To calculate the enthalpy change of the reduction reaction of Fe3+ to Fe2+, we can use the Nernst equation and the relationship between Gibbs free energy change G and enthalpy change H .First, we need to find the number of moles of Fe3+ ions in 1.44 g. The molar mass of Fe is 55.85 g/mol.moles of Fe3+ = 1.44 g / 55.85 g/mol = 0.0258 molNow, we can write the balanced redox reaction:Fe3+ + e- Fe2+ E = +0.77 V 2H+ + 2e- H2 g E = 0.00 V The balanced redox reaction is:Fe3+ + 2H+ + e- Fe2+ + H2 g The overall cell potential E is the difference between the reduction potential of the two half-reactions:E = E Fe3+/Fe2+ - E H+/H2 = 0.77 V - 0.00 V = 0.77 VNow, we can calculate the Gibbs free energy change G using the Nernst equation:G = -nFEwhere n is the number of moles of electrons transferred in this case, n = 1 , F is the Faraday constant 96,485 C/mol , and E is the overall cell potential.G = - 1 mol 96,485 C/mol 0.77 V = -74,293.45 J/molNow, we can use the relationship between Gibbs free energy change G and enthalpy change H at constant temperature:G = H - TSwhere T is the temperature in Kelvin 298 K and S is the entropy change. Since we are only interested in the enthalpy change H , we can rearrange the equation:H = G + TSUnfortunately, we don't have the entropy change S for this reaction. However, we can estimate the enthalpy change H using the Gibbs free energy change G as a rough approximation, assuming that the entropy change is not significant:H G = -74,293.45 J/molNow, we can calculate the enthalpy change for the 0.0258 mol of Fe3+ ions:H = -74,293.45 J/mol 0.0258 mol = -1,915.57 JThe enthalpy change of the reduction reaction of Fe3+ to Fe2+ when 1.44 g of Fe3+ ions are reduced using a standard hydrogen electrode at 298 K is approximately -1,915.57 J.