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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

Give the balanced cell equation and determine \(\mathscr{E}^{\circ}\) for the galvanic cells based on the following half-reactions. Standard reduction potentials are found in Table \(17-1\) a. \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O}\) \(\mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}\) b. \(2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2}\) \(\mathrm{Al}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Al}\)

The measurement of \(\mathrm{pH}\) using a glass electrode obeys the Nernst equation. The typical response of a pH meter at \(25.00^{\circ} \mathrm{C}\) is given by the equation $$ \mathscr{E}_{\text {meas }}=\mathscr{E}_{\text {ref }}+0.05916 \mathrm{pH} $$ where \(\mathscr{E}_{\text {ref }}\) contains the potential of the reference electrode and all other potentials that arise in the cell that are not related to the hydrogen ion concentration. Assume that \(\mathscr{E}_{\mathrm{ref}}=0.250 \mathrm{V}\) and that \(\mathscr{E}_{\text {meas }}=0.480 \mathrm{V}\) a. What is the uncertainty in the values of \(\mathrm{pH}\) and \(\left[\mathrm{H}^{+}\right]\) if the uncertainty in the measured potential is \(\pm 1 \mathrm{mV}\) \((\pm 0.001 \mathrm{V}) ?\) b. To what precision must the potential be measured for the uncertainty in \(\mathrm{pH}\) to be \(\pm 0.02 \mathrm{pH}\) unit?

Sketch the galvanic cells based on the following half-reactions. Show the direction of electron flow, show the direction of ion migration through the salt bridge, and identify the cathode and anode. Give the overall balanced equation, and determine \(\mathscr{C}^{\circ}\) for the galvanic cells. Assume that all concentrations are \(1.0 M\) and that all partial pressures are 1.0 atm. a. \(\mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.78 \mathrm{V}\) \(\mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2} \mathrm{O}_{2} \quad \quad \mathscr{E}^{\circ}=0.68 \mathrm{V}\) b. \(\mathrm{Mn}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Mn} \quad \mathscr{E}^{\circ}=-1.18 \mathrm{V}\) \(\mathrm{Fe}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe} \quad \mathscr{E}^{\circ}=-0.036 \mathrm{V}\)

Consider only the species (at standard conditions) $$ \mathrm{Br}^{-}, \quad \mathrm{Br}_{2}, \quad \mathrm{H}^{+}, \quad \mathrm{H}_{2}, \quad \mathrm{La}^{3+}, \quad \mathrm{Ca}, \quad \mathrm{Cd} $$ in answering the following questions. Give reasons for your answers. a. Which is the strongest oxidizing agent? b. Which is the strongest reducing agent? c. Which species can be oxidized by \(\mathrm{MnO}_{4}^{-}\) in acid? d. Which species can be reduced by \(\mathrm{Zn}(s) ?\)

Consider the following galvanic cell: a. Label the reducing agent and the oxidizing agent, and describe the direction of the electron flow. b. Determine the standard cell potential. c. Which electrode increases in mass as the reaction proceeds, and which electrode decreases in mass?

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