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Chapter 19: Problem 35

Balance the following oxidation-reduction equations. The reactions occur in acidic solution. a. \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+\mathrm{C}_{2} \mathrm{O}_{4}^{2-} \longrightarrow \mathrm{Cr}^{3+}+\mathrm{CO}_{2}\) b. \(\mathrm{Cu}+\mathrm{NO}_{3}^{-} \longrightarrow \mathrm{Cu}^{2+}+\mathrm{NO}\) c. \(\mathrm{MnO}_{2}+\mathrm{HNO}_{2} \longrightarrow \mathrm{Mn}^{2+}+\mathrm{NO}_{3}^{-}\) d. \(\mathrm{PbO}_{2}+\mathrm{Mn}^{2+}+\mathrm{SO}_{4}^{2-} \longrightarrow \mathrm{PbSO}_{4}+\mathrm{MnO}_{4}^{-}\) e. \(\mathrm{HNO}_{2}+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \longrightarrow \mathrm{Cr}^{3+}+\mathrm{NO}_{3}^{-}\).

Short Answer

Expert verified
Balance each redox equation using half-reactions and ensure electrons and atoms are conserved on both sides.

Step by step solution

01

Identify Oxidation and Reduction Half-Reactions for Reaction (a)

For the reaction \( \mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}+\mathrm{C}_{2}\mathrm{O}_{4}^{2-} \longrightarrow \mathrm{Cr}^{3+}+\mathrm{CO}_{2} \), identify the oxidation and reduction processes:- **Oxidation:** \( \mathrm{C}_{2}\mathrm{O}_{4}^{2-} \rightarrow \mathrm{CO}_{2} \)- **Reduction:** \( \mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} \rightarrow \mathrm{Cr}^{3+} \).
02

Balance Half-Reactions for Reaction (a)

Balance each half-reaction separately in medium:- **Oxidation:** \( \mathrm{C}_{2}\mathrm{O}_{4}^{2-} \rightarrow 2\mathrm{CO}_{2} + 2e^- \) - **Reduction:** \( \mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} + 14\mathrm{H}^+ + 6e^- \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_2\mathrm{O} \).
03

Balance Electrons and Combine Half-Reactions for Reaction (a)

Multiply the oxidation half-reaction by 3 to match the 6 electrons in the reduction reaction:- **Combined:** \( 3\left(\mathrm{C}_{2}\mathrm{O}_{4}^{2-} \rightarrow 2\mathrm{CO}_{2} + 2e^- \right)\) \( \mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} + 14\mathrm{H}^+ + 6e^- \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_2\mathrm{O} \)- **Net Reaction:** \( \mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} + 3\mathrm{C}_{2}\mathrm{O}_{4}^{2-} + 14\mathrm{H}^+ \rightarrow 2\mathrm{Cr}^{3+} + 6\mathrm{CO}_{2} + 7\mathrm{H}_2\mathrm{O} \).
04

Follow the Same Process for Reactions (b), (c), (d), and (e)

Repeat steps 1-3 for each reaction:- **(b)** \( \mathrm{Cu}+\mathrm{NO}_{3}^{-} \rightarrow \mathrm{Cu}^{2+}+\mathrm{NO} \) - Oxidation: \( \mathrm{Cu} \rightarrow \mathrm{Cu}^{2+} + 2e^- \) - Reduction: \( \mathrm{NO}_{3}^{-} + 4H^+ + 3e^- \rightarrow \mathrm{NO} + 2\mathrm{H}_2\mathrm{O} \) - Balanced: \( 3\mathrm{Cu} + 2\mathrm{NO}_{3}^{-} + 8\mathrm{H}^+ \rightarrow 3\mathrm{Cu}^{2+} + 2\mathrm{NO} + 4\mathrm{H}_2\mathrm{O} \)- **(c)** \( \mathrm{MnO}_{2}+\mathrm{HNO}_{2} \rightarrow \mathrm{Mn}^{2+}+\mathrm{NO}_{3}^{-} \) - Oxidation: \( \mathrm{HNO}_{2} \rightarrow \mathrm{NO}_{3}^{-} + 2e^- \) - Reduction: \( \mathrm{MnO}_{2} + 4H^+ + 2e^- \rightarrow \mathrm{Mn}^{2+} + 2\mathrm{H}_2\mathrm{O} \) - Balanced: \( \mathrm{HNO}_{2} + \mathrm{MnO}_{2} + 2\mathrm{H}^+ \rightarrow \mathrm{NO}_{3}^{-} + \mathrm{Mn}^{2+} + 2\mathrm{H}_2\mathrm{O} \)- **(d)** \( \mathrm{PbO}_{2} + \mathrm{Mn}^{2+} + \mathrm{SO}_{4}^{2-} \rightarrow \mathrm{PbSO}_{4} + \mathrm{MnO}_{4}^{-} \) - Oxidation: \( \mathrm{Mn}^{2+} \rightarrow \mathrm{MnO}_{4}^{-} + 5e^- \) - Reduction: \( \mathrm{PbO}_{2} + 2\mathrm{H}^+ + 2e^- \rightarrow \mathrm{PbSO}_{4} \) - Balanced: \( 2\mathrm{Mn}^{2+} + \mathrm{PbO}_{2} + 2\mathrm{H}_2\mathrm{SO}_{4} \rightarrow \mathrm{4H}_2\mathrm{O} + 2\mathrm{MnO}_{4}^{-} + \mathrm{Pb}\mathrm{SO}_{4} \)- **(e)** \( \mathrm{HNO}_{2} + \mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} \rightarrow \mathrm{Cr}^{3+} + \mathrm{NO}_{3}^{-} \) - Oxidation: \( 2\mathrm{HNO}_{2} \rightarrow 2\mathrm{NO}_{3}^{-} + 2e^- \) - Reduction: \( \mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} + 14\mathrm{H}^{+} + 6e^- \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_2\mathrm{O} \) - Balanced: \( 3\mathrm{HNO}_{2} + \mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} + 6\mathrm{H}^+ \rightarrow 2\mathrm{Cr}^{3+} + 3\mathrm{NO}_{3}^{-} + 3\mathrm{H}_2\mathrm{O} \).
05

