Question

Justify that the following reactions are redox reactions: (a) \(\mathrm{CuO}(\mathrm{s})+\mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{Cu}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (b) \(\mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{~s})+3 \mathrm{CO}(\mathrm{g}) \rightarrow 2 \mathrm{Fe}(\mathrm{s})+3 \mathrm{CO}_{2}(\mathrm{~g})\) (c) \(4 \mathrm{BCl}_{3}(\mathrm{~g})+\) 3LiAlH \(_{4}(\mathrm{~s}) \rightarrow 2 \mathrm{~B}_{2} \mathrm{H}_{6}(\mathrm{~g})+3 \mathrm{LiCl}(\mathrm{s})+3 \mathrm{AlCl}_{3}(\mathrm{~s})\) (d) \(2 \mathrm{~K}(\mathrm{~s})+\mathrm{F}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{~K}^{+} \mathrm{F}^{-}(\mathrm{s})\) (e) \(4 \mathrm{NH}_{3}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\)

Short answer

All reactions are redox because they involve the change in oxidation states, indicating transfer of electrons.

Step-by-step solution

Introduction to Redox Reactions

Redox reactions involve the transfer of electrons between species. Specifically, one species undergoes oxidation (loses electrons) while another species undergoes reduction (gains electrons). To determine if a reaction is redox, identify the changes in oxidation states of the involved elements.

Determine Oxidation States for Reaction (a)

In the reaction \( \text{CuO (s) + H}_2\text{ (g)} \rightarrow \text{Cu (s) + H}_2\text{O (g)} \):- \( \text{CuO} \): Cu is \(+2\), O is \(-2\).- \( \text{H}_2 \): Hydrogen is \(0\).- \( \text{Cu} \): Cu is \(0\).- \( \text{H}_2\text{O} \): O is \(-2\), each H is \(+1\).Cu is reduced from +2 to 0, and H is oxidized from 0 to +1, thus it is a redox reaction.

Evaluate Oxidation States for Reaction (b)

For \( \text{Fe}_2\text{O}_3\text{ (s) + 3CO (g)} \rightarrow 2\text{Fe (s) + 3CO}_2\text{ (g)} \):- \( \text{Fe}_2\text{O}_3 \): Fe is \(+3\), O is \(-2\).- \( \text{CO} \): C is \(+2\), O is \(-2\).- \( \text{Fe} \): Fe is \(0\).- \( \text{CO}_2 \): C is \(+4\), O is \(-2\).Fe goes from +3 to 0 (reduced) and C goes from +2 to +4 (oxidized). Hence, it is a redox reaction.

Check Oxidation States for Reaction (c)

In the reaction \( 4\text{BCl}_3\text{ (g) + 3LiAlH}_4\text{ (s)} \rightarrow 2\text{B}_2\text{H}_6\text{ (g) + 3LiCl (s) + 3AlCl}_3\text{ (s)} \):- \( \text{BCl}_3 \): B is \(+3\), Cl is \(-1\).- \( \text{LiAlH}_4 \): Li is \(+1\), Al is \(+3\), H is \(-1\).- \( \text{B}_2\text{H}_6 \): B is \(-3\), H is \(+1\).- \( \text{LiCl} \): Li is \(+1\), Cl is \(-1\).- \( \text{AlCl}_3 \): Al is \(+3\), Cl is \(-1\).B is reduced from +3 in \( \text{BCl}_3 \) to -3 in \( \text{B}_2\text{H}_6 \), while H goes from -1 to +1, indicating a redox reaction.

Analyze Oxidation Changes for Reaction (d)

For \( 2\text{K (s) + F}_2\text{ (g)} \rightarrow 2\text{K}^+\text{F}^-\text{ (s)} \):- \( \text{K} \): Initially 0, becomes \(+1\) in \( \text{K}^+ \).- \( \text{F}_2 \): Initially 0, becomes \(-1\) in \( \text{F}^- \).K is oxidized from 0 to +1, and F is reduced from 0 to -1, confirming a redox reaction.

Examine Oxidation States for Reaction (e)

In the reaction \( 4\text{NH}_3\text{ (g) + 5O}_2\text{ (g)} \rightarrow 4\text{NO (g) + 6H}_2\text{O (g)} \):- \( \text{NH}_3 \): N is \(-3\), H is \(+1\).- \( \text{O}_2 \): O is \(0\).- \( \text{NO} \): N is \(+2\), O is \(-2\).- \( \text{H}_2\text{O} \): H is \(+1\), O is \(-2\).N is oxidized from -3 to +2, and O is reduced from 0 to -2, indicating a redox reaction.

Key concepts

Oxidation States

Oxidation states, also known as oxidation numbers, are instrumental in understanding chemical reactions, particularly redox reactions. They help to track how electrons are distributed among atoms in molecules. An oxidation state is essentially a bookkeeping tool that indicates the degree of oxidation of an atom.

• Positive oxidation states reflect a loss of electrons.
• Negative oxidation states indicate a gain of electrons.
• Elements in their natural state have an oxidation state of zero. For example, in \(\text{O}_2\), oxygen has an oxidation state of 0.

By examining the changes in oxidation states before and after a reaction, we determine whether a redox reaction has occurred. For example, in the reaction between copper(II) oxide (\(\text{CuO}\)) and hydrogen (\(\text{H}_2\)), copper's oxidation state changes from \(+2\) to 0 (reduction) and hydrogen's from 0 to \(+1\) (oxidation). This is a clear indication of a redox process by electron transfer.

Electron Transfer

In redox reactions, electron transfer is the fundamental process that leads to changes in oxidation states. Electron transfer involves moving electrons from one species to another, which results in the oxidation of one element and the reduction of another.

• Oxidation is the loss of electrons; the oxidized element shows an increase in its oxidation state.
• Reduction is the gain of electrons; the reduced element shows a decrease in its oxidation state.

For instance, in the reduction of iron(III) oxide (\(\text{Fe}_2\text{O}_3\)) by carbon monoxide (\(\text{CO}\)), iron goes from \(+3\) in the oxide to 0 in elemental iron. Meanwhile, carbon in \(\text{CO}\) is oxidized from +2 to +4 in \(\text{CO}_2\). The electrons lost by carbon are the electrons gained by iron.

Oxidation and Reduction

Oxidation and reduction are two sides of a redox reaction and always occur simultaneously. They are often remembered through the mnemonic: OIL RIG ("Oxidation Is Loss, Reduction Is Gain" of electrons).

• In oxidation, a substance loses electrons, increasing its oxidation state.
• In reduction, a substance gains electrons, decreasing its oxidation state.

An example can be seen when lithium aluminum hydride (\(\text{LiAlH}_4\)) reacts with boron trichloride (\(\text{BCl}_3\)). Boron is reduced from \(+3\) in \(\text{BCl}_3\) to \(-3\) in \(\text{B}_2\text{H}_6\), while hydrogen is oxidized from \(-1\) to \(+1\). This complementary exchange of electrons confirms the nature of redox reactions, where one species is oxidized and another is reduced, preserving the balance of charge and electrons throughout the chemical process.