Cell Potential and Free Energy Free-energy change, G, is related to cell potenti

Question

Cell Potential and Free Energy Free-energy change, G, is related to cell potential, E, by the equation G=nFE where n is the number of moles of electrons transferred and F=96,500C/(mol e) is the Faraday constant. When E is measured in volts and can be determined from half-reaction potentials as given in the table below. G must be in joules since 1 J=1 CV. Reduction half-reaction E (V) Ag+(aq)+eAg(s) 0.80 Cu2+(aq)+2eCu(s) 0.34 Sn4+(aq)+4eSn(s) 0.15 2H+(aq)+2eH2(g) 0 Ni2+(aq)+2eNi(s) 0.26 Fe2+(aq)+2eFe(s) 0.45 Zn2+(aq)+2eZn(s) 0.76 Al3+(aq)+3eAl(s) 1.66 Mg2+(aq)+2eMg(s) 2.37

Part A Calculate the standard free-energy change at 25 C for the following reaction using the table in the introduction: Mg(s)+Fe2+(aq)Mg2+(aq)+Fe(s)

Part B Calculate the standard cell potential at 25 C for the reaction X(s)+2Y+(aq)X2+(aq)+2Y(s) where H = -759 kJ and S = -345 J/K .

please help

Explanation / Answer

Part A

Ecell = Eooxidation(Mg->Mg(+2) + Eoreduction(Fe+2->Fe)

=> 2.37 - 0.45

=> 1.88V

Part B

Delta G = Delta H - T Delta S

Delta G = -759 + 345 * 298/1000

Delta G = -656.19 kJ/mol

-656190 = -(2)(96500)Ecell

Ecell = 3.399V

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