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Chapter 18: Problem 22

Balance the following oxidation–reduction reactions that occur in basic solution using the half-reaction method. a. $\mathrm{PO}_{3}^{3-}(a q)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{PO}_{4}^{3-}(a q)+\mathrm{MnO}_{2}(s)$ b. $\operatorname{Mg}(s)+\mathrm{OCl}^{-}(a q) \rightarrow \mathrm{Mg}(\mathrm{OH})_{2}(s)+\mathrm{Cl}^{-}(a q)$ c. $\mathrm{H}_{2} \mathrm{CO}(a q)+\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}+(a q) \rightarrow$ $$\mathrm{HCO}_{3}(a q)+\mathrm{Ag}(s)+\mathrm{NH}_{3}(a q)$$

Short Answer

Expert verified
The short answer for the balanced reactions are: a. \(3 \mathrm{PO}_{3}^{3-} + 2\: \mathrm{MnO}_{4}^{-} + 14\: \mathrm{OH}^{-} \rightarrow 3 \mathrm{PO}_{4}^{3-} + 2\: \mathrm{MnO}_{2} + 7\: \mathrm{H}_{2}\mathrm{O}\) b. \(2 \mathrm{Mg}(s) + 2 \mathrm{OCl}^{-}(a q) + 4 \mathrm{OH}^{-}(a q) \rightarrow 2 \mathrm{Mg}(\mathrm{OH})_{2}(s) + 2 \mathrm{Cl}^{-}(a q)\) c. \(3 \mathrm{H}_{2} \mathrm{CO}(a q) + 2 \mathrm{Ag}(\mathrm{NH}_{3})_{2}^{+}(a q) \rightarrow 3 \mathrm{HCO}_{3}^{-}(a q) + 2 \mathrm{Ag}(s) + 6 \mathrm{NH}_{3}(a q)\)

Step by step solution

01

Determine the oxidation states

Assign oxidation states for all elements in the reactants and products. Here: P: +3 (in \(\mathrm{PO}_{3}^{3-}\)), +5 (in \(\mathrm{PO}_{4}^{3-}\)) Mn: +7 (in \(\mathrm{MnO}_{4}^{-}\)), +4 (in \(\mathrm{MnO}_{2}\))
02

Write half-reactions

Write the oxidation half-reaction and reduction half-reaction. Oxidation: \(\mathrm{PO}_{3}^{3-}\rightarrow \mathrm{PO}_{4}^{3-}\) Reduction: \(\mathrm{MnO}_{4}^{-}\rightarrow \mathrm{MnO}_{2}\)
03

Balance atoms

Balance atoms other than oxygen and hydrogen in each half-reaction. Oxidation: \(\mathrm{PO}_{3}^{3-}\rightarrow \mathrm{PO}_{4}^{3-}\) (P atoms balanced) Reduction: \(\mathrm{MnO}_{4}^{-}\rightarrow \mathrm{MnO}_{2}\) (Mn atoms balanced)
04

Balance oxygen atoms

Add water molecules to balance the number of oxygen atoms in each half-reaction. Oxidation: \(\mathrm{PO}_{3}^{3-}\rightarrow \mathrm{PO}_{4}^{3-} + \mathrm{H}_{2}\mathrm{O}\) Reduction: \(\mathrm{MnO}_{4}^{-}\rightarrow \mathrm{MnO}_{2} + 2\: \mathrm{H}_{2}\mathrm{O}\)
05

Balance hydrogen atoms

Add hydroxide ions (\(\mathrm{OH}^-\)) to balance the number of hydrogen atoms in each half-reaction. Oxidation: \(\mathrm{PO}_{3}^{3-} + 2\: \mathrm{OH}^{-}\rightarrow \mathrm{PO}_{4}^{3-} + \mathrm{H}_{2}\mathrm{O}\) Reduction: \(\mathrm{MnO}_{4}^{-} + 4\: \mathrm{OH}^{-}\rightarrow \mathrm{MnO}_{2} + 2\: \mathrm{H}_{2}\mathrm{O}\)
06

Balance charge

Balance the charges in each half-reaction by adding electrons. Oxidation: \(\mathrm{PO}_{3}^{3-} + 2\: \mathrm{OH}^{-} + 2\: e^{-}\rightarrow \mathrm{PO}_{4}^{3-} + \mathrm{H}_{2}\mathrm{O}\) Reduction: \(\mathrm{MnO}_{4}^{-} + 4\: \mathrm{OH}^{-}\rightarrow \mathrm{MnO}_{2} + 2\: \mathrm{H}_{2}\mathrm{O} + 3\: e^{-}\)
07

Equalize electrons and add half-reactions

Equalize the number of electrons in both half-reactions and add the half-reactions. Oxidation: \(3\cdot (\mathrm{PO}_{3}^{3-} + 2\: \mathrm{OH}^{-} + 2\: e^{-}\rightarrow \mathrm{PO}_{4}^{3-} + \mathrm{H}_{2}\mathrm{O})\) Reduction: \(2\cdot (\mathrm{MnO}_{4}^{-} + 4\: \mathrm{OH}^{-}\rightarrow \mathrm{MnO}_{2} + 2\: \mathrm{H}_{2}\mathrm{O} + 3\: e^{-})\) Adding the equalized half-reactions: \(3 \mathrm{PO}_{3}^{3-} + 6\: \mathrm{OH}^{-} + 2\: \mathrm{MnO}_{4}^{-} + 8\: \mathrm{OH}^{-} \rightarrow 3 \mathrm{PO}_{4}^{3-} + 3\: \mathrm{H}_{2}\mathrm{O} + 2\: \mathrm{MnO}_{2} + 4\: \mathrm{H}_{2}\mathrm{O}\)
08

Simplify and write balanced equation

Simplify the equation and write the final balanced equation. \(3 \mathrm{PO}_{3}^{3-} + 2\: \mathrm{MnO}_{4}^{-} + 14\: \mathrm{OH}^{-} \rightarrow 3 \mathrm{PO}_{4}^{3-} + 2\: \mathrm{MnO}_{2} + 7\: \mathrm{H}_{2}\mathrm{O}\) Similarly, analyze and balance reactions b. and c. using the same step-by-step procedure in base.

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Most popular questions from this chapter

An electrochemical cell consists of a silver metal electrode immersed in a solution with \(\left[\mathrm{Ag}^{+}\right]=1.00 M\) separated by a porous disk from a compartment with a copper metal electrode immersed in a solution of 10.00\(M \mathrm{NH}_{3}\) that also contains $2.4 \times 10^{-3} \mathrm{M} \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+} .$ The equilibrium between \(\mathrm{Cu}^{2+}\) and \(\mathrm{NH}_{3}\) is: $$\mathrm{Cu}^{2+}(\mathrm{aq})+4 \mathrm{NH}_{3}(\mathrm{aq}) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(\mathrm{aq}) \qquad K=1.0 \times 10^{13}$$ and the two cell half-reactions are: $$\begin{array}{rl}{\mathrm{Ag}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}} & {\mathscr{E}^{\circ}=0.80 \mathrm{V}} \\ {\mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}} & {\mathscr{E}^{\circ}=0.34 \mathrm{V}}\end{array}$$ Assuming \(\mathrm{Ag}^{+}\) is reduced, what is the cell potential at \(25^{\circ} \mathrm{C} ?\)

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