First, we need to balance the half-reactions to have the same number of electrons. In this case, both half-reactions involve 2 electrons, so they are already balanced.Next, we need to determine which half-reaction will be the reduction cathode and which will be the oxidation anode . The half-reaction with the higher standard reduction potential E will be the reduction, and the other will be the oxidation. In this case, Cu^2+ has a higher E +0.34 V than Sn^4+ +0.15 V , so Cu^2+ will be reduced, and Sn^4+ will be oxidized.Now we can write the overall cell reaction by combining the two half-reactions:Cu^2+ aq + Sn^2+ aq Cu s + Sn^4+ aq The cell diagram can be constructed as follows:Sn s | Sn^2+ aq || Cu^2+ aq | Cu s The anode is Sn s and the cathode is Cu s .To calculate the overall cell potential Ecell , we subtract the standard reduction potential of the anode from the standard reduction potential of the cathode:Ecell = Ecathode - EanodeEcell = +0.34 V - +0.15 V Ecell = +0.19 VSince the overall cell potential is positive, the reaction is spontaneous.Finally, we can calculate the standard Gibbs free energy change G using the following equation:G = -nFEcellwhere n is the number of moles of electrons transferred in this case, 2 , F is Faraday's constant 96,485 C/mol , and Ecell is the overall cell potential.G = - 2 mol 96,485 C/mol +0.19 V G = -36,644 J/molThe standard Gibbs free energy change for this electrochemical cell reaction is -36,644 J/mol.