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Balance this redox reaction (in acidic solution). $$ \mathrm{Cu}(\mathrm{s})+\mathrm{NO}_{3}^{-}(\mathrm{aq}) \longrightarrow \mathrm{Cu}^{2+}(\mathrm{aq})+\mathrm{NO}_{2}(\mathrm{~g}) $$

Short Answer

Expert verified
The balanced equation is: \( \mathrm{Cu} + 2 \mathrm{H}^{+} + \mathrm{NO}_{3}^{-} \rightarrow \mathrm{Cu}^{2+} + \mathrm{NO}_{2} + \mathrm{H}_{2}O \).

Step by step solution

01

Identify Oxidation and Reduction Half-Reactions

First, determine which elements are being oxidized and reduced. Copper (Cu) is oxidized from 0 to +2, and Nitrogen in \( \text{NO}_3^{-} \) is reduced from +5 in \( \text{NO}_3^{-} \) to +4 in \( \text{NO}_2 \).
02

Write Oxidation Half-Reaction

The oxidation half-reaction involves copper: \[ \mathrm{Cu} \rightarrow \mathrm{Cu}^{2+} + 2e^{-} \]
03

Write Reduction Half-Reaction

The reduction half-reaction involves nitrate: \[ \mathrm{NO}_{3}^{-} + 2H^{+} + e^{-} \rightarrow \mathrm{NO}_{2} + H_{2}O \]
04

Balance Electrons between Half-Reactions

Make sure the electrons lost in the oxidation half-reaction equals electrons gained in the reduction half-reaction. We have already balanced electrons as both reactions involve 2 electrons.
05

Combine Balanced Half-Reactions

Add the balanced half-reactions together, ensuring that electrons cancel out:\[ \mathrm{Cu} + 2 \mathrm{H}^{+} + \mathrm{NO}_{3}^{-} \rightarrow \mathrm{Cu}^{2+} + \mathrm{NO}_{2} + \mathrm{H}_{2}O \]
06

Verify Atoms and Charges Balance

Check that each element is balanced on both sides, and ensure the charges match. Here, we have: Cu, two H, one N, and three O atoms on both sides, with a net charge of +1 on each side.

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Key Concepts

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

Oxidation and Reduction
Understanding oxidation and reduction is essential for solving redox reactions. These processes describe the transfer of electrons between substances. In every redox reaction, oxidation refers to the loss of electrons by a molecule, atom, or ion.
On the other hand, reduction signifies the gain of electrons.
In our example, copper (Cu) is oxidized because it loses electrons, transforming from Cu (0) to Cu\(^{2+}\). This means it loses two electrons. Meanwhile, nitrogen in nitrate ion (\(\text{NO}_3^-\)) is reduced. It gains electrons, transitioning from an oxidation state of +5 to +4 in \(\text{NO}_2\).
Always remember that the substance that gets oxidized is the reducing agent (since it donates electrons), and the one getting reduced is the oxidizing agent (as it accepts electrons). Recognizing these can help in balancing complex reactions.
Half-Reactions
Half-reactions simplify the process of balancing redox reactions by isolating the oxidation and reduction processes. They allow us to see clearly the flow of electrons.
Each half-reaction focuses on one part of the redox equation. For oxidation, we write out the species losing electrons, while for reduction, the species gaining electrons.
In our specific redox equation:
  • The oxidation half-reaction is: \[ \text{Cu} \rightarrow \text{Cu}^{2+} + 2e^- \]
  • The reduction half-reaction is: \[ \text{NO}_3^- + 2H^+ + e^- \rightarrow \text{NO}_2 + H_2O \]
After writing them out, it becomes easier to see how electrons are transferred and balance each half-reaction's electron count. This is necessary before combining them to ensure that the number of electrons lost equals the number gained.
Acidic Solution Balancing
Balancing redox reactions in acidic solutions involves more than just balancing charges and atoms. It includes adding \( \text{H}^+ \) ions to reflect the acidic environment.
In an acidic solution,\( \text{H}^+ \) ions not only help in balancing hydrogen atoms but can also influence the overall charge balance of the reaction.
Take a look at our reduction half-reaction from above: \[ \text{NO}_3^- + 2H^+ + e^- \rightarrow \text{NO}_2 + H_2O \]
The \( \text{H}^+ \) ions are necessary to balance the hydrogen in the final product \( \text{H}_2O \). Also, in an acidic solution, the presence of \( \text{H}^+ \) helps maintain the electrical charge across the reaction. This aspect of balancing ensures that all atoms, including hydrogen and oxygen, as well as charges, are equal on both sides of the reaction. Always double-check your final equation to verify that both mass and charge are balanced.

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Most popular questions from this chapter

Two different isomers are known with the formula \(\left[\mathrm{Pt}(\mathrm{py})_{2} \mathrm{Cl}_{2}\right],\) where py represents pyridine, an uncharged monodentate ligand in which an \(\mathrm{N}\) atom bonds to the metal ion. There is, however, only one structure known for \(\left[\mathrm{Pt}(\mathrm{phen}) \mathrm{Cl}_{2}\right],\) where phen represents 1,10 -phenanthroline, an uncharged bidentate ligand (Question 104 ). Draw the structural formulas of all three molecules and explain why there are isomers in one case, but not the other.

An electrochemical cell is made by immersing a strip of chromium into a \(1.0-\mathrm{M}\) solution of \(\mathrm{Cr}^{3+}\) and a strip of gold into a \(1.0-\mathrm{M}\) solution of \(\mathrm{Au}^{3+} .\) The half-cells are connected by a salt bridge. A wire and light bulb complete the circuit. (a) Write the balanced chemical equation for the reaction that is product- favored. (b) Calculate the cell potential. (c) Draw a sketch of the cell and indicate the anode, cathode, and direction of electron flow.

Predict which element in the pair has the larger atomic radius. (a) Cu or Ag (b) \(\mathrm{T} \mathrm{i}\) or \(\mathrm{Cr}\) (c) W or \(\mathrm{Hg}\)

Determine whether each statement is true or false. If it is false, correct the statement. (a) In \(\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{4}\right]\), platinum has a \(4+\) charge and a coordination number of six. (b) In general, \(\mathrm{Cu}^{2+}\) is more stable than \(\mathrm{Cu}^{+}\) in aqueous solutions.

In aqueous solution, \(\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}\) is yellow, but aqueous \(\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2}\) is purple. Explain the difference in colors.

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