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Write balanced equations for the following halfreactions. Specify whether each is an oxidation or reduction. (a) \(\mathrm{Cr}(\mathrm{s}) \rightarrow \mathrm{Cr}^{3+}(\text { aq })\) (in acid) (b) \(\mathrm{AsH}_{3}(\mathrm{g}) \rightarrow \mathrm{As}(\mathrm{s})\) (in acid) (c) \(\mathrm{VO}_{3}^{-}(\mathrm{aq}) \rightarrow \mathrm{V}^{2+}(\mathrm{aq})\) (in acid) (d) \(\mathrm{Ag}(\mathrm{s}) \rightarrow \mathrm{Ag}_{2} \mathrm{O}(\mathrm{s})\) (in base)

Short Answer

Expert verified
(a) Oxidation, (b) Oxidation, (c) Reduction, (d) Oxidation.

Step by step solution

01

Understand the Problem

We need to write balanced half-reactions for the given equations and determine whether each represents an oxidation or a reduction process. Balancing will consider the appropriate acidic or basic media as specified.
02

Balance Chromium Half-Reaction (a)

Start with \( \mathrm{Cr} \rightarrow \mathrm{Cr}^{3+} \). Since it occurs in acidic solution, add \( 3\,\mathrm{e}^- \) to the right side to account for the loss of three electrons in forming \( \mathrm{Cr}^{3+} \). The balanced half-reaction is:\[ \mathrm{Cr} \rightarrow \mathrm{Cr}^{3+} + 3\, \mathrm{e}^- \]This is an oxidation reaction because electrons are lost.
03

Balance Arsenic Half-Reaction (b)

For \( \mathrm{AsH}_3 \rightarrow \mathrm{As} \), remove hydrogen by adding 3 protons and 3 electrons to the right side in an acidic solution. The balanced half-reaction is:\[ \mathrm{AsH}_3 \rightarrow \mathrm{As} + 3\, \mathrm{H}^+ + 3\, \mathrm{e}^- \]This is an oxidation reaction since electrons are released.
04

Balance Vanadium Half-Reaction (c)

Take \( \mathrm{VO}_3^- \rightarrow \mathrm{V}^{2+} \) and balance oxygen by adding water molecules to the right. Then balance the hydrogen by adding protons to the left, and finally balance charge with electrons. The balanced half-reaction is:\[ \mathrm{VO}_3^- + 4\, \mathrm{H}^+ + 3\, \mathrm{e}^- \rightarrow \mathrm{V}^{2+} + 3\, \mathrm{H}_2\mathrm{O} \]This is a reduction reaction because electrons are gained.
05

Balance Silver Half-Reaction (d)

For \( \mathrm{Ag} \rightarrow \mathrm{Ag}_2 \mathrm{O} \) in a basic solution, start by balancing oxygen with water on the right and hydrogen by adding \( \mathrm{OH}^- \) on the left to neutralize protons. Then balance electrons. The balanced half-reaction is:\[ 2\, \mathrm{Ag} + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{Ag}_2\mathrm{O} + 2\, \mathrm{H}^+ + 2\, \mathrm{e}^- \]This is an oxidation reaction since electrons are lost.

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Key Concepts

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

Oxidation
In chemistry, oxidation refers to the process whereby a substance loses electrons. When a material undergoes oxidation, it increases its oxidation state. This is a fundamental part of a larger class of reactions known as redox reactions, where one chemical species is oxidized and another is reduced.
Oxidation can be remembered using the acronym OIL RIG: "Oxidation Is Loss" of electrons. For instance, in the half-reaction \( \mathrm{Cr} \rightarrow \mathrm{Cr}^{3+} + 3e^- \), chromium loses three electrons, thus its oxidation state increases from 0 to +3. This confirms it is an oxidation process.
  • Loss of electrons
  • Increase in oxidation state
  • Often involves a combining with oxygen in chemical reactions
Understanding oxidation helps in deciphering redox reactions and identifying agents that cause or undergo changes in electron number.
Reduction
Reduction is the counter process to oxidation in redox reactions. It involves the gain of electrons by a molecule, atom, or ion. As a result, the oxidation state of the substance decreases. This can be remembered by the RIG part of the acronym OIL RIG: "Reduction Is Gain" of electrons.
For example, in the balanced half-reaction \( \mathrm{VO}_3^- + 4\, \mathrm{H}^+ + 3\, \mathrm{e}^- \rightarrow \mathrm{V}^{2+} + 3\, \mathrm{H}_2\mathrm{O} \), the vanadium ion gains three electrons, reducing its oxidation state from +5 to +2. This indicates a reduction reaction.
  • Gain of electrons
  • Decrease in oxidation state
  • Common in processes involving the donation of electrons to other chemicals
Identifying reduction is crucial in deciphering how substances interact and transform during chemical reactions.
Half-Reactions
Redox reactions are composed of two half-reactions: one for oxidation and another for reduction. Each half-reaction represents either the loss or gain of electrons, making it easier to understand and balance complex reactions.
In our examples, we write separate half-reactions for the processes. For instance, \( \mathrm{AsH}_3 \rightarrow \mathrm{As} + 3e^- + 3H^+ \) is an oxidation half-reaction where arsenic hydride loses electrons, while \( \mathrm{VO}_3^- + 4H^+ + 3e^- \rightarrow \mathrm{V}^{2+} + 3H_2O \) acts as a reduction half-reaction.
  • Helps isolate each step in a redox reaction
  • Essential for balancing and understanding electron transfer
Half-reactions are an essential schematic tool to systematically analyze the electron movements within a reaction, facilitating understanding and calculation.
Acidic Media
In redox reactions, acidic media refer to aqueous environments with an abundance of hydrogen ions (\(H^+\)). This environment influences how substances interact, particularly affecting which species are stable and how they react.
When balancing half-reactions in acidic media, such as \( \mathrm{AsH}_3 \rightarrow \mathrm{As} + 3\, \mathrm{H}^+ + 3\, \mathrm{e}^- \), protons are used to balance hydrogen atoms. Acidic conditions help in distributing charge and matter in reactions as protons need to be accounted for.
  • Frequent presence of \(H^+\) ions
  • Used to balance hydrogen atoms in reactions
  • Promotes stability for different chemical species
Understanding the role of acidic media in chemical reactions is fundamental for predicting and balancing redox processes.
Basic Media
Basic media describe conditions in a solution with significant hydroxide ions (\( \mathrm{OH}^- \)) concentration. These basic environments significantly impact the course and equilibrium of redox reactions.
When dealing with balancing in basic media, such as with \( 2 \mathrm{Ag} + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{Ag}_2\mathrm{O} + 2 \mathrm{H}^+ + 2 \mathrm{e}^- \), hydroxide ions are added to balance any additional hydrogen introduced by water molecules
  • Dominated by \(\mathrm{OH}^-\) ions
  • Utilized to balance excess hydrogen in half-reactions
  • Contributes to the neutralization of net charge in reactions
Comprehending the influences of basic media helps in mastering the balancing of chemical equations within alkaline environments.

