Chapter 17, Problem 39QAP

Use the following half-equations to write three spontaneous reactions. Justify your answers by calculating E° for the cells.
1. ${\text{MnO}}_4{}^{-}\left(aq\right)+8{\text{H}}^{+}\left(aq\right)+5e^{-}\to {\text{Mn}}^{2+}\left(aq\right)+4{\text{H}}_2\text{O}E°=+1.512\text{V}$
2. ${\text{O}}_2\left(g\right)+4{\text{H}}^{+}\left(aq\right)+4e^{-}\to 2{\text{H}}_2\text{O}E°=+1.229\text{V}$
3. ${\text{Co}}^{2+}\left(aq\right)+2e^{-}\to \text{Co}\left(s\right)E°=-0.282\text{V}$

Interpretation Introduction

Interpretation:

Using the given half-equations, the three spontaneous reactions needs to be determined by showing the value E° for the cells.

Concept introduction:

A reducing or a reductant is a species that loses electron/s and gets oxidized in the chemical reaction. The reducing agent is usually in one of its lower probable oxidation states, is recognized as the electron donor. Since, the reducing agent in the redox reaction loses electron/s, reducing agent gets oxidized.

An oxidizing agent is an agent which gains the electrons and get reduced within the chemical reaction. It is also recognized as electron acceptor; it is usually in one of its higher probable oxidation states so that it can reduce after accepting electron/s.

Spontaneity of a reaction is dependent on the free energy sign that is $\Delta G^o$. It should be negative for a reaction to be spontaneous.

Since,

$\Delta G=-nF{E}^o$

Here, n = number of electrons involved in reaction and F is faraday constant.

If the value of E° for a reaction is positive, then the reaction occurs spontaneous.

Answer to Problem 39QAP

The three spontaneous reactions are as follows:

(a) ${\text{4MnO}}_{\text{4}}{}^{\text{-}}\left(\text{aq}\right){\text{ + 12H}}^{\text{+}}\left(\text{aq}\right)\to {\text{4Mn}}^{\text{2+}}\left(\text{aq}\right){\text{+5O}}_{\text{2}}\left(\text{g}\right){\text{+6H}}_{\text{2}}{\text{O E}}^0=+0.283V$

(b) ${\text{2MnO}}_{\text{4}}{}^{\text{-}}\left(\text{aq}\right){\text{+16H}}^{\text{+}}\left(\text{aq}\right)\text{+5Co(s)}\to {\text{2Mn}}^{\text{2+}}\left(\text{aq}\right){\text{+5Co}}^{\text{2+}}\left(\text{g}\right){\text{+8H}}_{\text{2}}{\text{O E}}^0=+1.794V$

(c) $\text{2CO}\left(\text{s}\right){\text{+O}}_{\text{2}}\left(\text{g}\right){\text{+4H}}^{\text{+}}\text{(aq)}\to {\text{2Co}}^{\text{2+}}\left(\text{aq}\right){\text{+2H}}_{\text{2}}{\text{O E}}^o=+1.511V$

Explanation of Solution

Given Information:

The three half-equations are given as follows:

$\begin{array}{l}{\text{MnO}}_4^{-}(aq)+{\text{8H}}^{+}(aq)+5e^{-}\to {\text{Mn}}^{2+}(aq)+4{\text{H}}_{\text{2}}{\text{O E}}^0=+1.512\text{ V}\\ {\text{O}}_{\text{2}}(g)+4{\text{H}}^{+}(aq)+4e^{-}\to 2{\text{H}}_{\text{2}}{\text{O E}}^0=+1.229\text{ V}\\ {\text{Co}}^{2+}(aq)+2e^{-}\to \text{Co}(s){\text{ E}}^0=-0.282\text{ V}\end{array}$

Suppose cell consisting half-equation (2) as oxidation half-reaction and half-equation (1) as reduction half-reaction.

