Use the following half-equations to write three spontaneous reactions. Justify your answers by calculating \(E^{\circ}\) for the cells. (1) \(\mathrm{MnO}_{4}^{-}(a q)+8 \mathrm{H}^{+}(a q)+5 e^{-} \longrightarrow \mathrm{Mn}^{2+}(a q)+4 \mathrm{H}_{2} \mathrm{O}\) \(E^{\circ}=+1.512 \mathrm{~V}\) (2) \(\mathrm{O}_{2}(\mathrm{~g})+4 \mathrm{H}^{+}(\mathrm{aq})+4 e^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O} \quad E^{\circ}=+1.229 \mathrm{~V}\) (3) \(\mathrm{Co}^{2+}(a q)+2 e^{-} \longrightarrow \mathrm{Co}(s) \quad E^{\circ}=-0.282 \mathrm{~V}\)

When combining half-reactions, a spontaneous reaction will occur if the sum of the half-reactions' \(E^{\circ}\) values is positive. So, we should choose two half-equations with \(E^{\circ}\) values that, when added, result in a positive value.

To write the first spontaneous reaction, we will combine half-equations (1) and (3), ensuring that electrons are balanced: (1) \(\mathrm{MnO}_{4}^{-}(a q)+8 \mathrm{H}^{+}(a q)+5 e^{-} \longrightarrow \mathrm{Mn}^{2+}(a q)+4 \mathrm{H}_{2} \mathrm{O}\) (3) \(2(\mathrm{Co}^{2+}(a q)+2 e^{-} \longrightarrow \mathrm{Co}(s))\) Now, we add the two half-equations to create the spontaneous reaction: \(\mathrm{MnO}_{4}^{-}(a q)+8 \mathrm{H}^{+}(a q)+5 e^{-} + 2(\mathrm{Co}^{2+}(a q)+2 e^{-}) \longrightarrow \mathrm{Mn}^{2+}(a q)+4 \mathrm{H}_{2} \mathrm{O} + 2\mathrm{Co}(s)\) Notice that the number of electrons in the reactions (5 and 4) are balanced.

To calculate the standard cell potential for this reaction, add the values of the two half-equations: \(E^{\circ}_{\text{cell}} = E^{\circ}_{1} + E^{\circ}_{3} = (+1.512 \mathrm{~V}) + (-0.282 \mathrm{~V}) = +1.230 \mathrm{~V}\) The positive value of \(E^{\circ}_{\text{cell}}\) indicates a spontaneous reaction.

To write the second spontaneous reaction, we will combine half-equations (2) and (3), ensuring that electrons are balanced: (2) \(\mathrm{O}_{2}(\mathrm{~g})+4 \mathrm{H}^{+}(\mathrm{aq})+4 e^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}\) (3) \(2(\mathrm{Co}^{2+}(a q)+2 e^{-} \longrightarrow \mathrm{Co}(s))\) Now, we add the two half-equations to create the spontaneous reaction: \(\mathrm{O}_{2}(\mathrm{~g})+4 \mathrm{H}^{+}(\mathrm{aq})+4 e^{-} + 2(\mathrm{Co}^{2+}(a q)+2 e^{-}) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O} + 2\mathrm{Co}(s)\) The number of electrons in the reactions (4 and 4) are balanced.

To calculate the standard cell potential for this reaction, add the values of the two half-equations: \(E^{\circ}_{\text{cell}} = E^{\circ}_{2} + E^{\circ}_{3} = (+1.229 \mathrm{~V}) + (-0.282 \mathrm{~V}) = +0.947 \mathrm{~V}\) The positive value of \(E^{\circ}_{\text{cell}}\) indicates a spontaneous reaction.

Since there are only three half-equations, our third spontaneous reaction will combine half-equations (1) and (2): (1) \(\mathrm{MnO}_{4}^{-}(a q)+8 \mathrm{H}^{+}(a q)+5 e^{-} \longrightarrow \mathrm{Mn}^{2+}(a q)+4 \mathrm{H}_{2} \mathrm{O}\) (2) \(1.25(\mathrm{O}_{2}(\mathrm{~g})+4 \mathrm{H}^{+}(\mathrm{aq})+4 e^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O})\) Now, we add the two half-equations to create the spontaneous reaction: \(\mathrm{MnO}_{4}^{-}(a q)+8 \mathrm{H}^{+}(a q)+5 e^{-} + 1.25(\mathrm{O}_{2}(\mathrm{~g})+4 \mathrm{H}^{+}(\mathrm{aq})+4 e^{-}) \longrightarrow \mathrm{Mn}^{2+}(a q)+4 \mathrm{H}_{2} \mathrm{O} + 2.5 \mathrm{H}_{2} \mathrm{O}\) The number of electrons in the reactions (5 and 4) are not balanced. Therefore, we cannot create a third spontaneous reaction by combining these two half-equations.