Chapter 18, Problem 41QAP

Interpretation Introduction

(a)

Interpretation:

The given half reaction should be balanced

Concept Introduction:

The oxidation and reduction can be identified by change in oxidation state. If oxidation state of atom of an element increases, it undergoes oxidation and if it decreases, it undergoes reduction.

The general oxidation and reduction half reactions are as follows:

$\text{A}\left(\text{s}\right)\to {\text{A}}^{+}\left(\text{a}\text{q}\right)+\text{e}$

Here, $\text{A}\left(\text{s}\right)$ undergoes oxidation to form ${\text{A}}^{+}\left(\text{a}\text{q}\right)$ with release of 1 electron. The oxidation state of A increases from 0 to 1.

Similarly,

${\text{B}}^{+}\left(\text{a}\text{q}\right)+\text{e}\to \text{B}\left(\text{s}\right)$

Here, ${\text{B}}^{+}\left(\text{a}\text{q}\right)$ undergoes reduction to form $\text{B}\left(\text{s}\right)$ with addition of 1 electron. The oxidation state of A decreases from 1 to 0.

Answer to Problem 41QAP

$\text{Cu}\left(\text{s}\right)\to {\text{Cu}}^{2+}\left(\text{a}\text{q}\right)+2{\text{e}}^{-}$.

Explanation of Solution

The given half reaction is as follows:

$\text{Cu}\left(\text{s}\right)\to {\text{Cu}}^{2+}\left(\text{a}\text{q}\right)$

In the above reaction, oxidation state of copper is zero in solid state, it get oxidized to ${\text{Cu}}^{2+}$ in aqueous solution and oxidation state increases to + 2. In oxidation reaction, loss of electron/s takes place. Here, the change in oxidation state is + 2-0=2 thus, 2 electrons loss from elemental Cu.

Thus, to balance the above reaction, 2 electrons are added to right side of the reaction.

$\text{Cu}\left(\text{s}\right)\to {\text{Cu}}^{2+}\left(\text{a}\text{q}\right)+2{\text{e}}^{-}$.

Interpretation Introduction

(b)

Interpretation:

The given half reaction should be balanced

Concept Introduction:

The oxidation and reduction can be identified by change in oxidation state. If oxidation state of atom of an element increases, it undergoes oxidation and if it decreases, it undergoes reduction.

The general oxidation and reduction half reactions are as follows:

$\text{A}\left(\text{s}\right)\to {\text{A}}^{+}\left(\text{a}\text{q}\right)+\text{e}$

Here, $\text{A}\left(\text{s}\right)$ undergoes oxidation to form ${\text{A}}^{+}\left(\text{a}\text{q}\right)$ with release of 1 electron. The oxidation state of A increases from 0 to 1.

Similarly,

${\text{B}}^{+}\left(\text{a}\text{q}\right)+\text{e}\to \text{B}\left(\text{s}\right)$

Here, ${\text{B}}^{+}\left(\text{a}\text{q}\right)$ undergoes reduction to form $\text{B}\left(\text{s}\right)$ with addition of 1 electron. The oxidation state of A decreases from 1 to 0.

Answer to Problem 41QAP

${\text{Fe}}^{3+}\left(\text{a}\text{q}\right)+1{\text{e}}^{-}\to {\text{Fe}}^{2+}\left(\text{a}\text{q}\right)$.

Explanation of Solution

The given half reaction is as follows:

${\text{Fe}}^{3+}\left(\text{a}\text{q}\right)\to {\text{Fe}}^{2+}\left(\text{a}\text{q}\right)$

In the above reaction, oxidation state of iron is + 3 in reactant side, it gets reduced to ${\text{Fe}}^{2+}$ in product side and oxidation state decreases to + 2. In reduction reaction, gain of electron/s takes place. Here, the change in oxidation state is + 2-(+3) = -1 thus, 1 electron is gained by ${\text{Fe}}^{3+}$.

Thus, to balance the above reaction, 1 electron is added to left side of the reaction.

${\text{Fe}}^{3+}\left(\text{a}\text{q}\right)+1{\text{e}}^{-}\to {\text{Fe}}^{2+}\left(\text{a}\text{q}\right)$.

Interpretation Introduction

(c)

Interpretation:

The given half reaction should be balanced

Concept Introduction:

The oxidation and reduction can be identified by change in oxidation state. If oxidation state of atom of an element increases, it undergoes oxidation and if it decreases, it undergoes reduction.

