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Chapter 18: Problem 31

Balance the following oxidation-reduction reactions that occur in basic solution. a. \(\mathrm{Al}(s)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{MnO}_{2}(s)+\mathrm{Al}(\mathrm{OH})_{4}^{-}(a q)\) b. \(\mathrm{Cl}_{2}(g) \rightarrow \mathrm{Cl}^{-}(a q)+\mathrm{OCl}^{-}(a q)\) c. \(\mathrm{NO}_{2}^{-}(a q)+\mathrm{Al}(s) \rightarrow \mathrm{NH}_{3}(g)+\mathrm{AlO}_{2}^{-}(a q)\)

Short Answer

Expert verified
The short answer is: For reaction a, the balanced equation is: \( 3\mathrm{Al}(s) + 4\mathrm{MnO}_4^-(aq) + 12\mathrm{OH}^-(aq) \rightarrow 3\mathrm{Al}(\mathrm{OH})_{4}^-(aq) + 4\mathrm{MnO}_2(s) \)

Step by step solution

01

Identify the oxidation and reduction half-reactions

Oxidation half-reaction: \(\mathrm{Al}(s) \rightarrow \mathrm{Al}(\mathrm{OH})_{4}^{-}(a q)\) Reduction half-reaction: \(\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{MnO}_{2}(s)\)
02

Balance the half-reactions

Oxidation: \(\mathrm{Al}(s) \rightarrow \mathrm{Al}^{3+}+\cancel{4 e^-}+\mathrm{OH}^-\) Reduction: \(\mathrm{MnO}_{4}^{-}+\cancel{3 e^-} \rightarrow \mathrm{MnO}_{2}\)
03

Multiply half-reactions to make the number of electrons equal

The oxidation half-reaction already loses 4 electrons and the reduction half-reaction gains 3 electrons. Therefore, multiply the oxidation half-reaction by 3 and the reduction half-reaction by 4. Oxidation: \( 3\mathrm{Al}(s) \rightarrow 3\mathrm{Al}^{3+}+\cancel{12 e^-}+3\mathrm{OH}^-\) Reduction: \( 4\mathrm{MnO}_{4}^{-}+\cancel{12 e^-} \rightarrow 4\mathrm{MnO}_{2}\)
04

Add the half-reactions, cancel common terms, and balance mass and charge

\( 3\mathrm{Al}(s) + 4\mathrm{MnO}_{4}^{-}(a q) \rightarrow 3\mathrm{Al}^{3+} + 3\mathrm{OH}^- + 4\mathrm{MnO}_{2}(s)\)
05

Balance hydrogens and oxygens using water and hydroxide ions

Add 12 hydroxide ions to both sides: \( 3\mathrm{Al}(s) + 4\mathrm{MnO}_{4}^{-}(a q) + 12\mathrm{OH}^{-} \rightarrow 3\mathrm{Al}^{3+} + 15\mathrm{OH}^- + 4\mathrm{MnO}_{2}(s)\) Combine the aluminum and hydroxide ions, and cancel common terms: \( 3\mathrm{Al}(s) + 4\mathrm{MnO}_{4}^{-}(a q) + 12\mathrm{OH}^{-} \rightarrow 3\mathrm{Al}\mathrm{(OH)}_{4}^{-}(a q) + 4\mathrm{MnO}_{2}(s)\) The balanced equation for reaction a is: \( 3\mathrm{Al}(s) + 4\mathrm{MnO}_4^-(aq) + 12\mathrm{OH}^-(aq) \rightarrow 3\mathrm{Al}(\mathrm{OH})_{4}^-(aq) + 4\mathrm{MnO}_2(s) \)

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Most popular questions from this chapter

Consider only the species (at standard conditions) $$ \mathrm{Br}^{-}, \mathrm{Br}_{2}, \mathrm{H}^{+}, \quad \mathrm{H}_{2}, \quad \mathrm{La}^{3+}, \quad \mathrm{Ca}, \quad \mathrm{Cd} $$ in answering the following questions. Give reasons for your answers. a. Which is the strongest oxidizing agent? b. Which is the strongest reducing agent? c. Which species can be oxidized by \(\mathrm{MnO}_{4}^{-}\) in acid? d. Which species can be reduced by \(\mathrm{Zn}(s)\) ?

The Ostwald process for the commercial production of nitric acid involves the following three steps: $$ \begin{aligned} 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) & \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{NO}_{2}(g) \\ 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g) \end{aligned} $$ a. Which reactions in the Ostwald process are oxidationreduction reactions? b. Identify each oxidizing agent and reducing agent.

The compound with the formula \(\mathrm{TII}_{3}\) is a black solid. Given the following standard reduction potentials, $$ \begin{aligned} \mathrm{Tl}^{3+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Tl}^{+} & \mathscr{E}^{\circ}=1.25 \mathrm{~V} \\ \mathrm{I}_{3}^{-}+2 \mathrm{e}^{-} \longrightarrow 3 \mathrm{I}^{-} & \mathscr{E}^{\circ}=0.55 \mathrm{~V} \end{aligned} $$ would you formulate this compound as thallium(III) iodide or thallium(I) triiodide?

Assign oxidation numbers to all the atoms in each of the following: a. \(\mathrm{HNO}_{3}\) e. \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) i. \(\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) b. \(\mathrm{CuCl}_{2}\) f. \(\mathrm{Ag}\) j. \(\mathrm{CO}_{2}\) c. \(\mathrm{O}_{2}\) g. \(\mathrm{PbSO}_{4}\) k. \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Ce}\left(\mathrm{SO}_{4}\right)_{3}\) d. \(\mathrm{H}_{2} \mathrm{O}_{2}\) h. \(\mathrm{PbO}_{2}\) 1\. \(\mathrm{Cr}_{2} \mathrm{O}_{3}\)

The solubility product for \(\operatorname{CuI}(s)\) is \(1.1 \times 10^{-12}\). Calculate the value of \(\mathscr{E}^{\circ}\) for the half-reaction $$ \operatorname{CuI}(s)+\mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s)+\mathrm{I}^{-}(a q) $$
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