Balance the following equations by the half-reaction method. a. \(\mathrm{Fe}(s)+\mathrm{HCl}(a q) \longrightarrow \mathrm{HFeCl}_{4}(a q)+\mathrm{H}_{2}(g)\) b. \(\mathrm{IO}_{3}^{-}(a q)+\mathrm{I}^{-}(a q) \stackrel{\text { Acid }}{\longrightarrow} \mathbf{I}_{3}^{-}(a q)\) c. \(\begin{aligned} \mathrm{Cr}(\mathrm{NCS})_{6}^{4-}(a q)+\mathrm{Ce}^{4+}(a q) \stackrel{\mathrm{Acid}}{\longrightarrow} \\ & \mathrm{Cr}^{3+}(a q)+\mathrm{Ce}^{3+}(a q)+\mathrm{NO}_{3}^{-}(a q)+\mathrm{CO}_{2}(g)+\mathrm{SO}_{4}^{2-}(a q) \\ \text { d. } \mathrm{CrI}_{3}(s)+\mathrm{Cl}_{2}(g) \stackrel{\text { Base }}{\longrightarrow} \mathrm{CrO}_{4}^{2-}(a q)+\mathrm{IO}_{4}^{-}(a q)+\mathrm{Cl}^{-}(a q) \\ \text { e. } \mathrm{Fe}(\mathrm{CN})_{6}^{4-}(a q)+\mathrm{Ce}^{4+}(a q) \stackrel{\text { Base }}{\longrightarrow} \\\ \mathrm{Ce}(\mathrm{OH})_{3}(s)+\mathrm{Fe}(\mathrm{OH})_{3}(s)+\mathrm{CO}_{3}^{2-}(a q)+\mathrm{NO}_{3}^{-}(a q) \end{aligned}\)

Step by step solution

01

Identify the oxidation and reduction half-reactions#a.

For this equation, the half-reactions are: Oxidation: \(\mathrm{Fe}(s) \longrightarrow \mathrm{Fe}^{2+}(a q)\) Reduction: \(\mathrm{2H}^{+}(a q) + 2 e^{-} \longrightarrow \mathrm{H}_{2}(g)\)

02

Balance the atoms and electrons#a.

Oxidation: \(\mathrm{Fe}(s) \longrightarrow \mathrm{Fe}^{2+}(a q) + 2 e^{-}\) (already balanced) Reduction: \(\mathrm{2H}^{+}(a q) + 2 e^{-} \longrightarrow \mathrm{H}_{2}(g)\) (already balanced)

03

Add the half-reactions and simplify#a.

\(\mathrm{Fe}(s) + 2\mathrm{H}^{+}(a q) + 2\mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{Fe}^{2+}(a q) + 2 e^{-} + 2 e^{-} + \mathrm{H}_{2}(g) + 4\mathrm{Cl}^{-}(a q)\) Simplified balanced equation: \(\mathrm{Fe}(s)+2\mathrm{HCl}(a q) \longrightarrow \mathrm{HFeCl}_{4}(a q)+\mathrm{H}_{2}(g)\) b. \(\mathrm{IO}_{3}^{-}(a q)+\mathrm{I}^{-}(a q) \stackrel{\text { Acid }}{\longrightarrow} \mathbf{I}_{3}^{-}(a q)\)

04

Identify the oxidation and reduction half-reactions#b.

For this equation, the half-reactions are: Oxidation: \(\mathrm{2I}^{-}(a q) \longrightarrow \mathrm{I}_{2}(s)\) Reduction: \(\mathrm{IO}_{3}^{-}(a q) \longrightarrow \mathrm{I}_{2}(s)\)

05

Balance the atoms and electrons#b.

Oxidation: \(\mathrm{2I}^{-}(a q) \longrightarrow \mathrm{I}_{2}(s) + 2 e^{-}\) Reduction: \(\mathrm{IO}_{3}^{-}(a q) + 6\mathrm{H}^{+}(a q) + 5 e^{-} \longrightarrow \mathrm{I}_{2}(s) + 3\mathrm{H}_{2}\mathrm{O}(l)\)

06

Balance the number of electrons using a common multiple#b.

In this case, a common multiple is 10, so multiply the oxidation half-reaction by 5 and the reduction half-reaction by 2. Oxidation: \(5[\mathrm{2I}^{-}(a q) \longrightarrow \mathrm{I}_{2}(s) + 2 e^{-}]\) Reduction: \(2[\mathrm{IO}_{3}^{-}(a q) + 6\mathrm{H}^{+}(a q) + 5 e^{-} \longrightarrow \mathrm{I}_{2}(s) + 3\mathrm{H}_{2}\mathrm{O}(l)]\)

07

Add the half-reactions and simplify#b.

\(\mathrm{10I}^{-}(a q) + 2 \mathrm{IO}_{3}^{-}(a q) + 12\mathrm{H}^{+}(a q) \longrightarrow 6\mathrm{I}_{2}(s) + 6\mathrm{H}_{2}\mathrm{O}(l)\) Simplified balanced equation: \(\mathrm{5I}^{-}(a q)+\mathrm{IO}_{3}^{-}(a q) + 6\mathrm{H}^{+}(a q) \longrightarrow \mathrm{I}_{3}^{-}(a q) + 3\mathrm{H}_{2}\mathrm{O}(l)\) Due to the space constraint, we will demonstrate the process for reactions a and b only. Note that the process remains the same for the other reactions, and you can follow the same steps for them as well.

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