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Chapter 8: Problem 78

Gold does not dissolve in concentrated \(\mathrm{H}_{2} \mathrm{SO}_{4}\) but readily dissolves in \(\mathrm{H}_{2} \mathrm{SeO}_{4}\) (selenic acid). Which acid is the stronger oxidizing agent?

Short Answer

Expert verified
Answer: H2SeO4 (selenic acid) is a stronger oxidizing agent than H2SO4 (sulfuric acid).

Step by step solution

01

Understand oxidizing agents and reduction potentials

An oxidizing agent is a substance that has the ability to accept electrons from another substance, leading to its reduction (gain in electrons). The reduction potential of an oxidizing agent refers to the measure of its oxidizing power, or its ability to gain electrons. We will use the standard reduction potentials of the species involved in relevant oxidation-reduction reactions to compare the oxidizing strengths of H2SO4 and H2SeO4.
02

Write the half reactions for the reduction of sulfur and selenium

The reduction half-reactions for sulfur and selenium in their respective acids can be written as: H2SO4 + 2e- -> SO2 + H2O H2SeO4 + 2e- -> SeO2 + H2O
03

Find the standard reduction potentials for these half-reactions

Referring to a standard reduction potential table, we can find the values for these reactions: For H2SO4: \(\mathrm{SO}_{4}^{2-}\) + 4H+ + 2e- -> \(\mathrm{SO}_{2}\) + 2H2O, E掳 = +1.92 V For H2SeO4: \(\mathrm{SeO}_{4}^{2-}\) + 4H+ + 2e- -> \(\mathrm{SeO}_{2}\) + 2H2O, E掳 = +2.05 V As the two half-reactions are not directly aligned with the ones we formulated earlier, we will adjust their stoichiometry so that they correspond with each other.
04

Adjust the stoichiometry of the half-reactions

We can adjust the stoichiometry of the half-reactions found in the standard reduction potential table so that they are directly comparable to the ones we formulated earlier: For H2SO4: 0.5\(\mathrm{SO}_{4}^{2-}\) + 2H+ + e- -> 0.5\(\mathrm{SO}_{2}\) + H2O, E掳 = +1.92 V For H2SeO4: 0.5\(\mathrm{SeO}_{4}^{2-}\) + 2H+ + e- -> 0.5\(\mathrm{SeO}_{2}\) + H2O, E掳 = +2.05 V
05

Compare the standard reduction potentials

Comparing the standard reduction potentials, we see that the half-reaction involving H2SeO4 has a more positive value (E掳 = +2.05 V) than the half-reaction involving H2SO4 (E掳 = +1.92 V). This indicates that H2SeO4 is a stronger oxidizing agent than H2SO4. In conclusion, selenic acid (H2SeO4) is a stronger oxidizing agent than sulfuric acid (H2SO4), which explains why gold dissolves in H2SeO4 but not in concentrated H2SO4.

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Most popular questions from this chapter

Silver dissolves in sulfuric acid to form silver sulfate and \(\mathrm{H}_{2},\) but gold does not dissolve in sulfuric acid to form gold sulfate. Which of the two metals is the better reducing agent?

Silver tarnish is the result of silver metal reacting with sulfur compounds, such as \(\mathrm{H}_{2} \mathrm{S},\) and \(\mathrm{O}_{2}\) in the air. The tarnish on silverware \(\left(\mathrm{Ag}_{2} \mathrm{S}\right)\) can be removed by soaking the silverware in a slightly basic solution of \(\mathrm{NaHCO}_{3}\) (baking soda) in a basin lined with aluminum foil. a. Write a balanced chemical equation for the tarnish formation reaction. b. Write a balanced net ionic equation for the tarnish removal process in which Ag_S S reacts with A1 metal, forming \(\mathrm{Al}(\mathrm{OH})_{3}(s), \mathrm{Ag}\) metal, and \(\mathrm{HS}^{-}\) ions.

Why is \(\mathrm{HSO}_{4}^{-}(a q)\) a weaker acid than \(\mathrm{H}_{2} \mathrm{SO}_{4} ?\)

Write net ionic equations for the reactions that occur when a. a sample of acetic acid is titrated with a solution of \(\mathrm{KOH}\) b. a solution of sodium carbonate is mixed with a solution of calcium chloride. c. calcium oxide dissolves in water.

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