Review and Verify the Overall Balanced Equations

Go through each reaction (a to e) to ensure that the electrons and atoms are balanced on both sides for each equation. This step ensures that both mass and charge conservation principles are satisfied across all the reactions provided.

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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 crucial in chemistry to ensure that mass and charge are conserved in a reaction. Every chemical equation represents a transformation where atoms and charge are reorganized rather than created or destroyed.
  • Identify the number of atoms for each element on both sides of the equation.
  • Ensure the same number of each type of atom on both sides to respect the Law of Conservation of Mass.
  • For ions, ensure the total charge is balanced on both sides by considering the oxidation states.
Balancing starts by adjusting coefficients, which are the numbers placed in front of molecules or atoms to indicate their quantity. This method doesn't alter chemical identities but rather ensures a stoichiometric balance.
Redox Chemistry
Redox, or reduction-oxidation, chemistry involves the transfer of electrons between species. Key to this concept is understanding oxidation and reduction:
  • **Oxidation**: Loss of electrons, leading to an increased oxidation state.
  • **Reduction**: Gain of electrons, leading to a decreased oxidation state.
Oxidizing agents gain electrons and are reduced, while reducing agents lose electrons and are oxidized. Recognizing these agents in a reaction helps to predict and balance chemical equations. For instance, in the given equation \(\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}+\mathrm{C}_{2}\mathrm{O}_{4}^{2-}\rightarrow \mathrm{Cr}^{3+}+\mathrm{CO}_{2} \), \(\mathrm{C}_{2}\mathrm{O}_{4}^{2-} \) is oxidized, and \(\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} \) is reduced.
Acidic Solution Chemistry
In acidic solutions, certain key ions ensure reactions proceed as intended. Acidic conditions often provide protons \(\mathrm{H}^+\) which can drive certain reactions or facilitate balance.
  • Acids can increase solubility of certain reactants and promote electron transfer.
  • Acidic conditions can also change the oxidation states of entities involved in redox reactions.
For example, in the reduction half-reaction of \(\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} + 14\mathrm{H}^+ + 6e^- \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_2\mathrm{O} \), the presence of \(\mathrm{H}^+\) is crucial for balancing both charge and atoms.
Half-Reaction Method
The half-reaction method is a systematic approach to balancing redox reactions, particularly effective in acidic or basic solutions. It breaks down the process into more manageable parts:
  • **Identify the redox pairs**: Determine which components of the reaction are oxidized and reduced.
  • **Write separate half-reactions**: Each for oxidation and reduction, focusing on just one aspect (either electron loss or gain).
  • **Balance each half-reaction**: Using electrons, \(\mathrm{H}_2\mathrm{O}\), and \(\mathrm{H}^+\) for acidic mediums, ensure mass and charge balance.
  • **Combine the half-reactions**: Ensure electrons cancel out, indicating a correct total balance.
This method allows for a detailed understanding of the electron shifts and provides clarity in balancing complex redox reactions like those given in the exercise.

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Most popular questions from this chapter

Zinc reacts spontaneously with silver ion. $$\mathrm{Zn}(s)+2 \mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Zn}^{2+}(a q)+2 \mathrm{Ag}(s)$$ Describe a voltaic cell using this reaction. What are the half-reactions?

Some metals, such as iron, can be oxidized to more than one oxidation state. Obtain the balanced net ionic equations for the following oxidation-reduction reactions, in which nitric acid is reduced to nitric oxide, NO. a. Oxidation of iron metal to iron(II) ion by nitric acid. b. Oxidation of iron(II) ion to iron(III) ion by nitric acid. c. Oxidation of iron metal to iron(III) by nitric acid.

Explain the electrochemistry of rusting.

A voltaic cell has an iron rod in \(0.30 M\) iron(III) chloride solution for the cathode and a zinc rod in \(0.20 \mathrm{M}\) zinc sulfate solution for the anode. The half-cells are connected by a salt bridge. Write the notation for this cell.

A silver oxide-zinc cell maintains a fairly constant voltage during discharge \((1.60 \mathrm{~V})\). The button form of this cell is used in watches, hearing aids, and other electronic devices. The half-reactions are $$\mathrm{Zn}(s)+2 \mathrm{OH}^{-}(a q) \longrightarrow \mathrm{Zn}(\mathrm{OH})_{2}(s)+2 \mathrm{e}^{-}$$ \(\mathrm{Ag}_{2} \mathrm{O}(s)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Ag}(s)+2 \mathrm{OH}^{-}(a q)\) Identify the anode and the cathode reactions. What is the overall reaction in the voltaic cell?
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