Chapter 19, Problem 1PS

Write balanced equations for the following half-reactions. Specify whether each is an oxidation or reduction.

1. (a) Cr(s) → Cr³⁺(aq)⇄(in acid)
2. (b) AsH₃(g) → As(s)⇄(in acid)
3. (c) VO₃⁻(aq) → V²⁺(aq)⇄(in acid)
4. (d) Ag(s) → Ag₂O(s)⇄(in base)

Interpretation Introduction

Interpretation:    

The balanced equation for the following half reaction has to be identified and also specify whether the reaction is an oxidation or reduction reaction.

$\text{(a) Cr(s) }\to {\text{Cr}}^{\text{3+}}\text{(aq) (in acid)}$

Concept introduction:

Steps for balancing half –reactions in ACIDIC solution:

1. 1. Balance all atoms except H and O in half reaction.
2. 2. Balance O atoms by adding water to the side missing O atoms.
3. 3. Balance the H atoms by adding ${\text{H}}^{\text{+}}$ to the side missing H atoms.
4. 4. Balance the charge by adding electrons to side with more total positive charge.
5. 5. Make the number of electrons the same in both half reactions by multiplication, while avoiding fractional number of electrons.

Steps for balancing half –reactions in BASIC solution:

1. 1. Balance all atoms except H and O in half reaction.
2. 2. Balance O atoms by adding water to the side missing O atoms.
3. 3. Balance the H atoms by adding ${\text{H}}^{\text{+}}$ to the side missing H atoms.
4. 4. Balance the charge by adding electrons to side with more total positive charge.
5. 5. Make the number of electrons the same in both half reactions by multiplication, while avoiding fractional number of electrons.
6. 6. Add the same number of ${\text{OH}}^{\text{-}}$ groups as there are ${\text{H}}^{\text{+}}$ present to both sides of the equation.

Answer to Problem 1PS

Balanced equation: $\text{Cr(s) }\to {\text{ Cr}}^{\text{3+}}{\text{(aq) + 3e}}^{\text{-}}$

It is a oxidation reaction.

Explanation of Solution

The given reaction:

$\text{Cr(s) }\to {\text{Cr}}^{\text{3+}}\text{(aq)}$

Steps for balancing half –reactions in ACIDIC solution:

1. 1. Balance all atoms except H and O in half reaction.

   $\text{Cr(s) }\to {\text{Cr}}^{\text{3+}}\text{(aq)}$

2. 2. Balance O atoms by adding water to the side missing O atoms.

   $\text{Cr(s) }\to {\text{Cr}}^{\text{3+}}\text{(aq)}$

   No need to add water because oxygen atoms are already balanced.

3. 3. Balance the H atoms by adding ${\text{H}}^{\text{+}}$ to the side missing H atoms.

   $\text{Cr(s) }\to {\text{Cr}}^{\text{3+}}\text{(aq)}$

4. 4. Balance the charge by adding electrons to side with more total positive charge

   $\text{Cr(s) }\to {\text{ Cr}}^{\text{3+}}{\text{(aq) + 3e}}^{\text{-}}$

5. 5. Make the number of electrons the same in both half reactions by multiplication, while avoiding fractional number of electrons.

   $\text{Cr(s) }\to {\text{ Cr}}^{\text{3+}}{\text{(aq) + 3e}}^{\text{-}}$

   Therefore, the balanced half cell reaction is as follows.

   $\text{Cr(s) }\to {\text{ Cr}}^{\text{3+}}{\text{(aq) + 3e}}^{\text{-}}$

Oxidation state of chromium increases in the above reaction.

$0to+3$

Therefore, it is an oxidation reaction.

Interpretation Introduction

Interpretation:    

The balanced equation for the following half reaction has to be identified and also specify whether the reaction is an oxidation or reduction reaction.

${\text{(b) AsH}}_{\text{3}}\text{(g)}\to \text{ As(s) (in acid)}$

Concept introduction:

Steps for balancing half –reactions in ACIDIC solution:

1. 6. Balance all atoms except H and O in half reaction.
2. 7. Balance O atoms by adding water to the side missing O atoms.
3. 8. Balance the H atoms by adding ${\text{H}}^{\text{+}}$ to the side missing H atoms.
4. 9. Balance the charge by adding electrons to side with more total positive charge.
5. 10. Make the number of electrons the same in both half reactions by multiplication, while avoiding fractional number of electrons.

Steps for balancing half –reactions in BASIC solution:

1. 7. Balance all atoms except H and O in half reaction.
2. 8. Balance O atoms by adding water to the side missing O atoms.
3. 9. Balance the H atoms by adding ${\text{H}}^{\text{+}}$ to the side missing H atoms.
4. 10. Balance the charge by adding electrons to side with more total positive charge.
5. 11. Make the number of electrons the same in both half reactions by multiplication, while avoiding fractional number of electrons.
6. 12. Add the same number of ${\text{OH}}^{\text{-}}$ groups as there are ${\text{H}}^{\text{+}}$ present to both sides of the equation.

