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Balance the following redox equations by the half-reaction method: (a) \(\mathrm{Mn}^{2+}+\mathrm{H}_{2} \mathrm{O}_{2} \longrightarrow \mathrm{MnO}_{2}+\mathrm{H}_{2} \mathrm{O}\) (in basic solution) (b) \(\mathrm{Bi}(\mathrm{OH})_{3}+\mathrm{SnO}_{2}^{2-} \longrightarrow \mathrm{SnO}_{3}^{2-}+\mathrm{Bi}\) (in basic solution) (c) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+\mathrm{C}_{2} \mathrm{O}_{4}^{2-} \longrightarrow \mathrm{Cr}^{3+}+\mathrm{CO}_{2}\) (in acidic solution) (d) \(\mathrm{ClO}_{3}^{-}+\mathrm{Cl}^{-} \longrightarrow \mathrm{Cl}_{2}+\mathrm{ClO}_{2}\) (in acidic solution) (e) \(\mathrm{Mn}^{2+}+\mathrm{BiO}_{3}^{-} \longrightarrow \mathrm{Bi}^{3+}+\mathrm{MnO}_{4}^{-}\) (in acidic solution)

Short Answer

Expert verified
Balance each reaction by writing and balancing its half-reactions with water, H+/OH-, and electrons, ensuring charges and atoms are balanced.

Step by step solution

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01

Define the Redox Reaction Type

For each of the given reactions, determine whether they occur in acidic or basic conditions. This influences how you balance oxygen and hydrogen atoms.
02

Write the Half-Reactions for each component

Split each reaction into oxidation and reduction half-reactions. This involves identifying which species gain and which lose electrons.
03

Balance Atoms other than O and H

In each half-reaction, balance all elements other than oxygen and hydrogen first. This ensures the major atoms involved are balanced.
04

Balance Oxygens using Water

Balance oxygen atoms by adding H2O molecules to the side with fewer oxygen atoms. This is done for each half-reaction.
05

Balance Hydrogens using H+ or OH-

For acidic solutions, balance hydrogens by adding H+ ions to the side with fewer hydrogen atoms. In basic solutions, add OH- to both sides for every H+ added (forming water) to balance it.
06

Balance Charges using Electrons

Equalize the charge on both sides of the half-reactions by adding electrons. This step ensures the charge is the same on both sides, thus balancing it electrochemically.
07

Equalize the Number of Electrons Transferred

Multiply each half-reaction by the appropriate coefficient so the number of electrons on both half-reactions is the same, allowing them to be added together.
08

Add the Half-Reactions

Combine the oxidation and reduction half-reactions, ensuring to cancel out electrons and any other species that appear on both sides.
09

Verify Mass and Charge Balance

Check that both mass and charge are balanced in the final reaction. Both sides should have the same total mass and charge.

