Chapter 19, Problem 77A

Interpretation Introduction

(a)

Interpretation:

Balanced chemical equation is to be written using half reactions.

Concept introduction:

A redox reaction is balanced if number of atoms and net charge on both sides is equal

To balance the reaction using half reaction method, following rules are used:

• Assign the oxidation number to each element.
• Check which element is oxidized and which element is reduced. Write oxidation and reduction half reactions.
• Check the change in oxidation number for the elements which are oxidized and reduced.
• Adjust the coefficients so that number of electrons lost in oxidation is equal to electrons gained in reduction.
• Add the balanced half reactions.

Answer to Problem 77A

The balanced equation is:

$Fe+T{e}^{2+}\to Te+F{e}^{2+}$

Explanation of Solution

Oxidation half reaction:

$Fe\to F{e}^{2+}+2e^{-}$

Reduction half reaction:

$T{e}^{2+}+2e^{-}\to Te$

Number of electrons lost = electrons gained

Therefore adding both equatios,

$Fe+T{e}^{2+}\to Te+F{e}^{2+}$

Interpretation Introduction

(b)

Interpretation:

Balanced chemical equation is to be written using half reactions.

Concept introduction:

A redox reaction is balanced if number of atoms and net charge on both sides is equal.

To balance the reaction using half reaction method, following rules are used:

• Assign the oxidation number to each element.
• Check which element is oxidized and which element is reduced. Write oxidation and reduction half reactions.
• Check the change in oxidation number for the elements which are oxidized and reduced.
• Then add enough hydrogen ions and water molecules to the equation to balance hydrogen atoms on both sides.
• Adjust the coefficients so that number of electrons lost in oxidation is equal to electrons gained in reduction.
• Add the balanced half reactions.

Answer to Problem 77A

The balanced equation is:

$2Al+3I{O}_4{}^{-}+6H^{+}\to 2A{l}^{3+}+3I{O}_3{}^{-}+3H_{2}O$

Explanation of Solution

Oxidation half reaction:

$Al\to A{l}^{3+}+3e^{-}$

Reduction half reaction:

$I{O}_4{}^{-}+2e^{-}\to I{O}_3{}^{-}(acidicsolution)$

Balancing the charge by adding hydrogen ion,

$I{O}_4{}^{-}+2e^{-}+2H^{+}\to I{O}_3{}^{-}(acidicsolution)$

Balancing the other atoms:

$I{O}_4{}^{-}+2e^{-}+2H^{+}\to I{O}_3{}^{-}+H_{2}O(acidicsolution)$

Therefore adding both equations after multiplying oxidation half reaction by 2 and reduction half reaction by 3

$2Al+3I{O}_4{}^{-}+6H^{+}\to 2A{l}^{3+}+3I{O}_3{}^{-}+3H_{2}O$

Interpretation Introduction

(c)

Interpretation:

Balanced chemical equation is to be written using half reactions.

Concept introduction:

A redox reaction is balanced if number of atoms and net charge on both sides is equal.

To balance the reaction using half reaction method, following rules are used:

• Assign the oxidation number to each element.
• Check which element is oxidized and which element is reduced. Write oxidation and reduction half reactions.
• Check the change in oxidation number for the elements which are oxidized and reduced.
• Then add enough hydrogen ions and water molecules to the equation to balance hydrogen atoms on both sides.
• Adjust the coefficients so that number of electrons lost in oxidation is equal to electrons gained in reduction.
• Add the balanced half reactions.

Answer to Problem 77A

The balanced equation is:

$4I_2+N_{2}O+5H_{2}O\to 8I^{-}+2N{O}_3{}^{-}+10H^{+}$

Explanation of Solution

Oxidation half reaction:

$N_{2}O\to N{O}_3{}^{-}+4e^{-}$

Reduction half reaction:

$I_2+2e^{-}\to 2I^{-}(acidicsolution)$

Balancing oxidation half reaction by adding hydrogen ion,

$N_{2}O\to 2N{O}_3{}^{-}+8e^{-}+10H^{+}$

Balancing the other atoms:

$N_{2}O+5H_{2}O\to 2N{O}_3{}^{-}+8e^{-}+10H^{+},$

Therefore adding both equations after multiplying reduction half reaction by 4

$4I_2+N_{2}O+5H_{2}O\to 8I^{-}+2N{O}_3{}^{-}+10H^{+}$