Problem 2
Question
The redox reaction among the following is: [Main Jan. 07, 2020 (II)] (a) formation of ozone from atmospheric oxygen in the presence of sunlight (b) reaction of \(\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{3}\) with \(\mathrm{AgNO}_{3}\) (c) reaction of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) with \(\mathrm{NaOH}\) (d) combination of dinitrogen with dioxygen at \(2000 \mathrm{~K}\)
Step-by-Step Solution
Verified Answer
Option (d) is the redox reaction.
1Step 1: Understanding Redox Reactions
Redox reactions involve the transfer of electrons between two substances. In a redox reaction, one substance is oxidized (loses electrons) and the other is reduced (gains electrons). Let's analyze each option to identify which one is a redox reaction.
2Step 2: Analyzing Option (a)
The formation of ozone from atmospheric oxygen can be represented as: \(3\, \text{O}_2 \rightarrow 2\,\text{O}_3\). During this process, oxygen molecules are recombined in the presence of sunlight, but there is no electron transfer involved. Thus, it is not a redox reaction.
3Step 3: Analyzing Option (b)
The reaction of \([\text{Co(H}_2\text{O})_6]\text{Cl}_3\) with \(\text{AgNO}_3\) is a metathesis or double displacement reaction where ions are exchanged, but there is no change in oxidation states of any element. Therefore, it's not a redox reaction.
4Step 4: Analyzing Option (c)
The reaction between \(\text{H}_2\text{SO}_4\) and \(\text{NaOH}\) is a neutralization reaction: \(\text{H}_2\text{SO}_4 + 2\,\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\,\text{H}_2\text{O}\). There is no change in oxidation states of any involved species, hence it's not a redox reaction.
5Step 5: Analyzing Option (d)
The combination of dinitrogen and dioxygen at high temperature \(2000 K\) can be expressed as: \(\text{N}_2 + \text{O}_2 \rightarrow 2\,\text{NO}\). In this reaction, the oxidation state of nitrogen changes from 0 in \(\text{N}_2\) to +2 in \(\text{NO}\), and for oxygen, it changes from 0 in \(\text{O}_2\) to -2 in \(\text{NO}\). This indicates electron transfer, confirming it as a redox reaction.
Key Concepts
Electron TransferOxidation StatesChemical Reactions Analysis
Electron Transfer
In the realm of redox reactions, electron transfer is the core concept. When these reactions occur, electrons are moved from one atom or molecule to another. This movement signals a change in chemical states and marks a process where one substance loses electrons—a process known as oxidation—while another gains these electrons, which is called reduction.
- Oxidation: Loss of electrons
- Reduction: Gain of electrons
Oxidation States
Oxidation states, or oxidation numbers, help us track electron transfer in a chemical reaction. These numbers represent the number of electrons that an atom gains or loses during a reaction. In some sense, they help detail how electron-rich or electron-poor an atom becomes through the reaction.
- An increase in oxidation state: Indicates oxidation
- A decrease in oxidation state: Indicates reduction
Chemical Reactions Analysis
Analyzing chemical reactions involves identifying what type of reaction is taking place. There are several types, including synthesis, decomposition, single displacement, and, notably, redox. In the given exercise, we identified redox reactions by detecting changes in oxidation states and electron transfer principles.
When analyzing a reaction, consider:
When analyzing a reaction, consider:
- Structure of reactants and products
- Oxidation states of involved elements
- Type of bonds breaking and forming
Other exercises in this chapter
Problem 1
Oxidation number of potassium in \(\mathrm{K}_{2} \mathrm{O}, \mathrm{K}_{2} \mathrm{O}_{2}\) and \(\mathrm{KO}_{2}\), respectively, is: \(\quad\) [Main Jan. 07
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Given : \(\mathrm{Co}^{3+}+\mathrm{e}^{-} \rightarrow \mathrm{Co}^{2+} ; \mathrm{E}^{\circ}=+1.81 \mathrm{~V}\) \(\mathrm{Pb}^{4+}+2 \mathrm{e}^{-} \rightarrow
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An example of a disproportionation reaction is: [Main April 12, 2019 (I)] (a) \(2 \mathrm{MnO}_{4}+10 \mathrm{I}^{-}+16 \mathrm{H}^{+} \rightarrow 2 \mathrm{Mn}
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Given that \(\mathrm{E}_{\mathrm{O}_{2} / \mathrm{H}_{2} \mathrm{O}}^{\circ}=+1.23 \mathrm{~V} ;\) \(\mathrm{E}_{\mathrm{S}_{2} \mathrm{O}_{8}^{2-} / \mathrm{SO
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