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Chapter 18: Problem 30

Balance each of the following redox reactions in acid solution. (a) \(\mathrm{MnO}_{4}^{-}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(\mathrm{aq}) \rightarrow \mathrm{Mn}^{2+}(\mathrm{aq})+\mathrm{CO}_{2}(\mathrm{~g})\) (b) \(\mathrm{Cl}_{2}(\mathrm{~g})+\mathrm{Br}^{-}(\mathrm{aq}) \rightarrow \mathrm{Cl}^{-}(\mathrm{aq})+\mathrm{Br}_{2}(\ell)\) (c) \(\mathrm{Cu}(\mathrm{s})+\mathrm{NO}_{3}^{-}(\mathrm{aq}) \rightarrow \mathrm{NO}(\mathrm{g})+\mathrm{Cu}^{2+}(\mathrm{aq})\)

Short Answer

Expert verified
(a) \(2\mathrm{MnO}_{4}^{-} + 5\mathrm{H}_{2}\mathrm{C}_{2}\mathrm{O}_{4} + 6\mathrm{H}^{+} \rightarrow 2\mathrm{Mn}^{2+} + 10\mathrm{CO}_{2} + 8\mathrm{H}_{2}\mathrm{O}\). (b) \(\mathrm{Cl}_{2} + 2\mathrm{Br}^{-} \rightarrow 2\mathrm{Cl}^{-} + \mathrm{Br}_2\). (c) \(3\mathrm{Cu} + 2\mathrm{NO}_{3}^{-} + 8\mathrm{H}^{+} \rightarrow 3\mathrm{Cu}^{2+} + 2\mathrm{NO} + 4\mathrm{H}_{2}\mathrm{O}\).

Step by step solution

01

Assign Oxidation Numbers

For each reaction, identify the elements that change oxidation states. (a) In \(\mathrm{MnO}_{4}^{-}\), Mn has an oxidation state of +7, reduced to +2 in \(\mathrm{Mn}^{2+}\). In \(\mathrm{H}_{2}\mathrm{C}_{2}\mathrm{O}_{4}\), C has an oxidation state of +3, increasing to +4 in \(\mathrm{CO}_{2}\).(b) In \(\mathrm{Cl}_2\), each Cl has an oxidation state of 0, oxidized to -1 in \(\mathrm{Cl}^{-}\). \(\mathrm{Br}^{-}\) goes from -1 to 0 in \(\mathrm{Br}_2\).(c) Cu starts at 0 and goes to +2 in \(\mathrm{Cu}^{2+}\). The N in \(\mathrm{NO}_3^{-}\) goes from +5 to +2 in \(\mathrm{NO}\).
02

Split into Half-Reactions

Separate each reaction into oxidation and reduction half-reactions.(a) Oxidation: \(\mathrm{H}_{2}\mathrm{C}_{2}\mathrm{O}_{4} \rightarrow \mathrm{CO}_{2}\) Reduction: \(\mathrm{MnO}_{4}^{-} \rightarrow \mathrm{Mn}^{2+}\)(b) Oxidation: \(\mathrm{Br}^{-} \rightarrow \mathrm{Br}_{2}\) Reduction: \(\mathrm{Cl}_{2} \rightarrow \mathrm{Cl}^{-}\)(c) Oxidation: \(\mathrm{Cu} \rightarrow \mathrm{Cu}^{2+}\) Reduction: \(\mathrm{NO}_{3}^{-} \rightarrow \mathrm{NO}\)
03

Balance Each Half-Reaction

Balance atoms in each half-reaction, starting with elements other than O and H.(a) Oxidation: \(\mathrm{H}_{2}\mathrm{C}_{2}\mathrm{O}_{4} \rightarrow 2\mathrm{CO}_{2} + 2\mathrm{H}^{+} + 2e^-\) Reduction: \(\mathrm{MnO}_{4}^{-} + 8\mathrm{H}^{+} + 5e^- \rightarrow \mathrm{Mn}^{2+} + 4\mathrm{H}_{2}\mathrm{O}\)(b) Oxidation: \(2\mathrm{Br}^{-} \rightarrow \mathrm{Br}_{2} + 2e^-\) Reduction: \(\mathrm{Cl}_{2} + 2e^- \rightarrow 2\mathrm{Cl}^{-}\)(c) Oxidation: \(\mathrm{Cu} \rightarrow \mathrm{Cu}^{2+} + 2e^-\) Reduction: \(\mathrm{NO}_{3}^{-} + 4\mathrm{H}^{+} + 3e^- \rightarrow \mathrm{NO} + 2\mathrm{H}_{2}\mathrm{O}\)
04

