Chapter 13, Problem 13.37PAE

Consult a table of standard reduction potentials and determine which of the following reactions are spontaneous under standard electrochemical conditions.

(a) $\text{Mn}\left(\text{s}\right)+2{\text{H}}^{+}\left(\text{aq}\right)\to {\text{H}}_2\left(\text{g}\right)+{\text{Mn}}^{2+}\left(\text{aq}\right)$

(b) $\text{2}{\text{Al}}^{3+}\left(\text{aq}\right)+3{\text{H}}_2\left(\text{g}\right)\to 2\text{Al}\left(\text{s}\right)+6{\text{H}}^{+}\left(\text{aq}\right)$

(c) $\text{2}\text{Cr}{\left(\text{OH}\right)}_3\left(\text{s}\right)+6{\text{F}}^{-}\left(\text{aq}\right)\to 2\text{Cr}\left(\text{s}\right)+6{\text{OH}}^{-}\left(\text{aq}\right)+3{\text{F}}_2\left(\text{g}\right)$

(d) ${\text{Cl}}_2\left(\text{g}\right)+2{\text{Br}}^{-}\left(\text{aq}\right)+\to {\text{Br}}_2\left(\mathcal{l}\right)+2{\text{Cl}}^{-}\left(\text{aq}\right)$

Interpretation Introduction

To determine: To find out whether the reaction ${\text{Mn(s)+2H}}^{\text{+}}\text{(aq)}\to {\text{Mn}}^{\text{2+}}{\text{(aq)+H}}_{\text{2}}\text{(g)}$ is spontaneous under standard electrochemical condition.

Explanation of Solution

From the given reaction the manganese is oxidized and hydrogen is reduced. The potentials of manganese is 1.18V and potential of hydrogen is 0.00 V. By changing the sign of the standard reduction potential we can obtain oxidation potential.

Spontaneous reaction will have overall positive cell voltage. We can obtain half reaction of overall cell voltage by adding reduction and oxidation.

$\begin{array}{l}{\text{Overall Reaction: E}}^0\\ {\text{2H}}^{\text{+}}{\text{+2e}}^{\text{-}}\to {\text{H}}_2\text{ 0}\text{.000 V}\\ \text{Mn}\to {\text{Mn}}^{\text{2+}}+{\text{2e}}^{\text{-}}\text{ 1}\text{.18 V}\\ {\text{(2e}}^{\text{-}}\text{ gets cancelled)}\\ \text{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}\\ {\text{Mn + 2H}}^{\text{+}}\to {\text{Mn}}^{\text{2+}}+{\text{H}}_2\text{ 1}\text{.18V}\end{array}$

For spontaneity the positive cell voltage must be positive. Here, the positive cell voltage is 1.18 which is positive and thus this reaction is spontaneous.

Interpretation Introduction

To determine: To find out whether the reaction ${\text{2Al}}^{3+}{\text{(aq)+3H}}_2\text{(g)}\to {\text{2Al(s)+6H}}^{+}\text{(aq)}$ is spontaneous under standard electrochemical condition.

Explanation of Solution

From the given reaction the hydrogen is oxidized and aluminium is reduced. The potentials of aluminiumis -1.66 V and potential of hydrogen is 0.000 V. By changing the sign of the standard reduction potential we can obtain oxidation potential.

Spontaneous reaction will have overall positive cell voltage. We can obtain half reaction of overall cell voltage by adding reduction and oxidation.

$\begin{array}{l}{\text{Overall Reaction: E}}^0\\ {\text{2Al}}^{\text{3+}}{\text{+6e}}^{\text{-}}\to 2\text{Al - 1}\text{.66 V}\\ {\text{3H}}_2\to 6{\text{H}}^{\text{2+}}+6{\text{e}}^{\text{-}}\text{ 0}\text{.0 V}\\ {\text{(6e}}^{\text{-}}\text{ gets cancelled)}\\ \text{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}\\ {\text{2Al}}^{\text{3+}}{\text{ + 3H}}_2\to 2\text{Al}+6{\text{H}}^{\text{+}}\text{ -1}\text{.66 V}\end{array}$

For spontaneity the positive cell voltage must be positive. Here the positive cell voltage is -1.66 V which is negative and thus this reaction is non- spontaneous.

