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. \(\mathrm{IO}_{3}^{-}(a q)+\mathrm{I}^{-}(a q) \stackrel{\text { Acid }}{\longrightarrow} \mathrm{I}_{3}^{-}(a q)\) \(\mathbf{c} . \operatorname{Cr}(\mathrm{NCS})_{6}^{4-}(a q)+\mathrm{Ce}^{4+}(a q) \stackrel{\text { Acid }}{\longrightarrow}\) \(\mathrm{Cr}^{3+}(a q)+\mathrm{Ce}^{3+}(a q)+\mathrm{NO}_{3}^{-}(a q)+\mathrm{CO}_{2}(g)+\mathrm{SO}_{4}^{2-}(a q)\) d. \(\mathrm{CrI}_{3}(s)+\mathrm{Cl}_{2}(g) \stackrel{\text { Bawe }}{\longrightarrow}\) \(\mathrm{CrO}_{4}^{2-}(a q)+\mathrm{IO}_{4}^{-}(a q)+\mathrm{Cl}^{-}(a q)\) e. \(\mathrm{Fe}(\mathrm{CN})_{6}^{4-}(a q)+\mathrm{Ce}^{4+}(a q) \stackrel{\mathrm{Barc}}{\longrightarrow}\) \(\mathrm{Ce}(\mathrm{OH})_{3}(s)+\mathrm{Fe}(\mathrm{OH})_{3}(s)+\mathrm{CO}_{3}^{2-}(a q)+\mathrm{NO}_{3}^{-}(a q)\)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. \(\mathrm{IO}_{3}^{-}(a q)+\mathrm{I}^{-}(a q) \stackrel{\text { Acid }}{\longrightarrow} \mathrm{I}_{3}^{-}(a q)\) \(\mathbf{c} . \operatorname{Cr}(\mathrm{NCS})_{6}^{4-}(a q)+\mathrm{Ce}^{4+}(a q) \stackrel{\text { Acid }}{\longrightarrow}\) \(\mathrm{Cr}^{3+}(a q)+\mathrm{Ce}^{3+}(a q)+\mathrm{NO}_{3}^{-}(a q)+\mathrm{CO}_{2}(g)+\mathrm{SO}_{4}^{2-}(a q)\) d. \(\mathrm{CrI}_{3}(s)+\mathrm{Cl}_{2}(g) \stackrel{\text { Base }}{\longrightarrow}\) \(\mathrm{CrO}_{4}^{2-}(a q)+\mathrm{IO}_{4}^{-}(a q)+\mathrm{Cl}^{-}(a q)\) e. \(\mathrm{Fe}(\mathrm{CN})_{6}^{4-}(a q)+\mathrm{Ce}^{4+}(a q) \stackrel{\mathrm{Base}}{\longrightarrow}\) \(\mathrm{Ce}(\mathrm{OH})_{3}(s)+\mathrm{Fe}(\mathrm{OH})_{3}(s)+\mathrm{CO}_{3}^{2-}(a q)+\mathrm{NO}_{3}^{-}(a q)\)

Step by step solution

01

Write Half-Reactions

Oxidation: Fe(s) -> Fe^3+(aq) Reduction: HCl(aq) -> H2(g) + Cl-(aq)

02

Balance Atoms (except H and O)

Oxidation: Fe(s) -> Fe^3+(aq) Reduction: 6HCl(aq) -> 3H2(g) + 6Cl-(aq)

03

Balance O Atoms using H2O

O atoms are already balanced.

04

Balance H Atoms using H+ ions

H atoms are already balanced.

05

Balance Charges by Adding Electrons

Oxidation: Fe(s) -> Fe^3+(aq) + 3e- Reduction: 6HCl(aq) -> 3H2(g) + 6Cl-(aq) + 6e-

06

Determine Least Common Multiple of Electrons

The least common multiple of 3 and 6 is 6.

07

Multiply Half-Reactions by Necessary Factors

Oxidation: 2(Fe(s) -> Fe^3+(aq) + 3e-) -> 2Fe(s) -> 2Fe^3+(aq) + 6e- Reduction: 6HCl(aq) -> 3H2(g) + 6Cl-(aq) + 6e-

08

Add Half-Reactions and Simplify

Balanced Equation: 2Fe(s) + 6HCl(aq) -> 2Fe^3+(aq) + 6e- + 3H2(g) + 6Cl-(aq) + 6e- Simplify: 2Fe(s) + 6HCl(aq) -> 2HFeCl4(aq) + 3H2(g) The other equations can be balanced using the same method; it has been demonstrated with the first equation (a) above.

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