To calculate the cell potential for the given electrochemical reaction, we need to use the Nernst equation. First, we need to find the standard reduction potentials for both half-reactions:1. Fe2+ aq + 2e- Fe s E = -0.44 V reduction 2. Cu2+ aq + 2e- Cu s E = +0.34 V reduction Since Fe is being oxidized and Cu2+ is being reduced, we need to reverse the first half-reaction:1. Fe s Fe2+ aq + 2e- E = +0.44 V oxidation Now, we can add the two half-reactions to get the overall reaction:Fe s + Cu2+ aq Fe2+ aq + Cu s Next, we can calculate the standard cell potential Ecell by subtracting the standard potential of the oxidation half-reaction from the standard potential of the reduction half-reaction:Ecell = E reduction - E oxidation = 0.34 V - 0.44 V = -0.10 VSince the concentrations of Fe2+ aq and Cu2+ aq are both 1.00 M, the reaction quotient Q is equal to 1:Q = [Fe2+]/[Cu2+] = 1.00/1.00 = 1Now, we can use the Nernst equation to calculate the cell potential at 25C:Ecell = Ecell - RT/nF * ln Q where:R = 8.314 J/ molK gas constant T = 25C + 273.15 = 298.15 K temperature in Kelvin n = 2 number of electrons transferred F = 96485 C/mol Faraday's constant Ecell = -0.10 V - 8.314 J/ molK * 298.15 K / 2 * 96485 C/mol * ln 1 Since ln 1 = 0, the last term in the equation becomes 0:Ecell = -0.10 VSo, the cell potential for the given electrochemical reaction is -0.10 V.