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Chapter 22: Problem 44

An aqueous solution of \(\mathrm{SO}_{2}\) reduces (a) aqueous \(\mathrm{KMnO}_{4}\) to \(\mathrm{MnSO}_{4}(a q),(\mathbf{b})\) acidic aqueous \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) to aqueous \(\mathrm{Cr}^{3+}\) (c) aqueous \(\mathrm{Hg}_{2}\left(\mathrm{NO}_{3}\right)_{2}\) to mercury metal. Write balanced equations for these reactions.

Short Answer

Expert verified
The balanced equations for the three reactions are: 1. \(5 SO_{2} + 2 MnO_{4}^{-} + 2 H^{+} \rightarrow 5 SO_{4}^{2-} + 2 Mn^{2+} + H_{2}O\) 2. \(3 SO_{2} + 2 Cr_{2}O_{7}^{2-}+ 14 H^{+} \rightarrow 3 SO_{4}^{2-} + 4 Cr^{3+} + 7 H_{2}O\) 3. \(SO_{2} + 2 Hg^{2+} + 2 H^{+} \rightarrow SO_{4}^{2-} + 2 Hg\)

Step by step solution

01

In all three reactions, SO2 acts as a reducing agent, meaning it loses electrons. Thus, the SO2 species will be involved in the oxidation half-reaction. In each case, we need to find the oxidation state of the species involving KMnO4, K2Cr2O7, and Hg2(NO3)2. #Step 2: Write the oxidation half-reaction for SO2#

SO2 gets oxidized to SO4^2-. In this process, each S atom changes its oxidation state from +4 to +6, and loses 2 electrons: \(SO_{2} \rightarrow SO_{4}^{2-} + 2 e^-\) #Step 3: Write the reduction half-reaction for KMnO4 reaction#
02

KMnO4 gets reduced to MnSO4. In this process, each Mn atom changes its oxidation state from +7 to +2, and gains 5 electrons: \(MnO_{4}^{-} + 5 e^{-} \rightarrow Mn^{2+}\) #Step 4: Combine the oxidation and reduction half-reactions for the KMnO4 reaction#

Multiply the oxidation half-reaction by 5 and the reduction half-reaction by 2 to equalize the number of electrons, and then add them together: \(5(SO_{2} \rightarrow SO_{4}^{2-} + 2 e^{-})\) \(2(MnO_{4}^{-} + 5 e^{-} \rightarrow Mn^{2+})\) The final balanced equation is: \(5 SO_{2} + 2 MnO_{4}^{-} + 2 H^{+} \rightarrow 5 SO_{4}^{2-} + 2 Mn^{2+} + H_{2}O\) #Step 5: Write the reduction half-reaction for the K2Cr2O7 reaction#
03

K2Cr2O7 gets reduced to Cr^3+. In this process, each Cr atom changes its oxidation state from +6 to +3, and gains 3 electrons: \(Cr_{2}O_{7}^{2-} + 6 e^{-} \rightarrow 2 Cr^{3+}\) #Step 6: Combine the oxidation and reduction half-reactions for the K2Cr2O7 reaction#

Multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2 to equalize the number of electrons, and then add them together: \(3(SO_{2} \rightarrow SO_{4}^{2-} + 2 e^{-})\) \(2(Cr_{2}O_{7}^{2-} + 6 e^{-} \rightarrow 2 Cr^{3+})\) The final balanced equation is: \(3 SO_{2} + 2 Cr_{2}O_{7}^{2-}+ 14 H^{+} \rightarrow 3 SO_{4}^{2-} + 4 Cr^{3+} + 7 H_{2}O\) #Step 7: Write the reduction half-reaction for the Hg2(NO3)2 reaction#
04

Hg2(NO3)2 gets reduced to Hg (mercury metal). In this process, each Hg atom changes its oxidation state from +2 to 0, and gains 2 electrons: \(Hg^{2+} + 2 e^{-} \rightarrow Hg\) #Step 8: Combine the oxidation and reduction half-reactions for the Hg2(NO3)2 reaction#