Equalize Electrons Transferred

Multiply each half-reaction by appropriate coefficients to equalize the number of electrons transferred in both half-reactions.(a) Multiply Oxidation by 5 and Reduction by 2: Oxidation: \(5\mathrm{H}_{2}\mathrm{C}_{2}\mathrm{O}_{4} \rightarrow 10\mathrm{CO}_{2} + 10\mathrm{H}^{+} + 10e^-\) Reduction: \(2\mathrm{MnO}_{4}^{-} + 16\mathrm{H}^{+} + 10e^- \rightarrow 2\mathrm{Mn}^{2+} + 8\mathrm{H}_{2}\mathrm{O}\)(b) Not needed as both are already balanced in terms of electrons (2e- each).(c) Multiply Oxidation by 3 and Reduction by 2: Oxidation: \(3\mathrm{Cu} \rightarrow 3\mathrm{Cu}^{2+} + 6e^-\) Reduction: \(2(\mathrm{NO}_{3}^{-} + 4\mathrm{H}^{+} + 3e^- \rightarrow \mathrm{NO} + 2\mathrm{H}_{2}\mathrm{O})\)
05

Combine Half-Reactions and Simplify

Combine the half-reactions and cancel out common species on both sides.(a) \(2\mathrm{MnO}_{4}^{-} + 5\mathrm{H}_{2}\mathrm{C}_{2}\mathrm{O}_{4} + 6\mathrm{H}^{+} \rightarrow 2\mathrm{Mn}^{2+} + 10\mathrm{CO}_{2} + 8\mathrm{H}_{2}\mathrm{O}\)(b) \(\mathrm{Cl}_{2} + 2\mathrm{Br}^{-} \rightarrow 2\mathrm{Cl}^{-} + \mathrm{Br}_2\)(c) \(3\mathrm{Cu} + 2\mathrm{NO}_{3}^{-} + 8\mathrm{H}^{+} \rightarrow 3\mathrm{Cu}^{2+} + 2\mathrm{NO} + 4\mathrm{H}_{2}\mathrm{O}\)
06

Verify Atom and Charge Balance

Ensure that atoms and charges are balanced on both sides of each equation. This confirms that the reactions are correctly balanced in acidic solution.

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Key Concepts

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

Balancing Chemical Equations
Balancing chemical equations is an essential skill in chemistry. It ensures that the number of atoms for each element is the same on both sides of the equation. This is important because it reflects the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction. To balance chemical equations, follow these steps:

  • Identify the reactants and products: Write down the chemical formulas for all reactants and products involved in the reaction.
  • Separate into half-reactions: In cases of redox reactions, divide the reaction into two half-reactions, one for oxidation and one for reduction.
  • Balance atoms other than O and H: First balance atoms that are not oxygen or hydrogen.
  • Balance oxygen atoms: Use water molecules to balance oxygen atoms.
  • Balance hydrogen atoms: Add hydrogen ions (H+) to balance hydrogen atoms, especially in acidic solutions.
  • Balance the charge: Add electrons to the side that requires balancing of charge.
  • Equalize electrons in both half-reactions: By multiplying the half-reactions, ensure that the number of electrons lost in oxidation equals the number gained in reduction.
  • Combine and simplify: Combine the half-reactions back into a balanced equation, simplifying where possible, especially by canceling out species that appear on both sides.
  • Verify the balance: Check both atom and charge balance to confirm that the equation is correctly balanced.
Balancing equations in an acidic solution also means considering the acidic environment through the inclusion of H+ ions and making sure they are balanced in the final equation.
Oxidation States
Oxidation states or oxidation numbers are vital for understanding redox reactions. An oxidation state is an indicator of the degree of oxidation (loss of electrons) of an atom in a chemical compound. It is often represented as an integer that can be positive, negative, or zero, and it helps in determining which elements are oxidized and which are reduced in a reaction.

Here are some key points to consider:
  • Element oxidation states: Pure elements have an oxidation state of zero. For example, N2 or O2 have an oxidation state of 0.
  • Simple ions: The oxidation state is equal to the ion's charge. For instance, Cl- has an oxidation state of -1.
  • Oxygen and hydrogen: In most compounds, oxygen has an oxidation state of -2, while hydrogen is usually +1.
  • Tracking electron transfer: Changes in oxidation states indicate electron transfer, vital for balancing redox reactions. When an element's oxidation state increases, it is oxidized (loses electrons). Conversely, a decrease in oxidation state means reduction (gain of electrons).
  • Assigning oxidation states: Use known oxidation states and the rule that the sum of oxidation states in a compound must equal its overall charge to assign unknown oxidation states to other elements.
Mastering oxidation states helps predict a compound's behavior in redox reactions and aids in balancing those reactions.
Acidic Solution Reactions
Redox reactions in acidic solutions require special consideration due to the acidic environment, which often influences the balance of hydrogen ions (H+) alongside electrons and atoms. An acidic solution not only supports certain reactions but also often significantly alters the way a reaction is balanced.

