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. \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{I}^{-}(a q) \rightleftharpoons \mathrm{I}_{2}(a q)+\mathrm{Mn}^{2+}(a q)\) b. \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons \mathrm{F}_{2}(g)+\mathrm{Mn}^{2+}(a q)\)

First, we need to balance the redox reaction in aqueous (aq) solution between permanganate ion (MnO4-) and iodide ion (I-): \(MnO_{4}^{-}(aq) + I^{-}(aq) \rightleftharpoons I_{2}(aq) + Mn^{2+}(aq)\) Balance the reaction using the half-reaction method. 1. Oxidation Half-Reaction: \(I^- \rightarrow I_2\) Balance the iodine atoms: \(2I^- \rightarrow I_2\) Balance the electrons: \(2I^- \rightarrow I_2 + 2e^-\) 2. Reduction Half-Reaction: \(MnO_4^- \rightarrow Mn^{2+}\) Balance the manganese atoms: \(MnO_4^- \rightarrow Mn^{2+}\) Balance the oxygen atoms by adding water: \(MnO_4^- + 4H_2O \rightarrow Mn^{2+}\) Balance the hydrogen atoms by adding protons: \(MnO_4^- + 4H_2O + 8H^+ \rightarrow Mn^{2+}\) Balance the charge by adding electrons: \(MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O\) 3. Combine the half-reactions by multiplying each half-reaction by a factor so that the number of electrons cancelled match up. Oxidation: \(2(2I^- \rightarrow I_2 + 2e^-)\) Reduction: \((5e^- + MnO_4^- + 8H^+ \rightarrow Mn^{2+} + 4H_2O)\) 4. Add the balanced half-reactions and simplify: \(2(2I^-) + MnO_4^- + 8H^+ \rightarrow 2I_2 + Mn^{2+} + 4H_2O\) The balanced redox equation is then: \(2I^{-}(aq) + MnO_{4}^{-}(aq) + 8H^+ \rightleftharpoons 2I_2(aq) + Mn^{2+}(aq) + 4H_2O\)

To calculate the standard cell potential, E°, we use the formula: \(E°_{cell} = E°_{cathode} - E°_{anode}\) From Table 18.1, the standard reduction potentials (E°) are: \(E°(MnO_4^-/Mn^{2+})=1.51\:V\) \(E°(I_2/I^-) = 0.54\:V\) In this reaction, iodine is being oxidized (anode) and manganese is being reduced (cathode): \(E°_{cell} = E°(MnO_4^-/Mn^{2+}) - E°(I_2/I^-)\) \(E°_{cell} = 1.51\:V - 0.54\:V = 0.97\:V\) Since \(E°_{cell}\) is positive, the reaction is spontaneous under standard conditions.

Next, we need to balance the redox reaction in aqueous (aq) solution between permanganate ion (MnO4-) and fluoride ion (F-): \(MnO_{4}^{-}(aq) + F^{-}(aq) \rightleftharpoons F_{2}(g) + Mn^{2+}(aq)\) Follow the same steps as above to balance the reaction. 1. Oxidation Half-Reaction: \(F^- \rightarrow F_2\) Balance the fluorine atoms: \(2F^- \rightarrow F_2\) Balance the electrons: \(2F^- \rightarrow F_2 + 2e^-\) 2. Reduction Half-Reaction: \(MnO_4^- \rightarrow Mn^{2+}\) (same as in part a) \(MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O\) 3. Combine and simplify the half-reactions: \(10F^{-}(aq) + MnO_{4}^{-}(aq) + 16H^+ \rightleftharpoons 5F_2(g) + Mn^{2+}(aq) + 8H_2O\) The balanced redox equation is then: \(10F^{-}(aq) + MnO_{4}^{-}(aq) + 16H^+ \rightleftharpoons 5F_{2}(g) + Mn^{2+}(aq) + 8H_2O\)

Use the formula: \(E°_{cell} = E°_{cathode} - E°_{anode}\). From Table 18.1, the standard reduction potentials (E°) are: \(E°(MnO_4^-/Mn^{2+})=1.51\rm V\) \(E°(F_2/F^{-}) = 2.87\rm V\) In this reaction, fluoride is being oxidized (anode) and manganese is being reduced (cathode): \(E°_{cell} = E°(MnO_4^-/Mn^{2+}) - E°(F_2/F^-)\) \(E°_{cell} = 1.51\:V - 2.87\:V = -1.36\:V\) Since \(E°_{cell}\) is negative, the reaction is nonspontaneous under standard conditions.