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 B Calculate the standard cell potential at 25 C for the reaction X(s)+2Y+(aq)X2+(aq)+2Y(s) where H = -749 kJ and S = -205 J/K . Express your answer numerically in volts.

Explanation / Answer

G = H - T S

= -749 - 298 x (-0.205)

= -687.9 kJ

G = - n F E

n= 2 eletrons

-687.9 = -2 x 96485 x E

E = 0.00356 V

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