To calculate the equilibrium constant K for the reaction between copper and silver ions in a galvanic cell, we first need to determine the overall cell reaction and the cell potential Ecell .The cell potential can be calculated using the reduction potentials of the two half-reactions. The half-reaction with the higher reduction potential will proceed as a reduction gain of electrons at the cathode, while the other half-reaction will proceed as an oxidation loss of electrons at the anode.Since the reduction potential of Ag+ Ered = +0.80 V is higher than that of Cu2+ Ered = +0.34 V , Ag+ will be reduced at the cathode and Cu2+ will be oxidized at the anode. Therefore, the silver electrode will be the cathode and the copper electrode will be the anode.The overall cell reaction can be obtained by combining the two half-reactions:Oxidation anode : Cu s Cu2+ aq + 2e- reverse of the given Cu2+ half-reaction Reduction cathode : Ag+ aq + e- Ag s given Ag+ half-reaction Overall cell reaction: Cu s + 2Ag+ aq Cu2+ aq + 2Ag s Now we can calculate the cell potential Ecell using the reduction potentials:Ecell = Ecathode - Eanode = +0.80 V - +0.34 V = +0.46 VNext, we can use the Nernst equation to calculate 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 .Rearranging the equation to solve for K:K = exp nF * Ecell / RT Plugging in the values:K = exp 2 * 96485 * 0.46 V / 8.314 * 298 K = exp 89.2 K 4.9 10^38So, the equilibrium constant K for the reaction between copper and silver ions in a galvanic cell at room temperature 25C is approximately 4.9 10^38. The silver electrode will be the cathode, and the copper electrode will be the anode.