Free-energy change, G , is related to cell potential, E , by the equation G = n

Question

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.

Calculate the standard cell potential at 25 C for the reaction

X(s)+2Y+(aq)X2+(aq)+2Y(s)

where H = -667 kJ and S = -333 J/K .

Express your answer numerically in volts.

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

Explanation / Answer

G = H - TS = -667*1000 - 298 * -333 = -567766 J/mol

then

E°cell = G/(-nF)

E°cell = (-567766)/(-2*96500) = 2.941792 V

E°cell = 2.941792 V

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