To calculate the Gibbs free energy change G for the electrochemical reaction between a copper electrode in a 1 M copper II sulfate solution and a standard hydrogen electrode SHE in a 1 M hydrogen ion solution at 25C, we need to use the following equation:G = -nFEwhere n is the number of moles of electrons transferred in the reaction, F is the Faraday constant 96,485 C/mol , and E is the cell potential.First, we need to determine the cell potential E for the reaction. The cell potential can be calculated using the Nernst equation:E = E - RT/nF * ln Q where E is the standard cell potential, R is the gas constant 8.314 J/molK , T is the temperature in Kelvin 25C = 298 K , and Q is the reaction quotient.For the given reaction, the half-reactions are:1. Cu + 2e Cu reduction 2. 2H + 2e H oxidation The standard reduction potential E for Cu/Cu is +0.34 V, and for the SHE H/H , it is 0 V by definition. Since the copper electrode is being reduced and the hydrogen electrode is being oxidized, the overall cell potential E is:E = E Cu/Cu - E H/H = 0.34 V - 0 V = 0.34 VSince both the copper II sulfate and hydrogen ion solutions have a concentration of 1 M, the reaction quotient Q is:Q = [Cu][H]/[H] = 1 1 / 1 = 1Now we can calculate the cell potential E using the Nernst equation:E = E - RT/nF * ln Q = 0.34 V - 8.314 J/molK * 298 K / 2 * 96,485 C/mol * ln 1 Since ln 1 = 0, the cell potential E remains 0.34 V.Finally, we can calculate the Gibbs free energy change G using the equation:G = -nFE = - 2 mol 96,485 C/mol 0.34 V = -65,609.4 J/molSo, the Gibbs free energy change for the electrochemical reaction is -65,609.4 J/mol.