Question

Use the provided reduction potentials to calculate ArGº for the following balanced redox reaction: Pb2+(aq) + Cu(s) → Pb(s) + Cu2+(aq) E°(Pb2+/Pb) = -0.13 V and E°(Cu2+/Cu) = +0.34 V -41 kJ mol-1 +91 kJ mol-1 -21 kJ mol-1 -0.47 kJ mol-1 +46 kJ mol-1

Answer 1

Answer : The correct option is, +91 kJ/mole

Solution :

The balanced cell reaction will be,  

$Cu(s)+Pb^{2+}(aq)\rightarrow Cu^{2+}(aq)+Pb(s)$

Here copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. Lead (Pb) undergoes reduction by gain of electrons and thus act as cathode.

First we have to calculate the standard electrode potential of the cell.

$E^0_{[Pb^{2+}/Pb]}=-0.13V$

$E^0_{[Cu^{2+}/Cu]}=+0.34V$

$E^0_{cell}=E^0_{cathode}-E^0_{anode}$

$E^0_{cell}=E^0_{[Pb^{2+}/Pb]}-E^0_{[Cu^{2+}/Cu]}$

$E^0_{cell}=-0.13V-(0.34V)=-0.47V$

Now we have to calculate the standard Gibbs free energy.

Formula used :

$\Delta G^o=-nFE^o_{cell}$

where,

$\Delta G^o$ = standard Gibbs free energy = ?

n = number of electrons = 2

F = Faraday constant = 96500 C/mole

$E^o$ = standard e.m.f of cell = -0.47 V

Now put all the given values in this formula, we get the Gibbs free energy.

$\Delta G^o=-(2\times 96500\times (-0.47))=+90710J/mole=+90.71kJ/mole\approx +91kJ/mole$

Therefore, the standard Gibbs free energy is +91 kJ/mole