Question

Calculate the equilibrium constant for the reaction
$F{e}^{2+}+C{e}^{4+}\rightleftharpoons F{e}^{3+}+C{e}^{3+}$
Given $E^{\circ }(C{e}^{4+}/C{e}^{3+})=1.44V,E^{\circ }(F{e}^{3+}/F{e}^{2+})=0.68V$

Answer 1

$F{e}^{2+}+C{e}^{4+}\rightleftharpoons F{e}^{3+}+C{e}^{3+}$
$E^{\circ }=E^{\circ }(C{e}^{4+}/C{e}^{3+})-E^{\circ }(F{e}^{3+}/F{e}^{2+})$
$=1.44-0.68=0.76V$
$\because E^{\circ }=0.0592\log K$
$\Rightarrow \log K=\frac{E^{\circ }}{0.0592}=\frac{0.76}{0.0592}=12.83$
$\Rightarrow k=6.88\times 10^{12}$