Chapter 20: Problem 20
Complete and balance the following half-reactions. In each case indicate whether the half-reaction is an oxidation or a reduction. (a) \(\mathrm{Mo}^{3+}(a q) \longrightarrow \mathrm{Mo}(s)\) (acidic or basic solution) (b) \(\mathrm{H}_{2} \mathrm{SO}_{3}(a q)--\rightarrow \mathrm{SO}_{4}^{2-}(a q)\) (acidic solution) (c) \(\mathrm{NO}_{3}^{-}(a q)-\cdots \rightarrow \mathrm{NO}(g)\) (acidic solution) (d) \(\mathrm{O}_{2}(\mathrm{~g})-\longrightarrow \mathrm{H}_{2} \mathrm{O}(l)\) (acidic solution) (e) \(\mathrm{Mn}^{2+}(a q)-\rightarrow \rightarrow \mathrm{MnO}_{2}(s)\) (basic solution) (f) \(\mathrm{Cr}(\mathrm{OH})_{3}(s)-\cdots \mathrm{CrO}_{4}^{2-}(a q)\) (basic solution) (g) \(\mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)\) (basic solution)
Short Answer
Expert verified
(a) \(\mathrm{Mo}^{3+}(aq) + 3e^- \longrightarrow \mathrm{Mo}(s)\) (reduction)
(b) \(\mathrm{H}_{2}\mathrm{SO}_{3}(aq) + \mathrm{H}_{2}\mathrm{O}(l) + 2e^- \rightarrow \mathrm{SO}_{4}^{2-}(aq) + 4\mathrm{H}^{+}(aq)\) (oxidation)
(c) \(3e^- + \mathrm{NO}_{3}^{-}(aq) + 6\mathrm{H}^{+}(aq) \rightarrow \mathrm{NO}(g) + 2\mathrm{H}_{2}\mathrm{O}(l)\) (reduction)
(d) \(4e^- + \mathrm{O}_{2}(g) + 4\mathrm{H}^{+}(aq) \rightarrow 2\mathrm{H}_{2}\mathrm{O}(l)\) (reduction)
(e) \(\mathrm{Mn}^{2+}(aq) + 2\mathrm{H}_{2}\mathrm{O}(l) + 2e^- \rightarrow \mathrm{MnO}_{2}(s) + 4\mathrm{OH}^{-}(aq)\) (oxidation)
(f) \(3e^- + \mathrm{Cr}(\mathrm{OH})_{3}(s) + 6\mathrm{OH}^{-}(aq) \rightarrow \mathrm{CrO}_{4}^{2-}(aq) + 3\mathrm{H}_{2}\mathrm{O}(l)\) (oxidation)
(g) \(4e^- + \mathrm{O}_{2}(g) + 4\mathrm{OH}^{-}(aq) \rightarrow 2\mathrm{H}_{2}\mathrm{O}(l)\) (reduction)
Step by step solution
01
Balance elements
Mo is already balanced.
02
Balance O
There are no O atoms in this half-reaction.
03
Balance H
There are no H atoms in this half-reaction.
04
Balance the charge
Add 3 electrons to the left side to balance the +3 charge on the Mo3+ ion:
\[\mathrm{Mo}^{3+}(aq) + 3e^- \longrightarrow \mathrm{Mo}(s)\]
05
Determine oxidation or reduction
The oxidation number of Mo decreases from +3 to 0, so this is a reduction half-reaction.
(b) \(\mathrm{H}_{2} \mathrm{SO}_{3}(a q)--\rightarrow \mathrm{SO}_{4}^{2-}(a q)\) (acidic solution)
06
Balance elements
Balance S: already balanced.
07
Balance O
There are 3 O atoms on the left side and 4 O atoms on the right side. Add 1 H2O molecule to the left side to balance oxygen:
\[\mathrm{H}_{2}\mathrm{SO}_{3}(aq) + \mathrm{H}_{2}\mathrm{O}(l) \rightarrow \mathrm{SO}_{4}^{2-}(aq)\]
08
Balance H
There are now 4 H atoms on the left side and none on the right side. Add 4 H+ ions to the right side to balance hydrogen:
\[\mathrm{H}_{2}\mathrm{SO}_{3}(aq) + \mathrm{H}_{2}\mathrm{O}(l) \rightarrow \mathrm{SO}_{4}^{2-}(aq) + 4\mathrm{H}^{+}(aq)\]
09
Balance the charge
Add 2 electrons to the right side to balance the charge:
\[\mathrm{H}_{2}\mathrm{SO}_{3}(aq) + \mathrm{H}_{2}\mathrm{O}(l) + 2e^- \rightarrow \mathrm{SO}_{4}^{2-}(aq) + 4\mathrm{H}^{+}(aq)\]
10
Determine oxidation or reduction
The oxidation number of S increases from +4 to +6, so this is an oxidation half-reaction.
