To calculate the equilibrium constant K for the given electrochemical reaction, we first need to determine the standard cell potential E_cell for the reaction. The standard cell potential can be calculated using the standard reduction potentials of the half-reactions:E_cell = E_cathode - E_anodeIn this case, the reduction half-reaction is:$\ce{Fe^3+ + e^- -> Fe^2+}$ with E = +0.77VThe oxidation half-reaction is:$\ce{2I^- -> I2 + 2e^- }$ with E = +0.54VSince the reduction potential of the iron half-reaction is greater than the iodine half-reaction, the iron half-reaction will act as the cathode and the iodine half-reaction will act as the anode.E_cell = +0.77V - +0.54V = +0.23VNow, we can use the Nernst equation to relate the standard cell potential to the equilibrium constant. The Nernst equation is:E_cell = - RT/nF * ln K Where R is the gas constant 8.314 J/molK , T is the temperature in Kelvin 298 K , n is the number of moles of electrons transferred in the balanced redox reaction 2 moles in this case , and F is the Faraday constant 96,485 C/mol .Rearranging the Nernst equation to solve for K, we get:ln K = - nFE_cell / RT Plugging in the values:ln K = - 2 * 96,485 C/mol * 0.23V / 8.314 J/molK * 298 K ln K - 11.04To find K, we take the exponential of both sides:K = e^-11.04 1.63 10^-5 So, the equilibrium constant for the given electrochemical reaction at 298 K is approximately 1.63 10^-5 .