To determine the equilibrium constant for the given redox reaction, we first need to find the overall cell potential E for the reaction. The given standard reduction potentials are:Fe3+ + e- Fe2+; E = +0.771 VH2O2 + 2H+ + 2e- 2H2O; E = +1.78 VWe need to reverse the first half-reaction to match the overall redox reaction:Fe2+ Fe3+ + e-; E = -0.771 VNow, we can add the two half-reactions to get the overall redox reaction:Fe2+ + H2O2 + 2H+ Fe3+ + 2H2OThe overall cell potential E is the sum of the potentials of the two half-reactions:E = -0.771 V + 1.78 V = 1.009 VNow we can use the Nernst equation to find the equilibrium constant K :E = RT/nF * ln K Where:E = 1.009 V overall cell potential R = 8.314 J/ molK gas constant T = 25C = 298.15 K temperature n = 2 number of electrons transferred in the reaction F = 96485 C/mol Faraday's constant Rearrange the equation to solve for K:ln K = nF * E / RT K = e^ nF * E / RT Plugging in the values:K = e^ 2 * 96485 C/mol * 1.009 V / 8.314 J/ molK * 298.15 K K e^198,000 / 2470 e^80.2The equilibrium constant K is approximately e^80.2, which is a very large number. This indicates that the reaction strongly favors the formation of Fe3+ and H2O.