(c) \(\mathrm{NO}_{3}^{-}(a q)-\cdots \rightarrow \mathrm{NO}(g)\) (acidic solution)
11
Balance elements
Balance N: already balanced.
12
Balance O
Add 2 H2O molecules to the right side to balance oxygen:
\[\mathrm{NO}_{3}^{-}(aq) \rightarrow \mathrm{NO}(g) + 2\mathrm{H}_{2}\mathrm{O}(l)\]
13
Balance H
Add 6 H+ ions to the left side to balance hydrogen:
\[\mathrm{NO}_{3}^{-}(aq) + 6\mathrm{H}^{+}(aq) \rightarrow \mathrm{NO}(g) + 2\mathrm{H}_{2}\mathrm{O}(l)\]
14
Balance the charge
Add 3 electrons to the left side to balance the charge:
\[3e^- + \mathrm{NO}_{3}^{-}(aq) + 6\mathrm{H}^{+}(aq) \rightarrow \mathrm{NO}(g) + 2\mathrm{H}_{2}\mathrm{O}(l)\]
15
Determine oxidation or reduction
The oxidation number of N decreases from +5 to +2, so this is a reduction half-reaction.
(d) \(\mathrm{O}_{2}(\mathrm{~g})-\longrightarrow \mathrm{H}_{2} \mathrm{O}(l)\) (acidic solution)
16
Balance elements
Balance O: Multiply the right side by 2 to balance oxygen:
\[\mathrm{O}_{2}(g) \rightarrow 2\mathrm{H}_{2}\mathrm{O}(l)\]
17
Balance O
Oxygen is already balanced.
18
Balance H
Add 4 H+ ions to the left side to balance hydrogen:
\[\mathrm{O}_{2}(g) + 4\mathrm{H}^{+}(aq) \rightarrow 2\mathrm{H}_{2}\mathrm{O}(l)\]
19
Balance the charge
Add 4 electrons to the left side to balance the charge:
\[4e^- + \mathrm{O}_{2}(g) + 4\mathrm{H}^{+}(aq) \rightarrow 2\mathrm{H}_{2}\mathrm{O}(l)\]
20
Determine oxidation or reduction
The oxidation number of O decreases from 0 to -2, so this is a reduction half-reaction.
(e) \(\mathrm{Mn}^{2+}(a q)-\rightarrow \rightarrow \mathrm{MnO}_{2}(s)\) (basic solution)
21
Balance elements
Balance Mn: already balanced.
22
Balance O
There are no O atoms on the left side and 2 O atoms on the right side. Add 2 H2O molecules to the left side to balance oxygen:
\[\mathrm{Mn}^{2+}(aq) + 2\mathrm{H}_{2}\mathrm{O}(l) \rightarrow \mathrm{MnO}_{2}(s)\]
23
Balance H
Add 4 OH- ions to the right side to balance hydrogen:
\[\mathrm{Mn}^{2+}(aq) + 2\mathrm{H}_{2}\mathrm{O}(l) \rightarrow \mathrm{MnO}_{2}(s) + 4\mathrm{OH}^{-}(aq)\]
24
Balance the charge
Add 2 electrons to the left side to balance the charge:
\[\mathrm{Mn}^{2+}(aq) + 2\mathrm{H}_{2}\mathrm{O}(l) + 2e^- \rightarrow \mathrm{MnO}_{2}(s) + 4\mathrm{OH}^{-}(aq)\]
25
Determine oxidation or reduction
The oxidation number of Mn decreases from +2 to +4, so this is an oxidation half-reaction.
(f) \(\mathrm{Cr}(\mathrm{OH})_{3}(s)-\cdots \mathrm{CrO}_{4}^{2-}(a q)\) (basic solution)
26
Balance elements
Balance Cr: already balanced.
27
Balance O
Add 3 H2O molecules to the right side to balance oxygen:
\[\mathrm{Cr}(\mathrm{OH})_{3}(s) \rightarrow \mathrm{CrO}_{4}^{2-}(aq) + 3\mathrm{H}_{2}\mathrm{O}(l)\]
28
Balance H
Add 6 OH- ions to the left side to balance hydrogen:
\[\mathrm{Cr}(\mathrm{OH})_{3}(s) + 6\mathrm{OH}^{-}(aq) \rightarrow \mathrm{CrO}_{4}^{2-}(aq) + 3\mathrm{H}_{2}\mathrm{O}(l)\]
29
Balance the charge
Add 3 electrons to the left side to balance the charge:
\[3e^- + \mathrm{Cr}(\mathrm{OH})_{3}(s) + 6\mathrm{OH}^{-}(aq) \rightarrow \mathrm{CrO}_{4}^{2-}(aq) + 3\mathrm{H}_{2}\mathrm{O}(l)\]
30
Determine oxidation or reduction
The oxidation number of Cr increases from +3 to +6, so this is an oxidation half-reaction.
(g) \(\mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)\) (basic solution)
31
Balance elements
Balance O: Multiply the right side by 2, as done with (d)
32
Balance O
Oxygen is already balanced.
