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Chapter 17: Problem 30

Balance the following oxidation-reduction reactions that occur in acidic solution using the half-reaction method. a. \(\mathrm{Cu}(s)+\mathrm{NO}_{3}^{-}(a q) \rightarrow \mathrm{Cu}^{2+}(a q)+\mathrm{NO}(g)\) b. \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{Cl}^{-}(a q) \rightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{Cl}_{2}(g)\) c. \(\mathrm{Pb}(s)+\mathrm{PbO}_{2}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{PbSO}_{4}(s)\) d. \(\mathrm{Mn}^{2+}(a q)+\mathrm{NaBiO}_{3}(s) \rightarrow \mathrm{Bi}^{3+}(a q)+\mathrm{MnO}_{4}^{-}(a q)\) e. \(\mathrm{H}_{3} \mathrm{AsO}_{4}(a q)+\mathrm{Zn}(s) \rightarrow \mathrm{AsH}_{3}(g)+\mathrm{Zn}^{2+}(a q)\)

Short Answer

Expert verified
The balanced redox reactions in acidic solutions are: a. \(3\mathrm{Cu}(s) + 8\mathrm{H}^+(aq) + 2\mathrm{NO}_3^-(aq) \rightarrow 3\mathrm{Cu}^{2+}(aq) + 2\mathrm{NO}(g) + 4\mathrm{H}_2\mathrm{O}(l)\) b. \(14\mathrm{H}^+(aq) + \mathrm{Cr}_2\mathrm{O}_7^{2-}(aq) + 6\mathrm{Cl}^-(aq) \rightarrow 2\mathrm{Cr}^{3+}(aq) + 3\mathrm{Cl}_2(g) + 7\mathrm{H}_2\mathrm{O}(l)\) c. \(\mathrm{Pb}(s) + \mathrm{Pb}\mathrm{O}_2(s) + 2\mathrm{H}_2\mathrm{SO}_4(aq) \rightarrow 2\mathrm{Pb}\mathrm{SO}_4(s) + 2\mathrm{H}_2\mathrm{O}(l)\) d. \(2\mathrm{Mn}^{2+}(aq) + 5\mathrm{Na}\mathrm{Bi}\mathrm{O}_3(s) + 14\mathrm{H}_2\mathrm{O}(l) \rightarrow 5\mathrm{Bi}^{3+}(aq) + 2\mathrm{Mn}\mathrm{O}_4^-(aq) + 16\mathrm{H}^+(aq) + 5\mathrm{Na}^+(aq)\) e. \(2\mathrm{H}_3\mathrm{As}\mathrm{O}_4(aq) + 3\mathrm{Zn}(s) \rightarrow 2\mathrm{As}\mathrm{H}_3(g) + 3\mathrm{Zn}^{2+}(aq) + 4\mathrm{H}_2\mathrm{O}(l)\)

Step by step solution

01

Identify the half-reactions

The half-reactions are: Cu(s) -> Cu²⁺(aq) (Oxidation) NO₃⁻(aq) -> NO(g) (Reduction)
02

Balance atoms other than oxygen and hydrogen

Both half-reactions have the same number of metal atoms on both sides, so we proceed to the next step.
03

Balance oxygen atoms

Only the reduction half-reaction has oxygen atoms. Balance the oxygen atoms by adding 2H₂O to the product side. NO₃⁻(aq) -> NO(g) + 2H₂O(l)
04

Balance hydrogen atoms

In the balanced reduction half-reaction, there are 4 hydrogen atoms on the product side, so add 4H⁺ to the reactant side. 4H⁺(aq) + NO₃⁻(aq) -> NO(g) + 2H₂O(l)
05

Balance charge with electrons

For the oxidation half-reaction, add 2e⁻ to the product side to balance the charge. Cu(s) -> Cu²⁺(aq) + 2e⁻ For the reduction half-reaction, add 3e⁻ to the product side to balance the charge. 4H⁺(aq) + NO₃⁻(aq) + 3e⁻ -> NO(g) + 2H₂O(l)
06

Combine balanced half-reactions

Multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2 to have the same number of electrons in both half-reactions. Then, combine them together and simplify. 3[Cu(s) -> Cu²⁺(aq) + 2e⁻] 2[4H⁺(aq) + NO₃⁻(aq) + 3e⁻ -> NO(g) + 2H₂O(l)] This leads to the balanced redox reaction: 3Cu(s) + 8H⁺(aq) + 2NO₃⁻(aq) -> 3Cu²⁺(aq) + 2NO(g) + 4H₂O(l) **Now proceed in a similar fashion for the other reactions. I will provide the balanced redox reactions below:** b. 14H⁺(aq) + Cr₂O₇²⁻(aq) + 6Cl⁻(aq) -> 2Cr³⁺(aq) + 3Cl₂(g) + 7H₂O(l) c. Pb(s) + PbO₂(s) + 2H₂SO₄(aq) -> 2PbSO₄(s) + 2H₂O(l) d. 2Mn²⁺(aq) + 5NaBiO₃(s) + 14H₂O(l) -> 5Bi³⁺(aq) + 2MnO₄⁻(aq) + 16H⁺(aq) + 5Na⁺(aq) e. 2H₃AsO₄(aq) + 3Zn(s) -> 2AsH₃(g) + 3Zn²⁺(aq) + 4H₂O(l)

