Calculate \(\mathscr{E}^{\circ}\) values for the following cells. Which reactions are spontaneous as written (under standard conditions)? Balance the equations. Standard reduction potentials are found in Table \(18.1\). a. \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{I}^{-}(a q) \longrightarrow \mathrm{I}_{2}(a q)+\mathrm{Mn}^{2+}(a q)\) b. \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{F}^{-}(a q) \longrightarrow \mathrm{F}_{2}(g)+\mathrm{Mn}^{2+}(a q)\)

First, balance the atoms other than oxygen and hydrogen. Then, balance the oxygens by adding water molecules, and balance the hydrogens by adding H+ ions. Finally, balance the charges by adding electrons. Half-reaction 1 (oxidation): \( \mathrm{I}^{-}(a q) \longrightarrow \mathrm{I}_{2}(a q)\) Balance the iodine atoms: \(2\mathrm{I}^{-}(a q) \longrightarrow \mathrm{I}_{2}(a q)\) There are no oxygens or hydrogens to balance. Balance the charges by adding 2 electrons: \(2\mathrm{I}^{-}(a q) \longrightarrow \mathrm{I}_{2}(a q) + 2e^-\) Half-reaction 2 (reduction): \(\mathrm{MnO}_{4}^{-}(a q) \longrightarrow \mathrm{Mn}^{2+}(a q)\) Balance the manganese atoms: \(\mathrm{MnO}_{4}^{-}(a q) \longrightarrow \mathrm{Mn}^{2+}(a q)\) Balance the oxygens by adding 4 water molecules: \(\mathrm{MnO}_{4}^{-}(a q) \longrightarrow \mathrm{Mn}^{2+}(a q) + 4 \mathrm{H}_{2} \mathrm{O} (l)\) Balance the hydrogens by adding 8 H+ ions: \(\mathrm{MnO}_{4}^{-}(a q) + 8 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{Mn}^{2+}(a q) + 4 \mathrm{H}_{2} \mathrm{O} (l)\) Balance the charges by adding 5 electrons: \(5 e^- + \mathrm{MnO}_{4}^{-}(a q) + 8 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{Mn}^{2+}(a q) + 4 \mathrm{H}_{2} \mathrm{O} (l)\) Now we have balanced the half-reactions.

Using the standard reduction potentials from Table 18.1, we can calculate the standard emf ( \(\mathscr{E}^{\circ}\) ) for the cell. For this, we need to use the Nernst equation: \[\mathscr{E}^{\circ} = \mathscr{E}_{red}^{\circ} - \mathscr{E}_{ox}^{\circ}\] For the reduction half-reaction: \(\mathscr{E}_{red}^{\circ} = +1.51\:V\) For the oxidation half-reaction: \(\mathscr{E}_{ox}^{\circ} = +0.54\:V\) Now we can calculate the emf of the cell: \[\mathscr{E}^{\circ} = 1.51\:V - 0.54\:V = 0.97\:V\]

Since the standard emf is positive (\(\mathscr{E}^{\circ} = 0.97\:V\)), the reaction is spontaneous under standard conditions. b. $\mathrm{MnO}_{4}^{-}(a q)+\mathrm{F}^{-}(a q) \longrightarrow \mathrm{F}_{2}(g)+\mathrm{Mn}^{2+}(a q)$ For this part, we can use the balanced half-reactions from part (a) and just replace the iodine half-reaction with the fluorine half-reaction.

Half-reaction 1 (oxidation): \( \mathrm{F}^{-}(a q) \longrightarrow \mathrm{F}_{2}(g)\) Balance the fluorine atoms: \(2\mathrm{F}^{-}(a q) \longrightarrow \mathrm{F}_{2}(g)\) There are no oxygens or hydrogens to balance. Balance the charges by adding 2 electrons: \(2\mathrm{F}^{-}(a q) \longrightarrow \mathrm{F}_{2}(g) + 2e^-\) The reduction half-reaction is the same as in part (a).

For the reduction half-reaction: \(\mathscr{E}_{red}^{\circ} = +1.51\:V\) For the oxidation half-reaction: \(\mathscr{E}_{ox}^{\circ} = +2.87\:V\) Now we can calculate the emf of the cell: \[\mathscr{E}^{\circ} = 1.51\:V - 2.87\:V = -1.36\:V\]

Since the standard emf is negative (\(\mathscr{E}^{\circ} = -1.36\:V\)), the reaction is not spontaneous under standard conditions.