Write balanced equations for the following reactions in acid solution. (a) \(\mathrm{P}_{4}(s)+\mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{PH}_{3}(g)+\mathrm{Cl}_{2}(g)\) (b) \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{NO}_{2}^{-}(a q) \longrightarrow \mathrm{Mn}^{2+}(a q)+\mathrm{NO}_{3}^{-}(a q)\) (c) \(\mathrm{HBrO}_{3}(a q)+\mathrm{Bi}(s) \longrightarrow \mathrm{HBrO}_{2}(a q)+\mathrm{Bi}_{2} \mathrm{O}_{3}(s)\) (d) \(\mathrm{CrO}_{4}^{2-}(a q)+\mathrm{SO}_{3}^{2-}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{SO}_{4}^{2-}(a q)\)

Begin by assigning oxidation numbers for the elements in the given equation: \(\mathrm{P}_{4}(s)+\mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{PH}_{3}(g)+\mathrm{Cl}_{2}(g)\). This will help us balance the equation later on. For P4: P is 0 (elemental form) For Cl-: Cl is -1 For PH3: P is -3, H is +1 (sum is 0) For Cl2: Cl is 0 (elemental form)

Write the oxidation and reduction half-reactions. The oxidation half-reaction involves P4 and PH3, while the reduction half-reaction involves Cl- and Cl2. Oxidation: \(\mathrm{P}_{4}(s) \longrightarrow \mathrm{PH}_{3}(g)\) Reduction: \(\mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{Cl}_{2}(g)\)

Balance the half-reactions by first balancing the atoms other than oxygen and hydrogen. Then, balance any oxygen or hydrogen atoms using H2O and H+. Oxidation: To balance the phosphorus atoms, add a coefficient of 4 for PH3: \(\mathrm{P}_{4}(s) \longrightarrow 4\mathrm{PH}_{3}(g)\) To balance the hydrogen atoms, add a coefficient of 12 for H+: \(\mathrm{P}_{4}(s) + 12\mathrm{H}^{+}(a q) \longrightarrow 4\mathrm{PH}_{3}(g)\) Reduction: To balance the chlorine atoms, add a coefficient of 2 for Cl-: \(2\mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{Cl}_{2}(g)\)

Balance the electrons in each half-reaction, and then combine them to get the balanced equation. Oxidation: The P atoms each lose 3 electrons, so multiply by 12 e- and add them to the right side: \(\mathrm{P}_{4}(s) + 12\mathrm{H}^{+}(a q) \longrightarrow 4\mathrm{PH}_{3}(g) + 12\mathrm{e}^{-}\) Reduction: The Cl atoms each gain 1 electron, so multiply by 2 e- and add them to the left side: \(2\mathrm{Cl}^{-}(a q) + 2\mathrm{e}^{-} \longrightarrow \mathrm{Cl}_{2}(g)\) Combine the two half-reactions: \(\mathrm{P}_{4}(s) + 12\mathrm{H}^{+}(a q) + 2\mathrm{Cl}^{-}(a q) \longrightarrow 4\mathrm{PH}_{3}(g)+\mathrm{Cl}_{2}(g) + 12\mathrm{e}^{-} + 2\mathrm{e}^{-}\) Cancel out the electrons: \(\mathrm{P}_{4}(s) + 12\mathrm{H}^{+}(a q) + 2\mathrm{Cl}^{-}(a q) \longrightarrow 4\mathrm{PH}_{3}(g)+\mathrm{Cl}_{2}(g)\)

The final balanced equation in acid solution for this given reaction is: \(\mathrm{P}_{4}(s) + 12\mathrm{H}^{+}(a q) + 2\mathrm{Cl}^{-}(a q) \longrightarrow 4\mathrm{PH}_{3}(g)+\mathrm{Cl}_{2}(g)\). Follow the same steps to balance equations (b), (c), and (d) in an acidic solution.