Interpretation Introduction

To determine: To find out whether the reaction ${\text{2Cr(OH)}}_{{}^3}{\text{(s)+6F}}^{-}\text{(aq)}\to {\text{2Cr(s)+6OH}}^{-}{\text{(aq)+3F}}_2(\text{g)}$ is spontaneous under standard electrochemical condition.

Explanation of Solution

From the given reaction the ${\text{F}}^{\text{-}}$ is oxidized to ${\text{F}}_{\text{2}}$ and ${\text{Cr(OH)}}_3$ is reducedto $\text{Cr}$. The potentials of ${\text{F}}^{\text{-}}$ is -2.87 V and potential of ${\text{Cr(OH)}}_3$ is -1.30 V. By changing the sign of the standard reduction potential we can obtain oxidation potential.

Spontaneous reaction will have overall positive cell voltage. We can obtain half reaction of overall cell voltage by adding reduction and oxidation.

$\begin{array}{l}{\text{Overall Reaction: E}}^0\\ {\text{Cr(OH)}}_3{\text{+6e}}^{\text{-}}\to {\text{Cr+3OH}}^{\text{-}}\text{ -1}\text{.30 V }\\ {\text{+ 6F}}^{-}\to 3{\text{F}}_{\text{2}}+6{\text{e}}^{\text{-}}\text{ -2}\text{.87 V}\\ {\text{(6e}}^{\text{-}}\text{ gets cancelled)}\\ \text{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}\\ {\text{2Cr(OH)}}_3{\text{ + 6F}}^{-}\to \text{2Cr}+6{\text{OH}}^{\text{-}}\text{ + }3{\text{F}}_{\text{2}}\text{ -4}\text{.17}\text{V }\end{array}$

For spontaneity the positive cell voltage must be positive. Here the positive cell voltage is -4.17 V which is negative and thus this reaction is non- spontaneous.

Interpretation Introduction

To determine: To find out whether the reaction ${\text{Cl}}_2{\text{(g)+2Br}}^{-}\text{(aq)}\to {\text{Br}}_2{\text{(l)+2Cl}}^{-}\text{(aq)}$ is spontaneous under standard electrochemical condition.

Explanation of Solution

From the given reaction the ${\text{Br}}^{-}$ manganese is oxidized and ${\text{Cl}}_2$ hydrogen is reduced. The potentials of ${\text{Br}}^{-}$ is -1.08V and potential of ${\text{Cl}}_2$ is -1.360 V. By changing the sign of the standard reduction potential we can obtain oxidation potential.

Spontaneous reaction will have overall positive cell voltage. We can obtain half reaction of overall cell voltage by adding reduction and oxidation.

$\begin{array}{l}{\text{Overall Reaction: E}}^0\\ {\text{Cl}}_2{\text{+2e}}^{\text{-}}\to {\text{2Cl}}^{-}\text{ 1}\text{.36 V }\\ {\text{ 2Br}}^{-}\to {\text{Br}}_{\text{2}}+2{\text{e}}^{\text{-}}\text{ -1}\text{.08 V}\\ {\text{(2e}}^{\text{-}}\text{ gets cancelled)}\\ \text{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}\\ {\text{ Cl}}_2{\text{+2Br}}^{-}\to {\text{ Br}}_{\text{2}}{\text{+2Cl}}^{-}\text{ 0}\text{.28}\text{ V }\end{array}$

For spontaneity the positive cell voltage must be positive. Here the positive cell voltage is 0.28V which is positive and thus this reaction is spontaneous.

Conclusion

• The reaction ${\text{Mn(s)+2H}}^{\text{+}}\text{(aq)}\to {\text{Mn}}^{\text{2+}}{\text{(aq)+H}}_{\text{2}}\text{(g)}$ is spontaneous.
• The reaction ${\text{2Al}}^{3+}{\text{(aq)+3H}}_2\text{(g)}\to {\text{2Al(s)+6H}}^{+}\text{(aq)}$ is non-spontaneous.
• The reaction ${\text{2Cr(OH)}}_{{}^3}{\text{(s)+6F}}^{-}\text{(aq)}\to {\text{2Cr(s)+6OH}}^{-}{\text{(aq)+3F}}_2(\text{g)}$ is non-spontaneous.
• The reaction ${\text{Cl}}_2{\text{(g)+2Br}}^{-}\text{(aq)}\to {\text{Br}}_2{\text{(l)+2Cl}}^{-}\text{(aq)}$ is spontaneous.