Answer to Problem 1PS

The balanced reaction

${\text{AsH}}_{\text{3}}\text{(g) }\to {\text{As(s) + 3H}}^{\text{+}}+3e^{-}$:

It is an oxidation reaction.

Explanation of Solution

The given reaction:

${\text{AsH}}_{\text{3}}\text{(g) }\to \text{As(s)}$

Steps for balancing half –reactions in ACIDIC solution:

1. 1. Balance all atoms except H and O in half reaction.

   ${\text{AsH}}_{\text{3}}\text{(g) }\to \text{As(s)}$

2. 2. Balance O atoms by adding water to the side missing O atoms.

   ${\text{AsH}}_{\text{3}}\text{(g) }\to \text{As(s)}$

3. 3. Balance the H atoms by adding ${\text{H}}^{\text{+}}$ to the side missing H atoms.

   ${\text{AsH}}_{\text{3}}\text{(g) }\to {\text{As(s)+3H}}^{+}(aq)$

4. 4. Balance the charge by adding electrons to side with more total positive charge.

   ${\text{AsH}}_{\text{3}}\text{(g) }\to {\text{As(s) + 3H}}^{\text{+}}{\text{(aq) + 3e}}^{\text{-}}$

5. 5. Make the number of electrons the same in both half reactions by multiplication, while avoiding fractional number of electrons.

   ${\text{AsH}}_{\text{3}}\text{(g) }\to {\text{As(s) + 3H}}^{\text{+}}{\text{(aq) + 3e}}^{\text{-}}$

   Therefore, the balanced half cell reaction is as follows.

   ${\text{AsH}}_{\text{3}}\text{(g) }\to {\text{As(s) + 3H}}^{\text{+}}{\text{(aq) + 3e}}^{\text{-}}$

   From the above reaction Hydrogen atom oxidation state increases from 0 to +1.Therefore, it is an oxidation reaction.

Oxidation state of arsenic increases in the above reaction.

$-3to0$

Therefore, it is an oxidation reaction.

Interpretation Introduction

Interpretation:    

The balanced equation for the following half reaction has to be identified and also specify whether the reaction is an oxidation or reduction reaction.

${\text{(c) VO}}_{\text{3}}^{\text{-}}\text{(g) }\to {\text{ V}}^{\text{2+}}\text{(aq) (in acid)}$

Concept introduction:

Steps for balancing half –reactions in ACIDIC solution:

1. 11. Balance all atoms except H and O in half reaction.
2. 12. Balance O atoms by adding water to the side missing O atoms.
3. 13. Balance the H atoms by adding ${\text{H}}^{\text{+}}$ to the side missing H atoms.
4. 14. Balance the charge by adding electrons to side with more total positive charge.
5. 15. Make the number of electrons the same in both half reactions by multiplication, while avoiding fractional number of electrons.

Steps for balancing half –reactions in BASIC solution:

1. 13. Balance all atoms except H and O in half reaction.
2. 14. Balance O atoms by adding water to the side missing O atoms.
3. 15. Balance the H atoms by adding ${\text{H}}^{\text{+}}$ to the side missing H atoms.
4. 16. Balance the charge by adding electrons to side with more total positive charge.
5. 17. Make the number of electrons the same in both half reactions by multiplication, while avoiding fractional number of electrons.
6. 18. Add the same number of ${\text{OH}}^{\text{-}}$ groups as there are ${\text{H}}^{\text{+}}$ present to both sides of the equation.

Answer to Problem 1PS

The balanced half cell reaction:

${\text{VO}}_{\text{3}}^{\text{-}}{\text{(aq) +6H}}^{+}(aq)+3e^{-}\to {\text{V}}^{\text{2+}}{\text{(aq) + 3H}}_{\text{2}}\text{O (l)}$

It is a reduction reaction.

Explanation of Solution

The given reaction: ${\text{VO}}_{\text{3}}^{\text{-}}\text{(aq) }\to {\text{V}}^{\text{2+}}\text{(aq)}$

Steps for balancing half –reactions in ACIDIC solution:

1. 1. Balance all atoms except H and O in half reaction

   . ${\text{VO}}_{\text{3}}^{\text{-}}\text{(aq) }\to {\text{V}}^{\text{2+}}\text{(aq)}$

2. 2. Balance O atoms by adding water to the side missing O atoms.

   ${\text{VO}}_{\text{3}}^{\text{-}}\text{(aq) }\to {\text{V}}^{\text{2+}}{\text{(aq) + 3H}}_{\text{2}}\text{O (l)}$

3. 3. Balance the H atoms by adding ${\text{H}}^{\text{+}}$ to the side missing H atoms.

   ${\text{VO}}_{\text{3}}^{\text{-}}{\text{(aq)+6H}}^{+}\to {\text{V}}^{\text{2+}}{\text{(aq) + 3H}}_{\text{2}}\text{O (l)}$

4. 4. Balance the charge by adding electrons to side with more total positive charge.

   ${\text{VO}}_{\text{3}}^{\text{-}}{\text{(aq) +6H}}^{+}(aq)+3e^{-}\to {\text{V}}^{\text{2+}}{\text{(aq) + 3H}}_{\text{2}}\text{O (l)}$

5. 5. Make the number of electrons the same in both half  reactions by multiplication, while avoiding fractional number of electrons.

