Chemistry - Redox Reactions Question with Solution | TestHub

To determine spontaneity, calculate the standard cell potential ($E_{cell}^{\circ }$). A reaction is spontaneous if $E_{cell}^{\circ }>0$.

The relevant standard electrode potentials are:

$E^{\circ }({\mathrm{Sn}}^{4+}/{\mathrm{Sn}}^{2+})=+0.15\mathrm{V}$

$E^{\circ }({\mathrm{I}}_2/{\mathrm{I}}^{-})=+0.54\mathrm{V}$

For option C: ${\mathrm{Sn}}^{2+}+{\mathrm{I}}_2⟶{\mathrm{Sn}}^{4+}+2{\mathrm{I}}^{-}$

Oxidation: ${\mathrm{Sn}}^{2+}⟶{\mathrm{Sn}}^{4+}+2{\mathrm{e}}^{-}$ ($E_{ox}^{\circ }=-0.15\mathrm{V}$)

Reduction: ${\mathrm{I}}_2+2{\mathrm{e}}^{-}⟶2{\mathrm{I}}^{-}$ ($E_{red}^{\circ }=+0.54\mathrm{V}$)

$E_{cell}^{\circ }=E_{red}^{\circ }+E_{ox}^{\circ }=+0.54\mathrm{V}-0.15\mathrm{V}=+0.39\mathrm{V}$

Since $E_{cell}^{\circ }>0$, the reaction is spontaneous.