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 powerful tool for balancing redox reactions. It involves breaking down a complicated reaction into two simpler ones: an oxidation half-reaction and a reduction half-reaction. This approach allows you to focus separately on electron gain and loss, which is the essence of redox reactions.
  • **Split the Reactions**: Identify which elements undergo oxidation (loss of electrons) and reduction (gain of electrons).
  • **Write Two Half-Reactions**: Separate the identified oxidation and reduction processes into distinct half-reactions. This makes it easier to track the movement of electrons.
By using this method, you simplify the process, ensuring that all atoms and charges are balanced correctly. This breakdown also helps in visualizing the entire reaction more clearly, making the balancing more straightforward.
Balancing Equations
Balancing equations is crucial because it ensures the conservation of mass and charge. When balancing redox reactions, you focus on both these aspects:
  • **Balance Atoms**: Begin by balancing all atoms except for oxygen and hydrogen. This means making sure each type of atom on the reactant side has an equivalent amount on the product side.
  • **Balance Oxygen and Hydrogen**: In this step, balance oxygen by adding water (H₂O) molecules. Then, balance hydrogen using hydrogen ions (H⁺) in acidic solutions or hydroxide ions (OH⁻) in basic solutions.
  • **Balance Charges**: Add electrons to one side of the half-reaction to equalize the charges on both sides.
After balancing individual half-reactions, they are recombined to form the full equation, ensuring that both mass and charge are conserved.
Basic Solution
In a basic solution, the procedures for balancing redox reactions require special adaptations due to the presence of OH⁻ ions. Here’s how the process differs:
  • **Identifying the Basic Solution**: Recognize the basic condition, which necessitates the use of OH⁻ ions in the balancing process.
  • **Balancing with Water and Hydroxide**: First, balance oxygen with water molecules. Then, balance hydrogen by adding an appropriate quantity of OH⁻ to both sides of the half-reaction for each H⁺ ion, yielding water molecules on one side.
  • **Balancing Charges with Electrons**: Add electrons to balance the charges throughout the reaction, ensuring it is electrically neutral.
Following these steps ensures that the redox equation stays balanced under basic conditions, making it chemically and mathematically accurate.
Acidic Solution
Balancing a redox reaction in an acidic solution involves using H⁺ ions to maintain balance between elements and charges. Here’s how it’s approached:
  • **Recognizing the Acidic Environment**: Verify that the solution is acidic, signaling the use of H⁺ ions in the balancing.
  • **Balancing with Water and Hydrogen Ions**: Balance oxygen atoms by introducing H₂O molecules, then add H⁺ ions to balance the hydrogen atoms.
  • **Balancing with Electrons**: Adjust the charge of the half-reactions by adding electrons where necessary, ensuring that both sides are charge equalized.
This straightforward method keeps the redox reaction balanced both in terms of mass and charge, suitable for the acidic conditions specified.

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Most popular questions from this chapter

One of the half-reactions for the electrolysis of water is: $$2 \mathrm{H}^{+}(a q)+2 e^{-} \longrightarrow \mathrm{H}_{2}(g)$$ If \(0.845 \mathrm{~L}\) of \(\mathrm{H}_{2}\) is collected at \(25^{\circ} \mathrm{C}\) and \(782 \mathrm{mmHg}\), how many faradays of electricity had to pass through the solution?

The concentration of a hydrogen peroxide solution can be conveniently determined by titration against a standardized potassium permanganate solution in an acidic medium according to the following unbalanced equation: $$\mathrm{MnO}_{4}^{-}+\mathrm{H}_{2} \mathrm{O}_{2} \longrightarrow \mathrm{O}_{2}+\mathrm{Mn}^{2+}$$ (a) Balance this equation. (b) If \(36.44 \mathrm{~mL}\) of a 0.01652 \(M \mathrm{KMnO}_{4}\) solution is required to completely oxidize \(25.00 \mathrm{~mL}\) of an \(\mathrm{H}_{2} \mathrm{O}_{2}\) solution, calculate the molarity of the \(\mathrm{H}_{2} \mathrm{O}_{2}\) solution.

The equilibrium constant for the reaction: $$\operatorname{Sr}(s)+\mathrm{Mg}^{2+}(a q) \rightleftharpoons \mathrm{Sr}^{2+}(a q)+\mathrm{Mg}(s)$$ is \(2.69 \times 10^{12}\) at \(25^{\circ} \mathrm{C}\). Calculate \(E^{\circ}\) for a cell made up of \(\mathrm{Sr} / \mathrm{Sr}^{2+}\) and \(\mathrm{Mg} / \mathrm{Mg}^{2+}\) half-cells.

Explain why chlorine gas can be prepared by electrolyzing an aqueous solution of \(\mathrm{NaCl}\) but fluorine gas cannot be prepared by electrolyzing an aqueous solution of NaF.

A silver rod and a SHE are dipped into a saturated aqueous solution of silver oxalate \(\left(\mathrm{Ag}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right)\), at \(25^{\circ} \mathrm{C}\). The measured potential difference between the rod and the SHE is \(0.589 \mathrm{~V},\) the rod being positive. Calculate the solubility product constant for silver oxalate.

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