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Chapter 22: Problem 45

In aqueous solution, hydrogen sulfide reduces (a) \(\mathrm{Fe}^{3+}\) to \(\mathrm{Fe}^{2+}\), (b) \(\mathrm{Br}_{2}\) to \(\mathrm{Br}^{-}\), (c) \(\mathrm{MnO}_{4}^{-}\) to \(\mathrm{Mn}^{2+}\), (d) \(\mathrm{HNO}_{3}\) to \(\mathrm{NO}_{2} .\) In all cases, under appropriate conditions, the product is elemental sulfur. Write a balanced net ionic equation for each reaction.

Short Answer

Expert verified
The balanced net ionic equations for the four reactions are: (a) \(2H_{2}S + 2Fe^{3+} + 4e^{-} \rightarrow 2Fe^{2+} + 2S + 4H^{+}\) (b) \(H_{2}S + Br_{2} \rightarrow 2Br^{-} + S + 2H^{+}\) (c) \(5H_{2}S + 2MnO_{4}^{-} + 16H^{+} \rightarrow 2Mn^{2+} + 5S + 8H_{2}O + 10e^{-}\) (d) \(3H_{2}S + 4HNO_{3} + 2e^{-} \rightarrow 3NO_{2} + 3S + 4H_{2}O + 6H^{+}\)

Step by step solution

01

Write the unbalanced equation

Begin by writing down the unbalanced equation for this reaction: \(H_{2}S + Fe^{3+} \rightarrow Fe^{2+} + S\)
02

Balance the atoms

Balance the atoms in the equation, starting with elements that appear only once on each side, excluding hydrogen and oxygen: \(2H_{2}S + 2Fe^{3+} \rightarrow 2Fe^{2+} + 2S\)
03

Balance the charge

Make sure the charge is balanced on both sides of the equation: \(2H_{2}S + 2Fe^{3+} + 4e^{-} \rightarrow 2Fe^{2+} + 2S + 4H^{+}\) Now, the net ionic equation for the first reaction is: \(2H_{2}S + 2Fe^{3+} + 4e^{-} \rightarrow 2Fe^{2+} + 2S + 4H^{+}\) (b) Hydrogen sulfide reduces \(\mathrm{Br}_{2}\) to \(\mathrm{Br}^{-}\):
04

Write the unbalanced equation

Begin with the unbalanced equation for this reaction: \(H_{2}S + Br_{2} \rightarrow Br^{-} + S\)
05

Balance the atoms

Balance the atoms, again starting with elements that appear only once on each side: \(H_{2}S + Br_{2} \rightarrow 2Br^{-} + S + 2H^{+}\)
06

Balance the charge

Check that the charge is balanced on both sides of the equation: Now, the net ionic equation for the second reaction is: \(H_{2}S + Br_{2} \rightarrow 2Br^{-} + S + 2H^{+}\) (c) Hydrogen sulfide reduces \(\mathrm{MnO}_{4}^{-}\) to \(\mathrm{Mn}^{2+}\):
07

Write the unbalanced equation

Start with the unbalanced equation for this reaction: \(H_{2}S + MnO_{4}^{-} \rightarrow Mn^{2+} + S\)
08

Balance the atoms

Balance the atoms, starting with elements that appear only once on each side, excluding hydrogen and oxygen: \(5H_{2}S + 2MnO_{4}^{-} \rightarrow 2Mn^{2+} + 5S\)
09

Balance the oxygen atoms and charge

Add water molecules to balance the oxygen atoms, and add hydrogen ions and electrons to balance the charge on both sides of the equation: \(5H_{2}S + 2MnO_{4}^{-} + 16H^{+} \rightarrow 2Mn^{2+} + 5S + 8H_{2}O + 10e^{-}\) Now, the net ionic equation for the third reaction is: \(5H_{2}S + 2MnO_{4}^{-} + 16H^{+} \rightarrow 2Mn^{2+} + 5S + 8H_{2}O + 10e^{-}\) (d) Hydrogen sulfide reduces \(\mathrm{HNO}_{3}\) to \(\mathrm{NO}_{2}\):
10

Write the unbalanced equation

Write down the unbalanced equation for this reaction: \(H_{2}S + HNO_{3} \rightarrow NO_{2} + S\)
11

Balance the atoms

Balance the atoms, starting with elements that appear only once on each side, excluding hydrogen and oxygen: \(3H_{2}S + 4HNO_{3} \rightarrow 3NO_{2} + 3S\)
12

Balance the oxygen atoms and charge

Add water molecules to balance the oxygen atoms, and add hydrogen ions and electrons to balance the charge on both sides of the equation: \(3H_{2}S + 4HNO_{3} + 2e^{-} \rightarrow 3NO_{2} + 3S + 4H_{2}O + 6H^{+}\) Now, the net ionic equation for the fourth reaction is: \(3H_{2}S + 4HNO_{3} + 2e^{-} \rightarrow 3NO_{2} + 3S + 4H_{2}O + 6H^{+}\)

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Most popular questions from this chapter

Write a balanced equation for each of the following reactions. (You may have to guess at one or more of the reaction products, but you should be able to make a reasonable guess, based on your study of this chapter.) (a) Hydrogen selenide can be prepared by reaction of an aqueous acid solution on aluminum selenide. (b) Sodium thiosulfate is used to remove excess \(\mathrm{Cl}_{2}\) from chlorine-bleached fabrics. The thiosulfate ion forms \(\mathrm{SO}_{4}^{2-}\) and elemental sulfur, while \(\mathrm{Cl}_{2}\) is reduced to \(\mathrm{Cl}^{-}\).

The atomic and ionic radii of the first three group \(6 \mathrm{~A}\) elements are (a) Explain why the atomic radius increases in moving downward in the group. (b) Explain why the ionic radii are larger than the atomic radii. (c) Which of the three anions would you expect to be the strongest base in water? Explain. [Sections 22.5 and 22.6\(]\)

Write complete balanced half-reactions for (a) oxidation of nitrous acid to nitrate ion in acidic solution, (b) oxidation of \(\mathrm{N}_{2}\) to \(\mathrm{N}_{2} \mathrm{O}\) in acidic solution.

Write the Lewis structure for each of the following species, and indicate the structure of each: (a) \(\mathrm{SeO}_{3}{ }^{2-} ;\) (b) \(\mathrm{S}_{2} \mathrm{Cl}_{2} ;(\mathrm{c})\) chlorosulfonic acid, \(\mathrm{HSO}_{3} \mathrm{Cl}\) (chlorine is bonded to sulfur).

Write the chemical formula for each of the following compounds, and indicate the oxidation state of nitrogen in each: (a) sodium nitrite, (b) ammonia, (c) nitrous oxide, (d) sodium cyanide, (e) nitric acid, (f) nitrogen dioxide, (g) nitrogen, (h) boron nitride.
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