$\begin{array}{l}\text{Oxidation}:2{\mathrm{H}}_{2}O\to {\text{O}}_{\text{2}}\left(g\right)+4{\text{H}}^{+}\left(aq\right)+4e^{-} E^o{}_{ox}=-1.229V\hfill \\ \text{Reduction}:{\text{ MnO}}_4^{-}(aq)+{\text{8H}}^{+}(aq)+5e^{-}\to {\text{Mn}}^{2+}(aq)+4{\text{H}}_{\text{2}}\text{O }{E}^o{}_{red}=+1.512V\hfill \\ \hfill \end{array}$

Balancing the electrons in above two reaction and then adding the two equations to get overall cell reaction.

$\begin{array}{l}{\text{ 5×[2H}}_{\text{2}}\text{O}\to {\text{O}}_{\text{2}}\left(\text{g}\right){\text{+4H}}^{\text{+}}\left(\text{aq}\right){\text{+ 4e}}^{\text{-}}\text{]}\\ \sout{\begin{array}{l}{\text{4×[MnO}}_{\text{4}}\left(\text{aq}\right){\text{ + 8H}}^{\text{+}}\left(\text{aq}\right){\text{ + 5e}}^{\text{-}}\to {\text{Mn}}^{\text{2+}}\left(\text{aq}\right){\text{+4H}}_{\text{2}}\text{O]}\\ \underset{\_}{{\text{4MnO}}_{\text{4}}{}^{\text{-}}\left(\text{aq}\right){\text{ + 12H}}^{\text{+}}\left(\text{aq}\right)\to {\text{4Mn}}^{\text{2+}}\left(\text{aq}\right){\text{+5O}}_{\text{2}}\left(\text{g}\right){\text{+6H}}_{\text{2}}\text{O}}\end{array}}\end{array}$

The value of E° for a cell is the sum of standard reduction as well as standard oxidation potentials.

$\begin{array}{l}{E}^o=E^o{}_{red}+E^o{}_{ox}\hfill \\ =\left(+1.512V\right)+\left(-1.229V\right)\hfill \\ =+0.283V\hfill \end{array}$

Since, the value of E° for the cell is positive, the overall cell reaction is spontaneous.

Suppose cell consisting half-equation (3) as oxidation half-reaction and half-equation (1) as reduction half-reaction.

$\begin{array}{l}\text{Oxidation}:\text{ CO}\to {\text{CO}}^{\text{2+}}\left(\text{aq}\right){\text{+2e}}^{\text{-}} E^o{}_{ox}=\text{ +}0.282V\\ \text{Reduction}:{\text{ MnO}}_{\text{4}}{}^{\text{-}}\left(\text{aq}\right){\text{+8H}}^{\text{+}}\left(\text{aq}\right){\text{+5e}}^{\text{-}}\to {\text{Mn}}^{\text{2+}}\left(\text{aq}\right){\text{+4H}}_{\text{2}}\text{O }{E}^o{}_{red}=+1.512V\end{array}$

Balancing the electrons in above two reaction and then adding the two equations to get overall cell reaction.

$\begin{array}{l}\text{ 5×[Co}\to {\text{Co}}^{\text{2+}}\left(\text{aq}\right){\text{+2e}}^{\text{-}}\text{]}\\ \sout{\begin{array}{l}{\text{2×[MnO}}_{\text{4}}\left(\text{aq}\right){\text{ + 8H}}^{\text{+}}\left(\text{aq}\right){\text{ + 5e}}^{\text{-}}\to {\text{Mn}}^{\text{2+}}\left(\text{aq}\right){\text{+4H}}_{\text{2}}\text{O]}\\ \underset{\_}{{\text{2MnO}}_{\text{4}}{}^{\text{-}}\left(\text{aq}\right){\text{+16H}}^{\text{+}}\left(\text{aq}\right)\text{+5Co(s)}\to {\text{2Mn}}^{\text{2+}}\left(\text{aq}\right){\text{+5Co}}^{\text{2+}}\left(\text{g}\right){\text{+8H}}_{\text{2}}\text{O}}\end{array}}\end{array}$

The value of E° for a cell is the sum of standard reduction as well as standard oxidation potentials.

