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Chapter 4: Problem 55

Wlassify each of the following half-equations as oxidation or reduction and balance. (a) (acidic) \(\quad \mathrm{Mn}^{2+}(a q) \longrightarrow \mathrm{MnO}_{4}^{-}(a q)\) (b) (basic) \(\quad \mathrm{CrO}_{4}^{2-}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q)\) (c) (basic) \(\quad \mathrm{PbO}_{2}(s) \longrightarrow \mathrm{Pb}^{2+}(a q)\) (d) (acidic) \(\quad \mathrm{ClO}_{2}^{-}(a q) \longrightarrow \mathrm{ClO}^{-}(a q)\)

Short Answer

Expert verified
Question: Identify and balance the given half-equations as oxidation or reduction: (a) (acidic) ℳn^2+(aq) ➡ ℳnO₄^-(aq) (b) (basic) CrO₄^(2-)(aq) ➡ Cr^(3+)(aq) (c) (basic) PbO₂(s) ➡ Pb^(2+)(aq) (d) (acidic) ClO₂^-(aq) ➡ ClO^-(aq) Answer: (a) Oxidation: 5H+(aq) + Mn^2+(aq) + 5e^- ➡ MnO₄^-(aq) + 4H₂O(l) (b) Reduction: 2e^- + CrO₄^2-(aq) + 4H₂O(l) ➡ Cr^3+(aq) + 8OH^-(aq) (c) Reduction: 2e^- + PbO₂(s) ➡ Pb^2+(aq) + 2OH^-(aq) (d) Reduction: 2H+(aq) + ClO₂^-(aq) + e^- ➡ ClO^-(aq) + H₂O(l)

Step by step solution

01

Determine oxidation numbers

Assign oxidation numbers to the elements on both sides of the equation: Mn has an oxidation number of +2 on the left side and Mn in MnO4- has an oxidation number of +7.
02

Identify oxidation or reduction

Since the oxidation number of Mn increases from +2 to +7, this half-equation represents an oxidation process.
03

Balance the half-equation

Balance the atoms other than oxygen and hydrogen first. In this case, Mn is already balanced. Then, balance the oxygens by adding water molecules and balance hydrogen with H+: \(\mathrm{Mn}^{2+}(a q) \longrightarrow \mathrm{MnO}_{4}^{-}(a q) + 4 \mathrm{H}_{2}\mathrm{O}(l)\) and \(5 \mathrm{H}^{+}(a q) + \mathrm{Mn}^{2+}(a q) \longrightarrow \mathrm{MnO}_{4}^{-}(a q) + 4 \mathrm{H}_{2}\mathrm{O}(l)\). Finally, balance the charge by adding electrons: \(5 \mathrm{H}^{+}(a q) + \mathrm{Mn}^{2+}(a q) + 5 \mathrm{e}^{-} \longrightarrow \mathrm{MnO}_{4}^{-}(a q) + 4 \mathrm{H}_{2}\mathrm{O}(l)\). (b) (basic) \(\quad \mathrm{CrO}_{4}^{2-}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q)\)
04

Determine oxidation numbers

Assign oxidation numbers to the elements on both sides of the equation: Cr in CrO4^(2-) has an oxidation number of +6 and Cr^(3+) has an oxidation number of +3.
05

Identify oxidation or reduction

Since the oxidation number of Cr decreases from +6 to +3, this half-equation represents a reduction process.
06

Balance the half-equation

Balance the atoms other than oxygen and hydrogen first. In this case, Cr is already balanced. Then, balance the oxygens by adding water molecules and balance hydrogen by adding OH- ions: \(\mathrm{CrO}_{4}^{2-}(a q) + 4 \mathrm{H}_{2}\mathrm{O}(l) \longrightarrow \mathrm{Cr}^{3+}(a q)\) and \(\mathrm{CrO}_{4}^{2-}(a q) + 4 \mathrm{H}_{2}\mathrm{O}(l) \longrightarrow \mathrm{Cr}^{3+}(a q) + 8\mathrm{OH}^{-}(a q)\). Finally, balance the charge by adding electrons: \(2 \mathrm{e}^{-} + \mathrm{CrO}_{4}^{2-}(a q) + 4 \mathrm{H}_{2}\mathrm{O}(l) \longrightarrow \mathrm{Cr}^{3+}(a q) + 8\mathrm{OH}^{-}(a q)\). (c) (basic) \(\quad \mathrm{PbO}_{2}(s) \longrightarrow \mathrm{Pb}^{2+}(a q)\)
07

Determine oxidation numbers

Assign oxidation numbers to the elements on both sides of the equation: Pb in PbO2 has an oxidation number of +4 and Pb^(2+) has an oxidation number of +2.
08

Identify oxidation or reduction

Since the oxidation number of Pb decreases from +4 to +2, this half-equation represents a reduction process.
09

