Calculate the cell potential, G (J), and the equilibrium constant Keq for the following electrochemical cell and determine the spontaneous reaction direction.

Question

Calculate the cell potential, G (J), and the equilibrium constant Keq for the following electrochemical cell and determine the spontaneous reaction direction.

Pb(s) | Pb2+ (1.50 x 10-5 M) || Cr2O72-(0.150 M), Cr3+ (0.250 M), H+ (0.010 M) | Pt(s)

Answer 1

From a table of standard reduction potentials:

Pb²⁺ + 2e- ==> Pb .. Eº = -0.13 V

Cr₂O₇^2-(aq) + 14 H⁺(aq) + 6 e-  ==>  2Cr³⁺(aq) + 7 H₂O(l) .. Eº = 1.38 V

Cathode: Cr₂O₇^2-(aq) + 14 H⁺(aq) + 6 e-  ==>  2Cr³⁺(aq) + 7 H₂O(l)

Anode: 3Pb(s) ==> 3Pb²⁺ + 6e-

Overall: Cr₂O₇^2-(aq) + 14 H⁺(aq) + 3Pb(s) ==> 2Cr³⁺(aq) + 7 H₂O(l) + 3Pb²⁺(aq) ..

Eºcell = 1.38 + 0.13 = 1.51V

For the cell potential, use the Nernst equation:

Ecell = Eºcell - RT/nF ln Q

Ecell = 1.51 - (8.314)(298)/(6)(96485) ln Q

Ecell = 1.51 - 0.004280 ln Q and Q = [Cr³⁺]²[Pb²⁺]³ / [Cr₂O₇^2-][H⁺]¹⁴ = 1.41x10¹³ = Q and ln Q =30.27

Ecell = 1.51 - (0.004280)(30.27) = 1.51 - 0.13

Ecell = 1.38 V

∆G = -nFEcell = -(6)(96485)(1.38)

∆G = -798896 C-V

∆G = -798.9 kJ

-nFEºcell = -RT ln K

-(6)(96485)(1.51) = -(8.314)(298) ln K

-874154 = -2478 ln K

ln K = 352.8

K = 1.7x10¹⁵³

Reaction is spontaneous as written, ie Cr₂O₇^2-(aq) + 3Pb(s) ==> 2Cr³⁺(aq) + 3Pb²⁺(aq) .. since Ecell is positive, ∆G is negative and Keq is >> 1.