$\begin{array}{l}{E}^o=E^o{}_{red}+E^o{}_{ox}\hfill \\ =\left(+1.512V\right)+\left(+0.282V\right)\hfill \\ =+1.794V\hfill \end{array}$

Since, the value of E° for the cell is positive, the overall cell reaction is spontaneous.

Suppose cell consisting half-equation (3) as oxidation half-reaction and half-equation (1) as reduction half-reaction.

$\begin{array}{l}\text{Oxidation}:\text{ CO}\to {\text{CO}}^{\text{2+}}\left(\text{aq}\right){\text{+2e}}^{\text{-}} E^o{}_{ox}=\text{ +}0.282V\\ \text{Reduction}:{\text{ O}}_{\text{2}}\left(\text{g}\right){\text{+4H}}^{\text{+}}\left(\text{aq}\right){\text{+4e}}^{\text{-}}\to {\text{2H}}_{\text{2}}\text{O }{E}^o{}_{red}=+1.229V\end{array}$

Balancing the electrons in above two reaction and then adding the two equations to get overall cell reaction.

$\begin{array}{l}\text{ 2×[CO}\to {\text{Co}}^{\text{2+}}\left(\text{aq}\right){\text{+2e}}^{\text{-}}\text{]}\\ \sout{\begin{array}{l}{\text{O}}_{\text{2}}\left(\text{g}\right){\text{+4H}}^{\text{+}}\left(\text{aq}\right){\text{+4e}}^{\text{-}}\to {\text{2H}}_{\text{2}}\text{O}\\ \underset{\_}{\text{2CO}\left(\text{s}\right){\text{+O}}_{\text{2}}\left(\text{g}\right){\text{+4H}}^{\text{+}}\text{(aq)}\to {\text{2Co}}^{\text{2+}}\left(\text{aq}\right){\text{+2H}}_{\text{2}}\text{O}}\end{array}}\end{array}$

The value of E° for a cell is the sum of standard reduction as well as standard oxidation potentials.

$\begin{array}{l}{E}^o=E^o{}_{red}+E^o{}_{ox}\hfill \\ =\left(+1.229V\right)+\left(+0.282V\right)\hfill \\ =+1.511V\hfill \end{array}$

Since, the value of E° for the cell is positive, the overall cell reaction is spontaneous.

Conclusion

Thus, the three spontaneous reactions will be:

(a) ${\text{4MnO}}_{\text{4}}{}^{\text{-}}\left(\text{aq}\right){\text{ + 12H}}^{\text{+}}\left(\text{aq}\right)\to {\text{4Mn}}^{\text{2+}}\left(\text{aq}\right){\text{+5O}}_{\text{2}}\left(\text{g}\right){\text{+6H}}_{\text{2}}{\text{O E}}^0=+0.283V$

(b) ${\text{2MnO}}_{\text{4}}{}^{\text{-}}\left(\text{aq}\right){\text{+16H}}^{\text{+}}\left(\text{aq}\right)\text{+5Co(s)}\to {\text{2Mn}}^{\text{2+}}\left(\text{aq}\right){\text{+5Co}}^{\text{2+}}\left(\text{g}\right){\text{+8H}}_{\text{2}}{\text{O E}}^0=+1.794V$

(c) $\text{2CO}\left(\text{s}\right){\text{+O}}_{\text{2}}\left(\text{g}\right){\text{+4H}}^{\text{+}}\text{(aq)}\to {\text{2Co}}^{\text{2+}}\left(\text{aq}\right){\text{+2H}}_{\text{2}}{\text{O E}}^o=+1.511V$