Finding Standard Gibbs Free energy for half cell reaction.

[jtruth914]

The standard potential of the Cl−(aq.)|AgCl(s)|Ag(s) electrode in the vicinity of T = 298 K has been found to fit the expression

E$\theta$/V = -0.00558 + 2.6967x10^-3(T/K) - 8.2299x10^-6(T/K)^2 + 5.869x10^-9(T/K)^3

Find the standard Gibbs free energy, the standard enthalpy, and the standard entropy of the silver|silver chloride half cell reaction at 298 K.

My question is do I take the derivative of this equation to find entropy? I'm confused because it states it's a half cell, and I know that the equation (dEcell/dT)= DS/vF pertains to the complete reaction not the half cell.

 

[Nino MMP]

The standard Gibbs free energy of the silver|silver chloride half cell reaction at 298 K can be calculated using the following equation:G = -nFE = -zF(E/V)where n is the number of moles of electrons transferred in the reaction, z is the valence of the species being oxidized or reduced, F is Faraday's Constant (96,485 C/mol), and E/V is the standard potential of the electrode.For the silver|silver chloride half cell reaction, n=1, z=1, and E/V = -0.00558 + 2.6967x10^-3(T/K) - 8.2299x10^-6(T/K)^2 + 5.869x10^-9(T/K)^3. Thus, the standard Gibbs free energy at 298 K is:G = -1*96,485C/mol*(-0.00558 + 2.6967x10^-3(298K) - 8.2299x10^-6(298K)^2 + 5.869x10^-9(298K)^3)G = -8.541 kJ/molThe standard enthalpy of the silver|silver chloride half cell reaction at 298 K can be calculated as follows = G + T*Swhere S is the standard entropy of the reaction. The standard entropy of the silver|silver chloride half cell reaction at 298 K can be calculated using the following equation:S = -(dE/dT)V/Fwhere dE/dT is the derivative of the standard potential with respect to temperature, V is the volume of the cell, and F is Faraday's Constant.For the silver|silver chloride half cell reaction, dE/dT = 2.6967x10^-3 - 16.4598x10^-6(T/K) + 17.607x10^-9(T/K)^2, V = 1 L, and F = 96,485 C/mol. Thus, the standard entropy at 298 K is:S = -(2.6967x10^-3 - 16.4598x