Here are essential aspects of balancing these reactions:
  • Inclusion of H+ ions: Since the reaction takes place in an acidic medium, balancing involves adding H+ ions to one or both sides of the half-reactions. This balances hydrogen atoms and often helps close the charge balance gap.
  • Use of water: Water molecules (H2O) are used to balance oxygen atoms that do not have counter hydrogen from H+ additions.
  • Balancing technique: Balancing in acidic solutions often follows a pattern where after balancing all other elements, O and H are balanced by adding H2O and H+. Remember to check the charges and add the right number of electrons to balance them across both half-reactions.
  • Confirm overall balance: Even after balancing for H+ and using water, make it a step to verify that all atoms and the total charge are equal on both sides of the equation for the final check.
Balancing redox reactions in acidic solutions is about carefully considering the H+ contributions, which is critical for accurate and complete reaction equations.

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Most popular questions from this chapter

Balance each of the following redox reactions in acid solution. (a) \(\mathrm{Fe}(\mathrm{s})+\mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Ag}(\mathrm{s})+\mathrm{Fe}^{2+}(\mathrm{aq})\) (b) \(\mathrm{I}_{2}(\mathrm{aq})+\mathrm{S}_{2} \mathrm{O}_{3}^{2-}(\mathrm{aq}) \rightarrow \mathrm{I}^{-}(\mathrm{aq})+\mathrm{S}_{4} \mathrm{O}_{6}^{2-}(\mathrm{aq})\) (c) \(\mathrm{MnO}_{4}^{-}(\mathrm{aq})+\mathrm{Fe}^{2+}(\mathrm{aq}) \rightarrow \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{Mn}^{2+}(\mathrm{aq})\)

Sphalerite is the naturally occurring mineral zinc sulfide, from which zinc metal is extracted. The ore is heated in oxygen to form zinc oxide and sulfur dioxide, followed by the reaction of metal oxide with elemental carbon to form CO. (a) Write a balanced equation for each of these reactions. (b) For each reaction, identify the element that is oxidized and the one that is reduced.

Balance each of the following redox reactions in acid solution. (a) \(\mathrm{Sn}(\mathrm{s})+\mathrm{Fe}^{3+}(\mathrm{aq}) \rightarrow \mathrm{Sn}^{2+}(\mathrm{aq})+\mathrm{Fe}^{2+}(\mathrm{aq})\) (b) \(\mathrm{HAsO}_{3}^{2-}(\mathrm{aq})+\mathrm{I}_{2}(\mathrm{aq}) \rightarrow \mathrm{H}_{2} \mathrm{AsO}_{4}^{-}(\mathrm{aq})+\mathrm{I}^{-}(\mathrm{aq})\) (c) \(\mathrm{Cu}(\mathrm{s})+\mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Cu}^{2+}(\mathrm{aq})+\mathrm{Ag}(\mathrm{s})\)

Calculate the value of the solubility product constant for \(\mathrm{PbSO}_{4}\) from the half-cell potentials. $$ \begin{aligned} \mathrm{PbSO}_{4}(\mathrm{~s})+2 \mathrm{e}^{-} \rightarrow \mathrm{Pb}(\mathrm{s})+\mathrm{SO}_{4}^{2-}(\mathrm{aq}) & E^{\circ}=-0.356 \mathrm{~V} \\ \mathrm{~Pb}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Pb}(\mathrm{s}) & E^{\circ}=-0.126 \mathrm{~V} \end{aligned} $$

In the analytical technique called electrogravimetry, electrolysis is used to separate the analyte from a solution by depositing it on an inert electrode. The electrode is weighed before and after the experiment to find the mass of analyte deposited. A \(0.122-\mathrm{g}\) sample of a copper-zinc alloy was treated with concentrated sulfuric acid to produce a solution containing copper(II) and zinc(II) sulfates. The platinum cathode used in the electrolysis of this solution increased in mass by \(0.073 \mathrm{~g}\) after exhaustive electrolysis. (a) Which metal was deposited on the cathode during the electrolysis? Write the balanced equation for the electrolysis reaction. (b) What was the mass percentages of copper and zinc in the alloy sample?
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