Calculate the equilibrium constant for the reaction $\mathrm{S}_{2} \mathrm{O}_{8}{ }^{2-}{ }_{(\mat

Question

Calculate the equilibrium constant for the reaction $\mathrm{S}_{2} \mathrm{O}_{8}{ }^{2-}{ }_{(\mathrm{aq})}+2 \mathrm{Fe}^{2+}{ }_{(\mathrm{aq})} ightarrow 2 \mathrm{Fe}^{3+}{ }_{(\mathrm{aq})}+2 \mathrm{SO}_{4}{ }^{2-}{ }_{(\mathrm{aq})}$ at $25.0^{\circ} \mathrm{C}$, given that the standard cells potentials for the two half reactions at this temperature are $$ \begin{aligned} & \mathrm{S}_{2} \mathrm{O}_{8}{ }^{2-}(\mathrm{aq})+2 \mathrm{e}^{-} ightarrow 2 \mathrm{SO}_{4}{ }^{2-}(\mathrm{aq}) E^{\circ}=+2.08 \mathrm{~V} \ & \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{e}^{-} ightarrow \mathrm{Fe}^{2+}(\mathrm{aq}) \quad E^{\circ}=+0.77 \mathrm{~V} \end{aligned} $$

Professor PC Thomas Classes · Answer

$\mathrm{E}_{	ext {cell }}^{0}=\mathrm{E}_{	ext {cathode }}^{0}-\mathrm{E}_{	ext {anode }}^{0}=2.08-0.77=1.31 \mathrm{~V}$.\ $\Delta \mathrm{G}^{0}=-\mathrm{nFE}^{0}=-2.303 \mathrm{RT} \log \mathrm{K}$\ $\log \mathrm{K}=\frac{\mathrm{nFE}^{0}}{2.303 \mathrm{RT}}=\frac{2 	imes 96500 	imes 1.31}{2.303 	imes 8.314 	imes 298}=44.3$\ $\log \mathrm{K}=44+0.3 \quad 	herefore \mathrm{~K}=10^{44} 	imes 2$