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Chapter 9: Problem 26

Which of the following does not represent redox reactions? (a) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+2 \mathrm{OH}^{-} \longrightarrow \mathrm{CrO}_{4}^{2-}+\mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{SO}_{4}^{2-}+2 \mathrm{I}^{-}+2 \mathrm{H}^{+} \longrightarrow \mathrm{I}_{2}+\mathrm{SO}_{4}^{2-}+\mathrm{H}_{2} \mathrm{O}\) (c) \(2 \mathrm{Ca}(\mathrm{OH})_{2}+2 \mathrm{Cl}_{2} \longrightarrow \mathrm{Ca}(\mathrm{ClO})_{2}\) \(+\mathrm{CaCl}_{2}+2 \mathrm{H}_{2} \mathrm{O}\) (d) \(\mathrm{PCl}_{5} \longrightarrow \mathrm{PCl}_{3}+\mathrm{Cl}_{2}\)

Short Answer

Expert verified
Equation (a) does not represent a redox reaction.

Step by step solution

01

Define Redox Reactions

Redox (reduction-oxidation) reactions involve the transfer of electrons between two species. An increase in oxidation number indicates oxidation, while a decrease indicates reduction.
02

Analyze Equation (a)

Examine the oxidation states in \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+2 \mathrm{OH}^{-} \longrightarrow \mathrm{CrO}_{4}^{2-}+\mathrm{H}_{2} \mathrm{O}\). Chromium in \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) is in the +6 oxidation state and remains +6 in \(\mathrm{CrO}_{4}^{2-}\). There is no change in oxidation states, indicating it is not a redox reaction.
03

Analyze Equation (b)

For \(\mathrm{SO}_{4}^{2-}+2 \mathrm{I}^{-}+2 \mathrm{H}^{+} \longrightarrow \mathrm{I}_{2}+\mathrm{SO}_{4}^{2-}+\mathrm{H}_{2} \mathrm{O}\), check the iodine atoms. The oxidation state of \(\mathrm{I}^{-}\) is -1 and changes to 0 in \(\mathrm{I}_{2}\). This change indicates a redox reaction, as iodine gets oxidized.
04

Analyze Equation (c)

Consider \(2 \mathrm{Ca}(\mathrm{OH})_{2}+2 \mathrm{Cl}_{2} \longrightarrow \mathrm{Ca}(\mathrm{ClO})_{2}+\mathrm{CaCl}_{2}+2 \mathrm{H}_{2} \mathrm{O}\). The oxidation state of chlorine changes from 0 in \(\mathrm{Cl}_2\) to -1 in \(\mathrm{CaCl}_{2}\) and +1 in \(\mathrm{Ca}(\mathrm{ClO})_{2}\). This is a redox reaction involving changes in oxidation states.
05

Analyze Equation (d)

For \(\mathrm{PCl}_{5} \longrightarrow \mathrm{PCl}_{3}+\mathrm{Cl}_{2}\), observe the oxidation states of phosphorus and chlorine. In \(\mathrm{PCl}_{5}\), phosphorus is +5 and chlorine is -1. In \(\mathrm{PCl}_{3}\), phosphorus is +3, and newly formed \(\mathrm{Cl}_2\) indicates chlorine goes to 0. Here, both reduction and oxidation occur; thus, it is a redox reaction.
06

Identify Non-Redox Reaction

From steps 2-5, identify which equation does not involve any change in oxidation states. According to the analysis, equation (a) involves no change in oxidation states and thus is not a redox reaction.

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Key Concepts

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

Oxidation States
To understand redox reactions, we need to grasp the concept of oxidation states. An oxidation state, also known as an oxidation number, is a theoretical charge on an atom if all bonds were ionic. It helps track how many electrons an atom gains, loses, or shares in a compound.

Determining oxidation states requires a few guidelines:
  • Elements in their pure form have an oxidation state of zero. For example, \(\mathrm{Cl}_2\) or \(\mathrm{O}_2\) as diatomic molecules have zero states.
  • For ions, the oxidation state is the same as the ion's charge. For instance, \(\mathrm{Na}^+\) has an oxidation state of +1.
  • Oxygen usually has an oxidation state of -2, and hydrogen is +1 in compounds.
By assigning these states to all atoms in a chemical equation, you can spot any changes.
A change in an atom's oxidation state indicates electron transfer, thus identifying the nature of the reaction as redox if applicable.
Reduction-Oxidation
Reduction-oxidation, or redox, reactions are integral to chemistry. They involve electron transfer between substances, changing oxidation states.