33
Balance H
Add 4 OH- ions to the left side to balance hydrogen, since it is a basic solution.
\[\mathrm{O}_{2}(g) + 4\mathrm{OH}^{-}(aq) \rightarrow 2\mathrm{H}_{2}\mathrm{O}(l)\]
34
Balance the charge
Add 4 electrons to the left side to balance the charge:
\[4e^- + \mathrm{O}_{2}(g) + 4\mathrm{OH}^{-}(aq) \rightarrow 2\mathrm{H}_{2}\mathrm{O}(l)\]
35
Determine oxidation or reduction
As in (d), the oxidation number of O decreases from 0 to -2, so this is a reduction half-reaction.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Oxidation and Reduction
Understanding the core concepts of oxidation and reduction is fundamental in the study of redox reactions. These are chemical processes where the oxidation state of atoms changes. Oxidation is the loss of electrons by a molecule, atom, or ion, which results in an increase in oxidation state. Conversely, reduction is the gain of electrons, leading to a decrease in the oxidation state.
To remember this, you can use the mnemonic 'OIL RIG' — Oxidation Is Loss, Reduction Is Gain. In a redox process, one substance gets oxidized while another gets reduced. For instance, in the reaction where \(\mathrm{Mo}^{3+}(aq) \rightarrow \mathrm{Mo}(s)\), molybdenum ions gain electrons and are reduced, while in \(\mathrm{H}_{2} \mathrm{SO}_{3}(aq) \rightarrow \mathrm{SO}_{4}^{2-}(aq)\), sulfur is oxidized as its oxidation state increases.
To remember this, you can use the mnemonic 'OIL RIG' — Oxidation Is Loss, Reduction Is Gain. In a redox process, one substance gets oxidized while another gets reduced. For instance, in the reaction where \(\mathrm{Mo}^{3+}(aq) \rightarrow \mathrm{Mo}(s)\), molybdenum ions gain electrons and are reduced, while in \(\mathrm{H}_{2} \mathrm{SO}_{3}(aq) \rightarrow \mathrm{SO}_{4}^{2-}(aq)\), sulfur is oxidized as its oxidation state increases.
Half-Reaction Method
The half-reaction method is an effective way to balance redox reactions. It involves separating the oxidation and reduction processes into two equations, which are then balanced individually. This method is particularly useful for complex reactions.
For example, in the reduction half-reaction \(\mathrm{NO}_{3}^{-}(aq) \rightarrow \mathrm{NO}(g)\), we first balance nitrogen and then add water and hydrogen ions to balance oxygen and hydrogen respectively. Finally, electrons are added to equate the total charge on both sides. Each half-reaction is balanced similarly, focusing on mass and charge conservation.
For example, in the reduction half-reaction \(\mathrm{NO}_{3}^{-}(aq) \rightarrow \mathrm{NO}(g)\), we first balance nitrogen and then add water and hydrogen ions to balance oxygen and hydrogen respectively. Finally, electrons are added to equate the total charge on both sides. Each half-reaction is balanced similarly, focusing on mass and charge conservation.
Balancing Charge in Redox
In redox reactions, it is crucial to balance the charge because the number of electrons lost during oxidation must equal the number gained during reduction. A discrepancy in charge indicates that the reaction is not yet properly balanced and cannot occur as written.
For instance, in the reaction \(\mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)\) under basic conditions, after balancing the oxygen and hydrogen atoms, we notice an imbalance of charge which is rectified by adding four electrons to the left side. These electrons represent the reduction that occurs when oxygen gas is converted to water.
For instance, in the reaction \(\mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)\) under basic conditions, after balancing the oxygen and hydrogen atoms, we notice an imbalance of charge which is rectified by adding four electrons to the left side. These electrons represent the reduction that occurs when oxygen gas is converted to water.
Oxidation Numbers
The concept of oxidation numbers is instrumental in understanding redox reactions. Oxidation numbers, also known as oxidation states, provide insights into the degree of oxidation or reduction an atom undergoes during a chemical reaction. These numbers are assigned based on a set of rules, considering the elements' electronegativities and the molecular structure.
By comparing the oxidation numbers of sulfur in \(\mathrm{H}_{2} \mathrm{SO}_{3}(aq)\) and \(\mathrm{SO}_{4}^{2-}(aq)\), we can deduce that an oxidation reaction has taken place since the oxidation number of sulfur increases from +4 to +6. The assignment of oxidation numbers is a key step in identifying which atoms are oxidized and which are reduced in a redox reaction.
By comparing the oxidation numbers of sulfur in \(\mathrm{H}_{2} \mathrm{SO}_{3}(aq)\) and \(\mathrm{SO}_{4}^{2-}(aq)\), we can deduce that an oxidation reaction has taken place since the oxidation number of sulfur increases from +4 to +6. The assignment of oxidation numbers is a key step in identifying which atoms are oxidized and which are reduced in a redox reaction.
