Question

Calculate the equilibrium constant for the following electrochemical cell reaction at 25°C:Cu(s) + 2 Ag+(aq) --> Cu2+(aq) + 2 Ag(s)Given the standard reduction potentials are:Cu2+(aq) + 2 e- → Cu(s)      E°red = +0.34 VAg+(aq) + 1 e- → Ag(s)       E°red = +0.80 V

Answer 1

To calculate the equilibrium constant for the given electrochemical cell reaction, we first need to determine the cell potential  Ecell  and then use the Nernst equation to find the equilibrium constant  K .1. Write the half-reactions for the given reaction:Oxidation half-reaction  Cu is oxidized : Cu s   Cu2+ aq  + 2 e- Reduction half-reaction  Ag+ is reduced : Ag+ aq  + 1 e-  Ag s 2. Determine the standard reduction potentials for each half-reaction:Ered Cu2+/Cu  = +0.34 VEred Ag+/Ag  = +0.80 V3. Calculate the standard cell potential  Ecell :Since Cu is being oxidized, we need to reverse the sign of its reduction potential to get the oxidation potential:Eox Cu/Cu2+  = -0.34 VNow, we can calculate the standard cell potential:Ecell = Ered Ag+/Ag  - Eox Cu/Cu2+  = 0.80 V -  -0.34 V  = 1.14 V4. Use the Nernst equation to find the equilibrium constant  K :Ecell =  RT/nF  * ln K where R is the gas constant  8.314 J/molK , T is the temperature in Kelvin  25C = 298 K , n is the number of electrons transferred  2 in this case , and F is the Faraday constant  96,485 C/mol .Rearrange the equation to solve for K:K = exp nFEcell/RT Plug in the values:K = exp 2 * 96,485 C/mol * 1.14 V /  8.314 J/molK * 298 K  K  1.1  10^37So, the equilibrium constant for the given electrochemical cell reaction at 25C is approximately 1.1  10^37.