Fayalite, 2FeO.SiO_2 , is the only iron silicate compound formed by reaction of FeO with SiO_2 at a total pressure of 1 atm, and the standard Gibbs free energy change for the reaction
2FeO_{(s)}+SiO_{2(s)}=2FeO.SiO_{2(s)}is – 11,070 J at 1200 K. Calculate the EMF of the cell
Fe\mid SiO_{2}\mid 2FeO.SiO_2\mid CaO-ZrO_2\mid FeO\mid Feat 1200 K.
This is an oxygen concentration cell in which the cell reaction can be written as
O_2(higher pressure at the cathode)\rightarrow O_2(lower pressure at the anode) (i)
for which the EMF is
\varepsilon =-\frac{RT}{4f}\ln \frac{p_{O_2}(at the anode)}{p_{O_2}(at the cathode)} (ii)
The oxygen pressure at the electrodes are fixed by the chemical equilibrium
Fe+\frac{1}{2}O_2=FeO (iii)
for which
K_{(iii)} =\frac{a_{FeO}}{a_{Fe}p^{1/2}_{O_2}}At the cathode, the activity of FeO with respect to Fe-saturated pure FeO is unity, and at the anode, the activity of FeO is that occurring in 2FeO.SiO_2 saturated with Fe and SiO_2 . As
K_{(iii)}=\frac{a_{FeO(cathode)}}{a_{Fe(cathode)}p^{1/2}_{O_2(cathode)}} =\frac{a_{FeO(anode)}}{a_{Fe(anode)}p_{O_2(anode)}}and
a_{Fe(anode)}={a_{Fe(cathode)}}= a_{FeO(cathode)}=1\frac{p_{O_2(anode)}}{p_{O_2(cathode)}} =a^{2}_{FeO(anode)}
and hence, in Equation (ii),
\varepsilon =-\frac{RT}{4f} \ln a^{2}_{FeO(anode)}For 2FeO + SiO_2 = 2FeO· SiO_2 ,
\Delta G^{\circ }_{1200 K}=-11,070 J
=-8.3144\times 1200\ln \frac{a_{2FeO.SiO_2}}{a^{2}_{FeO}a_{SiO_2}}
Thus, at the anode, with a_{2FeO.SiO_2}=a_{SiO_2} =1 ,
a_{FeO(anode)}=0.574
and thus,
\varepsilon =-\frac{8.3144\times 1200}{4\times 96,487}\ln (0.574)^2
=0.0287 volts
Alternatively, the anode half-cell reaction can be written as
O^{2-} =\frac{1}{2}O_{2(eq.Fe/\underline{FeO} )}+2e^{-}and the cathode half-cell reaction can be written as
\frac{1}{2}O_{2(eq.Fe/FeO)}+2e^{-}=O^{2-}or, at the anode,
2Fe+2O^{2-}+SiO_{2}=2FeO.SiO_2+4e^-and, at the cathode,
2FeO+4e^-=2Fe+2O^{2-}summation of which gives the cell reaction as
2FeO+SiO_2=2FeO.SiO_2The Gibbs free energy change for the cell reaction is
\Delta G^\circ =-zf\varepsilon ^\circ =-11,070 J
and thus,
\varepsilon =\frac{-\Delta G^\circ }{4f}=\frac{11,070}{4\times 96,487} =0.0287 volts