The general oxidation and reduction half reactions are as follows:

$\text{A}\left(\text{s}\right)\to {\text{A}}^{+}\left(\text{a}\text{q}\right)+\text{e}$

Here, $\text{A}\left(\text{s}\right)$ undergoes oxidation to form ${\text{A}}^{+}\left(\text{a}\text{q}\right)$ with release of 1 electron. The oxidation state of A increases from 0 to 1.

Similarly,

${\text{B}}^{+}\left(\text{a}\text{q}\right)+\text{e}\to \text{B}\left(\text{s}\right)$

Here, ${\text{B}}^{+}\left(\text{a}\text{q}\right)$ undergoes reduction to form $\text{B}\left(\text{s}\right)$ with addition of 1 electron. The oxidation state of A decreases from 1 to 0.

Answer to Problem 41QAP

${\text{2Br}}^{-}\left(\text{a}\text{q}\right)\to {\mathrm{Br}}_2\left(\text{l}\right)+2{\text{e}}^{-}$.

Explanation of Solution

The given half reaction is as follows:

${\text{Br}}^{-}\left(\text{a}\text{q}\right)\to {\mathrm{Br}}_2\left(\text{l}\right)$

In the above reaction, oxidation state of bromine is -1 in reactant side, it gets oxidized to ${\text{Br}}_{\text{2}}$ in product side and oxidation state increases to 0. In oxidation reaction, loss of electron/s takes place. Here, the change in oxidation state is 0-(-1) = + 1 thus, 1 electron is lost by ${\text{Br}}^{-}$.

Thus, 1 electron is added to right side of the reaction.

${\text{Br}}^{-}\left(\text{a}\text{q}\right)\to {\mathrm{Br}}_2\left(\text{l}\right)+1{\text{e}}^{-}$

Now, to balance the atoms give coefficient 2 to ${\text{Br}}^{-}$ thus,

${\text{2Br}}^{-}\left(\text{a}\text{q}\right)\to {\mathrm{Br}}_2\left(\text{l}\right)+1{\text{e}}^{-}$

To balance the charge, 1 more electron is added to right side, thus, the balanced half reaction will be:

${\text{2Br}}^{-}\left(\text{a}\text{q}\right)\to {\mathrm{Br}}_2\left(\text{l}\right)+2{\text{e}}^{-}$.

Interpretation Introduction

(d)

Interpretation:

The given half reaction should be balanced

Concept Introduction:

The oxidation and reduction can be identified by change in oxidation state. If oxidation state of atom of an element increases, it undergoes oxidation and if it decreases, it undergoes reduction.

The general oxidation and reduction half reactions are as follows:

$\text{A}\left(\text{s}\right)\to {\text{A}}^{+}\left(\text{a}\text{q}\right)+\text{e}$

Here, $\text{A}\left(\text{s}\right)$ undergoes oxidation to form ${\text{A}}^{+}\left(\text{a}\text{q}\right)$ with release of 1 electron. The oxidation state of A increases from 0 to 1.

Similarly,

${\text{B}}^{+}\left(\text{a}\text{q}\right)+\text{e}\to \text{B}\left(\text{s}\right)$

Here, ${\text{B}}^{+}\left(\text{a}\text{q}\right)$ undergoes reduction to form $\text{B}\left(\text{s}\right)$ with addition of 1 electron. The oxidation state of A decreases from 1 to 0.

Answer to Problem 41QAP

${\text{Fe}}^{2+}\left(\text{a}\text{q}\right)+2{\text{e}}^{-}\to \text{Fe}\left(\text{s}\right)$.

Explanation of Solution

The given half reaction is as follows:

${\text{Fe}}^{2+}\left(\text{a}\text{q}\right)\to \text{Fe}\left(\text{s}\right)$

In the above reaction, oxidation state of iron is + 2 in reactant side, it gets reduced to $\text{Fe}$ in product side and oxidation state decreases to 0. In reduction reaction, gain of electron/s takes place. Here, the change in oxidation state is 0-(+2) = -2 thus, 2 electrons are gained by ${\text{Fe}}^{2+}$.

Thus, to balance the above reaction, 2 electrons are added to left side of the reaction.

${\text{Fe}}^{2+}\left(\text{a}\text{q}\right)+2{\text{e}}^{-}\to \text{Fe}\left(\text{s}\right)$.