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Most popular questions from this chapter

S. Balance the following equations. (a) \(\mathrm{Zn}(\mathrm{s})+\mathrm{VO}^{2+}(\mathrm{aq}) \rightarrow\) \(\mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{V}^{3+}(\mathrm{aq}) \quad\) (acid solution) (b) \(\mathrm{Zn}(\mathrm{s})+\mathrm{VO}_{3}^{-}(\mathrm{aq}) \rightarrow\) \(\mathrm{V}^{2+}(\mathrm{aq})+\mathrm{Zn}^{2+}(\mathrm{aq}) \quad\) (acid solution) (c) \(\mathrm{Zn}(\mathrm{s})+\mathrm{ClO}^{-}(\mathrm{aq}) \rightarrow\) \(\mathrm{Zn}(\mathrm{OH})_{2}(\mathrm{s})+\mathrm{Cl}^{-}(\mathrm{aq}) \quad\) (basic solution) (d) CIO-(aq) \(+\left[\mathrm{Cr}(\mathrm{OH})_{4}\right]^{-}(\mathrm{aq}) \rightarrow\) \(\mathrm{Cl}^{-}(\mathrm{aq})+\mathrm{CrO}_{4}^{2-}(\mathrm{aq})\) (basic solution)

A The amount of oxygen, \(\mathrm{O}_{2}\), dissolved in a water sample at \(25^{\circ} \mathrm{C}\) can be determined by titration. The first step is to add solutions of \(\mathrm{MnSO}_{4}\) and NaOH to the water to convert the dissolved oxygen to \(\mathrm{MnO}_{2}\). A solution of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) and \(\mathrm{KI}\) is then added to convert the \(\mathrm{MnO}_{2}\) to \(\mathrm{Mn}^{2+},\) and the iodide ion is converted to \(\mathrm{I}_{2}\). The \(\mathrm{I}_{2}\) is then titrated with standardized \(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3-}\) (a) Balance the equation for the reaction of \(\mathrm{Mn}^{2+}\) ions with \(\mathrm{O}_{2}\) in basic solution. (b) Balance the equation for the reaction of \(\mathrm{MnO}_{2}\) with \(\mathbf{I}^{-}\) in acid solution. (c) Balance the equation for the reaction of \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) with \(\mathrm{I}_{2}\) (d) Calculate the amount of \(\mathbf{O}_{2}\) in \(25.0 \mathrm{mL}\) of water if the titration requires \(2.45 \mathrm{mL}\) of \(0.0112 \mathrm{M} \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}\) solution.

An old method of measuring the current flowing in a circuit was to use a "silver coulometer." The current passed first through a solution of \(\mathrm{Ag}^{+}(\mathrm{aq})\) and then into another solution containing an electroactive species. The amount of silver metal deposited at the cathode was weighed. From the mass of silver, the number of atoms of silver was calculated. Since the reduction of a silver ion requires one electron, this value equaled the number of electrons passing through the circuit. If the time was noted, the average current could be calculated. If, in such an experiment, 0.052 g of Ag is deposited during \(450 \mathrm{s},\) what was the current flowing in the circuit?

Predict the products formed in the electrolysis of an aqueous solution of \(\mathrm{CdSO}_{4}\)

A hydrogen-oxygen fuel cell operates on the simple reaction $$ \mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\ell) $$ If the cell is designed to produce 1.5 A of current and if the hydrogen is contained in a 1.0 -L. tank at 200 atm pressure at \(25^{\circ} \mathrm{C},\) how long can the fuel cell operate before the hydrogen runs out? (Assume there is an unlimited supply of \(\mathbf{O}_{2}\).)

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