Question

For the electrochemical reaction  +  ↔  + , the cell potential  is 0.60 V at 25°C. If the standard reduction potential of 2+ is 0.34 V and that of 2+ is -0.74 V, calculate the equilibrium constant () for the reaction at 25°C.

Answer 1

To calculate the equilibrium constant  Kc  for the reaction, we first need to determine the standard cell potential  E  for the reaction. The standard cell potential can be calculated using the standard reduction potentials of the two half-reactions:E cell  = E cathode  - E anode Given the standard reduction potentials of A2+  0.34 V  and D2+  -0.74 V , we can identify the cathode and anode half-reactions:Cathode: A2+ + 2e-  A  E = 0.34 V Anode: D  D2+ + 2e-  E = 0.74 V, since the reduction potential is -0.74 V Now we can calculate the standard cell potential:E cell  = 0.34 V -  -0.74 V  = 1.08 VNext, we can use the Nernst equation to relate the cell potential  E  at 25C to the standard cell potential  E  and the reaction quotient  Q :E = E -  RT/nF  * ln Q At equilibrium, E = 0.60 V, and Q = Kc. We can rearrange the Nernst equation to solve for Kc:0.60 V = 1.08 V -  RT/nF  * ln Kc Let's plug in the values for R  8.314 J/molK , T  298 K , n  2, since there are 2 electrons transferred in the reaction , and F  96,485 C/mol :0.60 V = 1.08 V -   8.314 J/molK * 298 K  /  2 * 96,485 C/mol   * ln Kc Solving for Kc:ln Kc  =  1.08 V - 0.60 V  *  2 * 96,485 C/mol  /  8.314 J/molK * 298 K ln Kc   10.61Now, we can find Kc by taking the exponent of both sides:Kc = e^10.61   40500So, the equilibrium constant  Kc  for the reaction at 25C is approximately 40,500.