Redox reactions have two parts:
  • Oxidation: Loss of electrons, resulting in an increase in oxidation state. For example, \(\mathrm{I}^-\) to \(\mathrm{I}_2\), where iodine's state grows from -1 to 0.
  • Reduction: Gain of electrons, resulting in a decrease in oxidation state.
Both processes happen simultaneously in a redox reaction. For instance, in \(\mathrm{PCl}_5 \ ightarrow \mathrm{PCl}_3 + \mathrm{Cl}_2\), phosphorus is reduced from +5 to +3 while chlorine is oxidized from -1 to 0.
To identify a redox reaction:- Check the initial and final oxidation states of reactants and products.- Look for shifts in states that indicate electron transfer.
Electron Transfer
At the heart of redox reactions is electron transfer. This transfer can be thought of as the main driver that fuels the changes within the reaction.

In each redox reaction:
  • Oxidizing agents accept electrons. They themselves are reduced.
  • Reducing agents donate electrons. They themselves are oxidized.
Identifying the electron transfer helps in determining which component acts as the oxidizing or reducing agent.
For example, in the reaction \(\mathrm{SO}_4^{2-} + 2 \mathrm{I}^- + 2 \mathrm{H}^+ \ ightarrow \mathrm{I}_2 + \mathrm{SO}_4^{2-} + \mathrm{H}_2\mathrm{O}\), iodine acts as the reducing agent as it donates electrons. By understanding these roles, you can predict how molecules might react and transform in varying chemical environments.

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Most popular questions from this chapter

On electrolysis of acidified water, if volume of hydrogen liberated is \(5.6 \mathrm{~cm}^{3}\), then the volume of oxygen liberated equal to (a) \(1.4 \mathrm{~cm}^{3}\) (b) \(2.8 \mathrm{~cm}^{3}\) (c) \(8.2 \mathrm{~cm}^{3}\) (d) \(5.6 \mathrm{~cm}^{3}\)

On the basis of the information available from the reaction, \(\frac{4}{3} \mathrm{Al}+\mathrm{O}_{2} \longrightarrow{ }_{3}^{2} \mathrm{Al}_{2} \mathrm{O}_{3}\) \(\Delta \mathrm{G}=-827 \mathrm{~kJ} \mathrm{~mol}^{-1}\) of \(\mathrm{O}_{2}\) The minimum emf, required to carry out an electrolysis of \(\mathrm{Al}_{2} \mathrm{O}_{3}\) is \(\left(F=96500 \mathrm{C} \mathrm{mol}^{-1}\right)\) (a) \(6.42 \mathrm{~V}\) (b) \(8.56 \mathrm{~V}\) (c) \(2.14 \mathrm{~V}\) (d) \(4.28 \mathrm{~V}\)

Among the following, identify the species with an atom in \(+6\) oxidation state (a) \(\mathrm{MnO}_{4}\) (b) \(\mathrm{Cr}(\mathrm{CN})_{6}^{3}\) (c) \(\mathrm{NiF}_{6}^{2-}\) (d) \(\mathrm{CrO}_{2} \mathrm{Cl}_{2}\)

The emf of a Daniell cell at \(298 \mathrm{~K}\) is \(E_{\mathrm{i}}\) \(\mathrm{Zn}\left|\mathrm{ZnSO}_{4} \| \mathrm{CuSO}_{4}\right| \mathrm{Cu}\) \((0.01 \mathrm{M}) \quad(1.0 \mathrm{M})\) when the concentration of \(\mathrm{ZnSO}_{4}\) is \(1.0 \mathrm{M}\) and that of \(\mathrm{CuSO}_{4}\) is \(0.01 \mathrm{M}\), the emf changed to \(E_{2}\). What is the relationship between \(E_{1}\) and \(E_{2} ?\) (a) \(E_{1}=E_{2}\) (b) \(E_{2} \neq E_{1}\) (c) \(E_{1}>E_{2}\) (d) \(E_{1}

The oxidation number of phosphorous in ATP (adenosine triphosphate) is (a) 2 (b) \(\underline{3}\) (c) 4 (d) 5
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