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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) -> Cu虏鈦�(aq) (Oxidation) NO鈧冣伝(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 2H鈧侽 to the product side. NO鈧冣伝(aq) -> NO(g) + 2H鈧侽(l)
04

Balance hydrogen atoms

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

Balance charge with electrons

For the oxidation half-reaction, add 2e鈦� to the product side to balance the charge. Cu(s) -> Cu虏鈦�(aq) + 2e鈦� For the reduction half-reaction, add 3e鈦� to the product side to balance the charge. 4H鈦�(aq) + NO鈧冣伝(aq) + 3e鈦� -> NO(g) + 2H鈧侽(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) -> Cu虏鈦�(aq) + 2e鈦籡 2[4H鈦�(aq) + NO鈧冣伝(aq) + 3e鈦� -> NO(g) + 2H鈧侽(l)] This leads to the balanced redox reaction: 3Cu(s) + 8H鈦�(aq) + 2NO鈧冣伝(aq) -> 3Cu虏鈦�(aq) + 2NO(g) + 4H鈧侽(l) **Now proceed in a similar fashion for the other reactions. I will provide the balanced redox reactions below:** b. 14H鈦�(aq) + Cr鈧侽鈧嚶测伝(aq) + 6Cl鈦�(aq) -> 2Cr鲁鈦�(aq) + 3Cl鈧�(g) + 7H鈧侽(l) c. Pb(s) + PbO鈧�(s) + 2H鈧係O鈧�(aq) -> 2PbSO鈧�(s) + 2H鈧侽(l) d. 2Mn虏鈦�(aq) + 5NaBiO鈧�(s) + 14H鈧侽(l) -> 5Bi鲁鈦�(aq) + 2MnO鈧勨伝(aq) + 16H鈦�(aq) + 5Na鈦�(aq) e. 2H鈧傾sO鈧�(aq) + 3Zn(s) -> 2AsH鈧�(g) + 3Zn虏鈦�(aq) + 4H鈧侽(l)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Half-Reaction Method
The half-reaction method is a systematic process used to balance oxidation-reduction (redox) reactions. This is particularly useful when complex reactions occur, as it breaks the reaction into two simpler parts, each handling either the oxidation or reduction process. Here's how it works:
  • Identify Half-Reactions: Start by breaking down the full redox reaction into two half-reactions: one for oxidation and one for reduction. Oxidation involves the loss of electrons, while reduction involves the gain of electrons.
  • Balance Atoms: Initially, focus on balancing all atoms except hydrogen and oxygen. This lays the groundwork for further balancing.
  • Balance Oxygen with Water: For any imbalanced oxygen atoms, add water molecules to ensure both sides have equal oxygen numbers.
  • Balance Hydrogen with Protons: Hydrogen atoms are then balanced by introducing hydrogen ions, particularly in acidic solutions.
  • Balance Charge with Electrons: Add electrons to either the reactant or product side of each half-reaction to balance the charge. Remember, electrons lost in oxidation must equal electrons gained in reduction.
  • Combine Half-Reactions: Multiply the half-reactions by appropriate coefficients to ensure the electrons are the same. Finally, add them together to achieve a balanced equation.
By following these steps methodically, redox reactions can be accurately balanced, reflecting both mass and charge conservation.
Balancing Chemical Equations
Balancing chemical equations is a fundamental skill in chemistry necessary for accurate reaction representation. It ensures that the number of each type of atom, and the electrical charge, is consistent on both sides of the equation. Here's how you can master it:
  • List Elements: Write down all the elements involved in the reaction. Track the number of each atom present in both the reactant and product sides.
  • Count and Compare: Compare these numbers to identify which atoms are unbalanced. A straightforward checklist helps to pinpoint errors quickly.
  • Adjust Coefficients: Use coefficients, the numbers placed before compounds, to balance these atoms. Start with elements that appear in only one reactant and one product.
  • Maintain Balance: Continuously check and adjust, changing coefficients rather than subscripts. Maintaining the identity of compounds is essential.
  • Verify Charges: In redox reactions, be sure that the total charge is balanced as well. This extra step is crucial where charged species are involved.
Balancing equations ensures that the law of conservation of mass holds, reflecting that mass can neither be created nor destroyed in a chemical reaction. It's a step that cannot be skipped for accurate chemistry work.
Acidic Solution Reactions
Chemical reactions in acidic solutions require specific attention due to the presence of additional hydrogen ions. Understanding how to manipulate these ions is key to balancing reactions in such environments. Here's what you need to know:
  • Identify Acidity: Recognize that reactions specified 'in acidic solutions' mean that hydrogen ions (H鈦�) are readily available and can be used freely for balancing.
  • Utilize H鈦� and H鈧侽: You can use these ions to balance hydrogen and oxygen atoms not naturally balancing out in the initial steps. Add H鈦� when you need hydrogen atoms and H鈧侽 when dealing with oxygen balancing.
  • Consider Acid-Base Behavior: Be aware that acidic conditions might cause additional shifts in reaction equilibrium or engage dynamically with reactants, affecting substance stability.
  • Check the System鈥檚 Properties: The behavior of chemicals can change in acidic versus basic environments, so always keep the nature of the solution in consideration for a plausible reaction mechanism.
Understanding these principles helps you accurately adjust reactions in the acidic environment, ensuring that both atom and electron balance are correctly maintained. Enhancing your skills with acidic solution reactions can significantly improve your chemical balancing accuracy.