Multiply the oxidation half-reaction by 1 (no need to change) and the reduction half-reaction by 2 to equalize the number of electrons, and then add them together: \(SO_{2} \rightarrow SO_{4}^{2-} + 2 e^{-}\) \(2(Hg^{2+} + 2 e^{-} \rightarrow Hg)\) The final balanced equation is: \(SO_{2} + 2 Hg^{2+} + 2 H^{+} \rightarrow SO_{4}^{2-} + 2 Hg\) The balanced equations for the three reactions are: 1. \(5 SO_{2} + 2 MnO_{4}^{-} + 2 H^{+} \rightarrow 5 SO_{4}^{2-} + 2 Mn^{2+} + H_{2}O\) 2. \(3 SO_{2} + 2 Cr_{2}O_{7}^{2-}+ 14 H^{+} \rightarrow 3 SO_{4}^{2-} + 4 Cr^{3+} + 7 H_{2}O\) 3. \(SO_{2} + 2 Hg^{2+} + 2 H^{+} \rightarrow SO_{4}^{2-} + 2 Hg\)

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Most popular questions from this chapter

Write a balanced equation for the reaction of each of the following compounds with water: (a) \(\mathrm{PCl}_{5}(s),\) (b) \(\mathrm{CO}_{2}(g)\) (c) \(\mathrm{K}_{2} \mathrm{O}_{2}(s)\) (d) \(\mathrm{Mg}_{3} \mathrm{P}_{2}(s)\) (e) \(\operatorname{LiAlH}_{4}(s)\) (f) \(\mathrm{Cl}_{2} \mathrm{O}(g)\) (g) \(\mathrm{NO}_{2}(g)\)

Complete and balance the following equations: (a) \(\mathrm{CO}_{2}(g)+\mathrm{OH}^{-}(a q) \longrightarrow\) (b) \(\mathrm{NaHCO}_{3}(s)+\mathrm{H}^{+}(a q) \longrightarrow\) (c) \(\mathrm{CaO}(s)+\mathrm{C}(s) \stackrel{\Delta}{\longrightarrow}\) (d) $\mathrm{C}(s)+\mathrm{H}_{2} \mathrm{O}(g) \stackrel{\Delta}{\longrightarrow}$ (e) \(\mathrm{CuO}(s)+\mathrm{CO}(g) \longrightarrow\)

Using \(\Delta G_{f}^{\circ}\) for \(\mathrm{NO}\) and \(\mathrm{NO}_{2},\) from Appendix \(\mathrm{C},\) calculate the equilibrium constant for the oxidation of \(\mathrm{NO}\) to \(\mathrm{NO}_{2}\) at \(298.0 \mathrm{~K}\) as described in equation \(22.38 .\)

Which of the following statements are true? (a) Both nitrogen and phosphorus can form a pentafluoride compound. (b) Although CO is a well-known compound, SiO does not exist under ordinary conditions. (c) \(\mathrm{Cl}_{2}\) is easier to oxidize than \(\mathrm{I}_{2}\). (d) At room temperature, the stable form of oxygen is \(\mathrm{O}_{2}\), whereas that of sulfur is \(S_{8}\)

Write balanced equations for each of the following reactions. (a) When mercury(II) oxide is heated, it decomposes to form \(\mathrm{O}_{2}\) and mercury metal. ( \(\mathbf{b}\) ) When copper(II) nitrate is heated strongly, it decomposes to form copper(II) oxide, nitrogen dioxide, and oxygen. (c) Lead(II) sulfide, \(\mathrm{PbS}(s)\) reacts with ozone to form \(\mathrm{PbSO}_{4}(s)\) and \(\mathrm{O}_{2}(g) .\) (d) When heated in air, \(\mathrm{ZnS}(s)\) is converted to \(\mathrm{ZnO}\). (e) Potassium peroxide reacts with \(\mathrm{CO}_{2}(g)\) to give potassium carbonate and \(\mathrm{O}_{2}\). (f) Oxygen is converted to ozone in the upper atmosphere.
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