Question

A copper electrode is placed in a 1.0 M Cu(NO3)2 solution while a silver electrode is placed in a 1.0 M AgNO3 solution. If the cell potential is measured to be 0.74 V at standard state conditions, what is the equilibrium constant for the reaction Cu(s) + 2Ag+(aq) ⇌ Cu2+(aq) + 2Ag(s)?

Answer 1

To find the equilibrium constant  K  for the reaction, we can use the Nernst equation and the relationship between the standard cell potential  E  and the equilibrium constant. First, let's write the half-cell reactions:Oxidation  anode : Cu s   Cu aq  + 2eReduction  cathode : 2Ag aq  + 2e  2Ag s Now, let's find the standard cell potential  E  for the reaction:E cell  = E cathode  - E anode We are given the cell potential  E cell   as 0.74 V. To find the standard reduction potentials for the half-cell reactions, we can look them up in a table of standard reduction potentials:E Cu/Cu  = +0.34 VE Ag/Ag  = +0.80 VNow we can find E cell :E cell  = E Ag/Ag  - E Cu/Cu  = 0.80 V - 0.34 V = 0.46 VNow we can use the relationship between the standard cell potential and the equilibrium constant:E cell  =  RT/nF  * ln K Where:E cell  = 0.46 VR = 8.314 J/ molK   gas constant T = 298 K  standard temperature n = 2  number of electrons transferred in the reaction F = 96485 C/mol  Faraday's constant Rearrange the equation to solve for K:K = e^nFE cell /RT Plugging in the values:K = e^2 * 96485 C/mol * 0.46 V /  8.314 J/ molK  * 298 K  K  2.04  10The equilibrium constant for the reaction Cu s  + 2Ag aq   Cu aq  + 2Ag s  is approximately 2.04  10.