Problem 26
Question
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) \(\mathrm{PCl}_{5} \longrightarrow \mathrm{PCl}_{3}+\mathrm{Cl}_{2}\)
Step-by-Step Solution
Verified Answer
Equation (a) does not represent a redox reaction.
1Step 1: 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.
2Step 2: 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.
3Step 3: 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.
4Step 4: 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.
5Step 5: 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.
6Step 6: 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.
Key Concepts
Oxidation StatesReduction-OxidationElectron Transfer
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:
A change in an atom's oxidation state indicates electron transfer, thus identifying the nature of the reaction as redox if applicable.
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.
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:
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.
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.
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:
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.
In each redox reaction:
- Oxidizing agents accept electrons. They themselves are reduced.
- Reducing agents donate electrons. They themselves are oxidized.
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.
Other exercises in this chapter
Problem 24
A, B and C have the oxidation numbers of \(+6,-2\) and \(-1\) respectively, the possible molecular formula when these atoms combine will be (a) \(\mathrm{A}_{2}
View solution Problem 25
Oxidation numbers of carbon in \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}, \mathrm{CH}_{4}\) and diamond respectively are (a) \(+3,4\) and \(+4\) (b) \(+3,-
View solution Problem 27
If \(\mathrm{Cl}_{2}\) is passed through hot \(\mathrm{NaOH}\), oxidation number of \(\mathrm{Cl}\) changes from (a) \(-1\) to 0 (b) 0 to \(-1\) (c) 0 to \(+5\)
View solution Problem 28
\(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+6 \mathrm{I}^{-}+14 \mathrm{H}^{+} \longrightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O}+3 \mathrm{I}_{2}\) Equiv
View solution