e-    Eo = - 0.74 V

reduction : 3 Cu+2 (aq) --------------> 3 Cu (s)      Eo = 0.34 V

Eo cell = E red - E ox

            = 0.34 - (- 0.74)

            = 1.08 V

Ecell = Eocell - 0.05916 / 6 log [Cr+3^2 / Cu+2^3]

         = 1.08 - 0.05916 / 6 log [(5.82×10^-4)^2 / (1.23)^3]

         = 1.15 V

cell potential = Ecell = 1.15 V