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Most popular questions from this chapter

In the electrolysis of an aqueous solution of \(\mathrm{Na}_{2} \mathrm{SO}_{4},\) what reactions occur at the anode and the cathode (assuming standard conditions)? $$\begin{array}{lr} \mathrm{S}_{2} \mathrm{O}_{8}^{2-}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{SO}_{4}^{2-} & 80^{\circ} \\ \mathrm{O}_{2}+4 \mathrm{H}^{+}+4 \mathrm{e}^{-} \longrightarrow_{2 \mathrm{H}_{2} \mathrm{O}} & 2.01 \mathrm{V} \\ 2 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2}+2 \mathrm{OH}^{-} & -0.83 \mathrm{V} \\ \mathrm{Na}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Na} & -2.71 \mathrm{V} \end{array}$$

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 chemist wishes to determine the concentration of \(\mathrm{CrO}_{4}^{2-}\) electrochemically. A cell is constructed consisting of a saturated calomel electrode (SCE; see Exercise 115 ) and a silver wire coated with \(\mathrm{Ag}_{2} \mathrm{CrO}_{4} .\) The \(8^{\circ}\) value for the following half-reaction is \(0.446 \mathrm{V}\) relative to the standard hydrogen electrode: $$\mathrm{Ag}_{2} \mathrm{CrO}_{4}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Ag}+\mathrm{CrO}_{4}^{2-}$$ a. Calculate \(\mathscr{C}_{\text {cell }}\) and \(\Delta G\) at \(25^{\circ} \mathrm{C}\) for the cell reaction when \(\left[\mathrm{CrO}_{4}^{2-}\right]=1.00 \mathrm{mol} / \mathrm{L}\) b. Write the Nernst equation for the cell. Assume that the SCE concentrations are constant. c. If the coated silver wire is placed in a solution (at \(25^{\circ} \mathrm{C}\) ) in which \(\left[\mathrm{CrO}_{4}^{2-}\right]=1.00 \times 10^{-5} \mathrm{M},\) what is the expected cell potential? d. The measured cell potential at \(25^{\circ} \mathrm{C}\) is \(0.504 \mathrm{V}\) when the coated wire is dipped into a solution of unknown \(\left[\mathrm{CrO}_{4}^{2-}\right] .\) What is \(\left[\mathrm{CrO}_{4}^{2-}\right]\) for this solution? e. Using data from this problem and from Table \(17-1,\) calculate the solubility product \(\left(K_{\mathrm{sp}}\right)\) for \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\)

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: \(\mathrm{ClO}_{3}^{-}+3 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{HClO}_{2}+\mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.21 \mathrm{V}\) \(\mathrm{HClO}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{HClO}+\mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.65 \mathrm{V}\)

Gold metal will not dissolve in either concentrated nitric acid or concentrated hydrochloric acid. It will dissolve, however, in aqua regia, a mixture of the two concentrated acids. The products of the reaction are the \(\mathrm{AuCl}_{4}^{-}\) ion and gaseous NO. Write a balanced equation for the dissolution of gold in aqua regia.
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