               ${\text{VO}}_{\text{3}}^{\text{-}}{\text{(aq) +6H}}^{+}(aq)+3e^{-}\to {\text{V}}^{\text{2+}}{\text{(aq) + 3H}}_{\text{2}}\text{O (l)}$

Therefore, the balanced half cell reaction is as follows.

${\text{VO}}_{\text{3}}^{\text{-}}{\text{(aq) +6H}}^{+}(aq)+3e^{-}\to {\text{V}}^{\text{2+}}{\text{(aq) + 3H}}_{\text{2}}\text{O (l)}$

Oxidation state of V in ${\text{VO}}_{\text{3}}^{\text{-}}$

$\begin{array}{l}{\text{VO}}_{\text{3}}^{\text{-}}\\ \text{x+3(-2)=-1}\\ \text{x = -1+6}\\ \text{x = +5}\end{array}$

Oxidation state of chromium decreases in the above reaction.

$+5to+2$

Therefore, it is a reduction reaction.

Interpretation Introduction

Interpretation:    

The balanced equation for the following half reaction has to be identified and also specify whether the reaction is an oxidation or reduction reaction.

$\text{(d) Ag(s) }\to {\text{ Ag}}_{\text{2}}\text{O(s) (in base)}$

Concept introduction:

Steps for balancing half –reactions in ACIDIC solution:

1. 16. Balance all atoms except H and O in half reaction.
2. 17. Balance O atoms by adding water to the side missing O atoms.
3. 18. Balance the H atoms by adding ${\text{H}}^{\text{+}}$ to the side missing H atoms.
4. 19. Balance the charge by adding electrons to side with more total positive charge.
5. 20. Make the number of electrons the same in both half reactions by multiplication, while avoiding fractional number of electrons.

Steps for balancing half –reactions in BASIC solution:

1. 19. Balance all atoms except H and O in half reaction.
2. 20. Balance O atoms by adding water to the side missing O atoms.
3. 21. Balance the H atoms by adding ${\text{H}}^{\text{+}}$ to the side missing H atoms.
4. 22. Balance the charge by adding electrons to side with more total positive charge.
5. 23. Make the number of electrons the same in both half reactions by multiplication, while avoiding fractional number of electrons.
6. 24. Add the same number of ${\text{OH}}^{\text{-}}$ groups as there are ${\text{H}}^{\text{+}}$ present to both sides of the equation.

Answer to Problem 1PS

The balanced half cell reaction:

${\text{2Ag(s) + 2OH}}^{\text{-}}\text{(aq) }\to {\text{Ag}}_{\text{2}}{\text{O(s) + H}}_{\text{2}}{\text{O(l) + 2e}}^{\text{-}}$

It is a oxidation reaction.

Explanation of Solution

The given reaction:

$\text{Ag(s) }\to {\text{Ag}}_{\text{2}}\text{O(s)}$

Steps for balancing half –reactions in BASIC solution:

1. 1. Balance all atoms except H and O in half reaction.

   $\text{2Ag(s) }\to {\text{Ag}}_{\text{2}}\text{O(s)}$

2. 2. Balance O atoms by adding water to the side missing O atoms.

   ${\text{2Ag(s) + 2OH}}^{-}\text{(aq) }\to {\text{Ag}}_{\text{2}}{\text{O(s)+H}}_{2}O(l)$

3. 3. Balance the H atoms by adding ${\text{H}}^{\text{+}}$ to the side missing H atoms.

   ${\text{2Ag(s) + 2OH}}^{-}\text{(aq) }\to {\text{Ag}}_{\text{2}}{\text{O(s)+H}}_{2}O(l)$

4. 4. Balance the charge by adding electrons to side with more total positive charge.

   ${\text{2Ag(s) + OH}}^{-}\text{(aq) }\to {\text{Ag}}_{\text{2}}{\text{O(s)+H}}_{\text{2}}\text{O(l})+2e^{-}$

5. 5. Make the number of electrons the same in both half  reactions by multiplication, while avoiding fractional number of electrons.

   ${\text{2Ag(s) + OH}}^{-}\text{(aq) }\to {\text{Ag}}_{\text{2}}{\text{O(s)+H}}_{\text{2}}\text{O(l})+2e^{-}$

6. 6. Add the same number of ${\text{OH}}^{\text{-}}$ groups as there are ${\text{H}}^{\text{+}}$ present to both sides of the equation.

             ${\text{2Ag(s) + OH}}^{-}\text{(aq) }\to {\text{Ag}}_{\text{2}}{\text{O(s)+H}}_{\text{2}}\text{O(l})+2e^{-}$

Therefore, the balanced half reaction is as follows.

${\text{2Ag(s) + OH}}^{-}\text{(aq) }\to {\text{Ag}}_{\text{2}}{\text{O(s)+H}}_{\text{2}}\text{O(l})+2e^{-}$

Oxidation state of silver decreases in the above reaction:

$0to-1$

Therefore, it is an oxidation reaction.