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Most popular questions from this chapter

An electrochemical cell consists of a nickel metal electrode immersed in a solution with \(\left[\mathrm{Ni}^{2+}\right]=1.0 \mathrm{M}\) separated by a porous disk from an aluminum metal electrode immersed in a solution with \(\left[\mathrm{Al}^{3+}\right]=1.0 \mathrm{M} .\) Sodium hydroxide is added to the aluminum compartment, causing \(\mathrm{Al}(\mathrm{OH})_{3}(s)\) to precipitate. After precipitation of Al(OH) \(_{3}\) has ceased, the concentration of \(\mathrm{OH}^{-}\) is \(1.0 \times 10^{-4} \mathrm{M}\) and the measured cell potential is \(1.82 \mathrm{V}\). Calculate the \(K_{\mathrm{sp}}\) value for \(\mathrm{Al}(\mathrm{OH})_{3}\) $$ \mathrm{Al}(\mathrm{OH})_{3}(s) \rightleftharpoons \mathrm{Al}^{3+}(a q)+3 \mathrm{OH}^{-}(a q) \quad K_{\mathrm{sp}}=? $$

The ultimate electron acceptor in the respiration process is molecular oxygen. Electron transfer through the respiratory chain takes place through a complex series of oxidationreduction reactions. Some of the electron transport steps use iron-containing proteins called cytochromes. All cytochromes transport electrons by converting the iron in the cytochromes from the +3 to the +2 oxidation state. Consider the following reduction potentials for three different cytochromes used in the transfer process of electrons to oxygen (the potentials have been corrected for \(\mathrm{pH}\) and for temperature): $$\begin{aligned} &\text { cytochrome } \mathrm{a}\left(\mathrm{Fe}^{3+}\right)+\mathrm{e}^{-} \longrightarrow \text { cytochrome } \mathrm{a}\left(\mathrm{Fe}^{2+}\right)\ &\mathscr{E}^{\circ}=0.385 \mathrm{V}\\\ &\text { cytochrome } \mathbf{b}\left(\mathrm{Fe}^{3+}\right)+\mathrm{e}^{-} \longrightarrow \text { cytochrome } \mathrm{b}\left(\mathrm{Fe}^{2+}\right)\ &\mathscr{E}^{\circ}=0.030 \mathrm{V}\\\ &\text { cytochrome } c\left(\mathrm{Fe}^{3+}\right)+\mathrm{e}^{-} \longrightarrow \text { cytochrome } \mathrm{c}\left(\mathrm{Fe}^{2+}\right)\ &\mathscr{E}^{\circ}=0.254 \mathrm{V} \end{aligned}$$ In the electron transfer series, electrons are transferred from one cytochrome to another. Using this information, determine the cytochrome order necessary for spontaneous transport of electrons from one cytochrome to another, which eventually will lead to electron transfer to \(\mathrm{O}_{2}\)

A disproportionation reaction involves a substance that acts as both an oxidizing and a reducing agent, producing higher and lower oxidation states of the same element in the products. Which of the following disproportionation reactions are spontaneous under standard conditions? Calculate \(\Delta G^{\circ}\) and \(K\) at \(25^{\circ} \mathrm{C}\) for those reactions that are spontaneous under standard conditions. a. \(2 \mathrm{Cu}^{+}(a q) \rightarrow \mathrm{Cu}^{2+}(a q)+\mathrm{Cu}(s)\) b. \(3 \mathrm{Fe}^{2+}(a q) \rightarrow 2 \mathrm{Fe}^{3+}(a q)+\mathrm{Fe}(s)\) c. \(\mathrm{HClO}_{2}(a q) \rightarrow \mathrm{ClO}_{3}^{-}(a q)+\mathrm{HClO}(a q) \quad\) (unbalanced) Use the half-reactions: \(\mathrm{ClO}_{3}^{-}+3 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{HClO}_{2}+\mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.21 \mathrm{V}\) \(\mathrm{HClO}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{HClO}+\mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.65 \mathrm{V}\)

Is the following statement true or false? Concentration cells work because standard reduction potentials are dependent on concentration. Explain.

The blood alcohol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) level can be determined by titrating a sample of blood plasma with an acidic potassium dichromate solution, resulting in the production of \(\mathrm{Cr}^{3+}(a q)\) and carbon dioxide. The reaction can be monitored because the dichromate ion \(\left(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\right)\) is orange in solution, and the \(\mathrm{Cr}^{3+}\) ion is green. The unbalanced redox equation is $$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{CO}_{2}(g)$$ If \(31.05 \mathrm{mL}\) of \(0.0600 M\) potassium dichromate solution is required to titrate \(30.0 \mathrm{g}\) blood plasma, determine the mass percent of alcohol in the blood.
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