Free-energy change, \\Delta G^\\circ, is related to cell potential, E^\\circ, by

Question

Free-energy change, Delta G^circ, is related to cell potential, E^circ, by the equation

Delta G^ circ = -nFE^circ
where n is the number of moles of electrons transferred and F =96,500; m C/(mol ~e^-) is the Faraday constant. When E^circ is measured in volts and can be determined from half-reaction potentials as given in the table below. Delta G^ circ must be in joules since 1 m ~J = 1 ~C cdot V.

Reduction half-reaction E^circ ( m V)
m Ag^+{(aq)}+e^- ightarrow Ag{(s)} 0.80
m Cu^{2+}{(aq)}+2e^- ightarrow Cu{(s)} 0.34
m Sn^{4+} {(aq)} + 4 e^- ightarrow m Sn {(s)} 0.15
m 2H^+ {(aq)} + 2 e^- ightarrow m H_2 {(g)} 0
m Ni^{2+}{(aq)}+2e^- ightarrow Ni{(s)} -0.26
m Fe^{2+}{(aq)}+2e^- ightarrow Fe{(s)} -0.45
m Zn^{2+}{(aq)}+2e^- ightarrow Zn{(s)} -0.76
m Al^{3+} {(aq)} + 3 e^- ightarrow m Al {(s)} -1.66
m Mg^{2+} {(aq)} + 2 e^- ightarrow m Mg {(s)} -2.37

Part A
Calculate the standard free-energy change at 25 ^ circ m C for the following reaction using the table in the introduction:

m Mg {(s)}+ m Fe^{2+}{(aq)} ightarrow m Mg^{2+}{(aq)}+ m Fe {(s)}
Express your answer numerically in joules.

Part B
Calculate the standard cell potential at 25 ^ circ m C for the reaction

m X{(s)}+ m 2Y^{+}{(aq)} ightarrow m X^{2+}{(aq)}+ m 2Y{(s)}

where Delta H^circ = -941 m kJ and Delta S^circ = -371 m J/K .
Express your answer numerically in volts.

Explanation / Answer

A) Eo cell = E cathode - E anode = -0.45 - (-2.37) = 1.92

Now. DeltaGo = -nFEo = - 2 x 96500 x 1.92 J = 370.56 kJ

B) Delta Go = deltaH - T deltaS

DeltaGo = -941000 - 298 x (-371) J = -830.442 kJ

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.