Balance the half-equation

Balance the atoms other than oxygen and hydrogen first. In this case, Pb is already balanced. Then, balance the oxygens by adding water molecules and balance hydrogen by adding OH- ions: \(\mathrm{PbO}_{2}(s) \longrightarrow \mathrm{Pb}^{2+}(a q) + 2\mathrm{OH}^{-}(a q)\). Finally, balance the charge by adding electrons: \(2 \mathrm{e}^{-} + \mathrm{PbO}_{2}(s) \longrightarrow \mathrm{Pb}^{2+}(a q) + 2\mathrm{OH}^{-}(a q)\). (d) (acidic) \(\quad \mathrm{ClO}_{2}^{-}(a q) \longrightarrow \mathrm{ClO}^{-}(a q)\)
10

Determine oxidation numbers

Assign oxidation numbers to the elements on both sides of the equation: Cl in ClO2- has an oxidation number of +3 and Cl in ClO- has an oxidation number of +1.
11

Identify oxidation or reduction

Since the oxidation number of Cl decreases from +3 to +1, this half-equation represents a reduction process.
12

Balance the half-equation

Balance the atoms other than oxygen and hydrogen first. In this case, Cl is already balanced. Then, balance the oxygens by adding water molecules and balance hydrogen with H+: \(\mathrm{ClO}_{2}^{-}(a q) \longrightarrow \mathrm{ClO}^{-}(a q) + \mathrm{H}_{2}\mathrm{O}(l)\) and \(\mathrm{ClO}_{2}^{-}(a q) + 2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{ClO}^{-}(a q) + \mathrm{H}_{2}\mathrm{O}(l)\). Finally, balance the charge by adding electrons: \(2 \mathrm{H}^{+}(a q) + \mathrm{ClO}_{2}^{-}(a q) + \mathrm{e}^{-} \longrightarrow \mathrm{ClO}^{-}(a q) + \mathrm{H}_{2}\mathrm{O}(l)\).

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Most popular questions from this chapter

Consider the following balanced redox reaction in basic medium. \(3 \mathrm{Sn}^{2+}(a q)+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+4 \mathrm{H}_{2} \mathrm{O} \longrightarrow\) \(3 \mathrm{Sn}^{4+}(a q)+\mathrm{Cr}_{2} \mathrm{O}_{3}(s)+8 \mathrm{OH}^{-}(a q)\) (a) What is the oxidizing agent? (b) What species has the element that increases its oxidation number? (c) What species contains the element with the highest oxidation number? (d) If the reaction were to take place in acidic medium, what species would not be included in the reaction?

A solution contains both iron(II) and iron(III) ions. A \(50.00-\mathrm{mL}\) sample of the solution is titrated with \(35.0 \mathrm{~mL}\) of \(0.0280 \mathrm{M} \mathrm{KMnO}_{4}\), which oxidizes \(\mathrm{Fe}^{2+}\) to \(\mathrm{Fe}^{3+} .\) The permanganate ion is reduced to manganese(II) ion. Another \(50.00-\mathrm{mL}\) sample of the solution is treated with zinc, which reduces all the \(\mathrm{Fe}^{3+}\) to \(\mathrm{Fe}^{2+}\). The resulting solution is again titrated with \(0.0280 \mathrm{M}\) \(\mathrm{KMnO}_{4} ;\) this time \(48.0 \mathrm{~mL}\) is required. What are the concentrations of \(\mathrm{Fe}^{2+}\) and \(\mathrm{Fe}^{3+}\) in the solution?

Write a balanced net ionic equation for each of the following acid-base reactions in water. (a) acetic acid \(\left(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)\) with strontium hydroxide (b) diethylamine, \(\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{NH}\), with sulfuric acid (c) aqueous hydrogen cyanide (HCN) with sodium hydroxide

Stomach acid is approximately \(0.020 \mathrm{M} \mathrm{HCl}\). What volume of this acid is neutralized by an antacid tablet that weighs \(330 \mathrm{mg}\) and contains \(41.0 \% \mathrm{Mg}(\mathrm{OH})_{2}, 36.2 \% \mathrm{NaHCO}_{3}\), and \(22.8 \% \mathrm{NaCl} ?\) The reactions in- volved are $$ \begin{gathered} \mathrm{Mg}(\mathrm{OH})_{2}(s)+2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{Mg}^{2+}(a q)+2 \mathrm{H}_{2} \mathrm{O} \\ \mathrm{HCO}_{3}^{-}(a q)+\mathrm{H}^{+}(a q) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O} \end{gathered} $$

Consider the following generic equation $$ \mathrm{OH}^{-}(a q)+\mathrm{HB}(a q) \longrightarrow \mathrm{B}^{-}(a q)+\mathrm{H}_{2} \mathrm{O} $$ For which of the following pairs would this be the correct prototype equation for the acid-base reaction in solution? If it is not correct, write the proper equation for the acid-base reaction between the pair. (a) hydrochloric acid and pyridine, \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}\) (b) sulfuric acid and rubidium hydroxide (c) potassium hydroxide and hydrofluoric acid (d) ammonia and hydriodic acid (e) strontium hydroxide and hydrocyanic acid
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