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Chapter 20: Problem 44

From each of the following pairs of substances, use data in Appendix \(\mathrm{E}\) to choose the one that is the stronger oxidizing agent: (a) \(\mathrm{Cl}_{2}(g)\) or \(\mathrm{Br}_{2}(l)\) (b) \(\mathrm{Zn}^{2+}(a q)\) or \(\mathrm{Cd}^{2+}(a q)\) (c) \(\mathrm{Cl}^{-}(a q)\) or \(\mathrm{ClO}_{3}^{-}(a q)\) (d) \(\mathrm{H}_{2} \mathrm{O}_{2}(a q)\) or \(\mathrm{O}_{3}(g)\)

Short Answer

Expert verified
In summary, using the reduction potentials found in Appendix E, the stronger oxidizing agents are: (a) \(Cl_{2}(g)\) (b) \(Cd^{2+}(aq)\) (c) \(ClO_{3}^{-}(aq)\) (d) \(O_{3}(g)\)

Step by step solution

01

(a) Cl2(g) or Br2(l)

\ First, we need to find the reduction potential for both substances in the pair. In Appendix E, we find the following half-reactions and their respective standard reduction potentials (E°): 1. \(\mathrm{Cl_{2}(g)} + 2 \mathrm{e^{-}} \to 2 \mathrm{Cl^{-}(aq)}\), \(E_{1}^\circ = +1.36V\) 2. \(\mathrm{Br_{2}(l)} + 2 \mathrm{e^{-}} \to 2 \mathrm{Br^{-}(aq)}\), \(E_{2}^\circ = +1.07V\) Now compare the reduction potentials (E°). Since \(E_{1}^\circ > E_{2}^\circ\), Cl2(g) is the stronger oxidizing agent.
02

(b) Zn2+(aq) or Cd2+(aq)

\ Similarly, find the reduction potential for both substances in the pair: 1. \(\mathrm{Zn^{2+}(aq)} + 2 \mathrm{e^{-}} \to\mathrm{Zn(s)}\), \(E_{3}^\circ = -0.76V\) 2. \(\mathrm{Cd^{2+}(aq)} + 2 \mathrm{e^{-}} \to\mathrm{Cd(s)}\), \(E_{4}^\circ = -0.40V\) Comparing the reduction potentials, we find that \(E_{4}^\circ > E_{3}^\circ\), so Cd2+(aq) is the stronger oxidizing agent.
03

(c) Cl-(aq) or ClO3-(aq)

\ Find the reduction potential for both substances in the pair: 1. \(2 \mathrm{Cl}^{-}(a q) \to \mathrm{Cl}_{2}(g) + 2 \mathrm{e^{-}}\), \(E_{5}^\circ = -1.36V\) 2. \(6 \mathrm{H}^{+}(a q)+2 \mathrm{ClO}_{3}^{-}(a q)+6 e^{-} \to 3\mathrm{H}_{2}\mathrm{O}(l) + 2 \mathrm{Cl}^{-}(a q)\), \(E_{6}^\circ = +1.47V\) Comparing the reduction potentials, we find that \(E_{6}^\circ > E_{5}^\circ\), so ClO3⁻(aq) is the stronger oxidizing agent.
04

(d) H2O2(aq) or O3(g)

\ Finally, find the reduction potential for both substances in the pair: 1. \(\mathrm{H}_{2} \mathrm{O}_{2}(a q) + 2\mathrm{H}^{+}(a q) +2 \mathrm{e^{-}} \to 2 \mathrm{H}_{2}\mathrm{O}(l)\), \(E_{7}^\circ = +1.77V\) 2. \(\mathrm{O}_{3}(g) + 2 \mathrm{H}^{+}(a q)+2 \mathrm{e^{-}} \to \mathrm{O}_{2}(g)+ \mathrm{H}_{2}\mathrm{O}(l)\), \(E_{8}^\circ = +2.07V\) Comparing the reduction potentials, we find that \(E_{8}^\circ > E_{7}^\circ\), so O3(g) is the stronger oxidizing agent. In summary, the stronger oxidizing agents are: (a) Cl2(g) (b) Cd2+(aq) (c) ClO3⁻(aq) (d) O3(g)

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Most popular questions from this chapter

Indicate whether each of the following statements is true or false: (a) If something is reduced, it is formally losing electrons. (b) A reducing agent gets oxidized as it reacts. (c) An oxidizing agent is needed to convert \(\mathrm{CO}\) into \(\mathrm{CO}_{2}\).

Given the following reduction half-reactions: $$ \begin{aligned} \mathrm{Fe}^{3+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Fe}^{2+}(a q) & E_{\mathrm{red}}^{\circ}=+0.77 \mathrm{~V} \\ \mathrm{~S}_{2} \mathrm{O}_{6}^{2-}(a q)+4 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{SO}_{3}(a q) & E_{\mathrm{red}}^{\circ}=+0.60 \mathrm{~V} \\ \mathrm{~N}_{2} \mathrm{O}(g)+2 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{N}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & E_{\mathrm{red}}^{\circ}=-1.77 \mathrm{~V} \\ \mathrm{VO}_{2}^{+}(a q)+2 \mathrm{H}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{VO}^{2+}+\mathrm{H}_{2} \mathrm{O}(l) & E_{\mathrm{red}}^{\circ}=+1.00 \mathrm{~V} \end{aligned} $$ (a) Write balanced chemical equations for the oxidation of $\mathrm{Fe}^{2+}(a q)\( by \)\mathrm{S}_{2} \mathrm{O}_{6}^{2-}(a q),\( by \)\mathrm{N}_{2} \mathrm{O}(a q),\( and by \)\mathrm{VO}_{2}^{+}(a q)\( (b) Calculate \)\Delta G^{\circ}\( for each reaction at \)298 \mathrm{~K}$. (c) Calculate the equilibrium constant \(K\) for each reaction at \(298 \mathrm{~K}\).

Magnesium is obtained by electrolysis of molten \(\mathrm{MgCl}_{2}\). (a) Why is an aqueous solution of \(\mathrm{MgCl}_{2}\) not used in the electrolysis? (b) Several cells are connected in parallel by very large copper bars that convey current to the cells. Assuming that the cells are \(96 \%\) efficient in producing the desired products in electrolysis, what mass of \(\mathrm{Mg}\) is formed by passing a current of 97,000 A for a period of 24 h?

Indicate whether the following balanced equations involve oxidation-reduction. If they do, identify the elements that undergo changes in oxidation number. (a) $2 \mathrm{AgNO}_{3}(a q)+\mathrm{CoCl}_{2}(a q) \longrightarrow 2 \mathrm{AgCl}(s)+ \mathrm{Co}\left(\mathrm{NO}_{3}\right)_{2}(a q)$ (b) $2 \mathrm{PbO}_{2}(s) \longrightarrow 2 \mathrm{PbO}(s)+\mathrm{O}_{2}(g)$ (c) $2 \mathrm{H}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{NaBr}(s) \longrightarrow \mathrm{Br}_{2}(l)+\mathrm{SO}_{2}(g)+ \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)$

(a) What is an electrolytic cell? (b) The negative terminal of a voltage source is connected to an electrode of an electrolytic cell. Is the electrode the anode or the cathode of the cell? Explain. (c) The electrolysis of water is often done with a small amount of sulfuric acid added to the water. What is the role of the sulfuric acid? (d) Why are active metals such as Al obtained by electrolysis using molten salts rather than aqueous solutions?
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