Chapter 17, Problem 73QAP

Consider the reaction below at 25°C:
$2{\text{MnO}}_4{}^{-}\left(aq\right)+16{\text{H}}^{+}\left(aq\right)+10{\text{Br}}^{-}\left(aq\right)\to 2{\text{Mn}}^{2+}\left(aq\right)+5{\text{Br}}_2\left(l\right)+8{\text{H}}_2\text{O}$
Use Table 17.1 to answer the following questions. Support your answers with calculations.
(a) Is the reaction spontaneous at standard conditions?
(b) Is the reaction spontaneous at a pH of 2.00 with all other ionic species at 0.100 M?
(c) Is the reaction spontaneous at a pH of 5.00 with all other ionic species at 0.100 M?
(d) At what pH is the reaction at equilibrium with all other ionic species at 0.100 M?

Interpretation Introduction

(a)

Interpretation:

The spontaneity of the reaction at standard conditions needs to be determined.

Concept introduction:

The standard electrode potential $\left(E^o\right)$ of the cell is given by,

$E^{\text{ο}}\text{ = }{E}^o{}_{\text{cathode}}-E^o{}_{\text{anode}}$

where,

$\begin{array}{c}{E}^o{}_{\text{anode}}\to \text{ standard reduction potential of anode}\\ E^o{}_{\text{cathode}}\to \text{ standard reduction potential of cathode}\end{array}$

Answer to Problem 73QAP

The reaction is spontaneous $\left(E^o\text{ is positive}\right)$.

Explanation of Solution

The cell reaction is as follows.

${\text{2MnO}}_{\text{4}}{}^{-}\left(\text{aq}\right){\text{ + 16H}}^{\text{+}}\left(\text{aq}\right){\text{ + 10Br}}^{-}\left(\text{aq}\right)\to {\text{ 2Mn}}^{\text{2+}}\left(\text{aq}\right){\text{ + 5Br}}_{\text{2}}\left(\text{l}\right){\text{ + 8H}}_{\text{2}}\text{O}$

Step 1: Write the two half-cell reactions.

Oxidation half-reaction (Anode): ${\text{2Br}}^{-}\left(\text{aq}\right){\text{ + 2e}}^{-}\to {\text{ Br}}_{\text{2}}\left(\text{l}\right)$

Reduction half-reaction (Cathode): ${\text{MnO}}_{\text{4}}{}^{-}\left(\text{aq}\right){\text{ + 8H}}^{\text{+}}\left(\text{aq}\right)\to {\text{ Mn}}^{\text{2+}}\left(\text{aq}\right){\text{ + 4H}}_{\text{2}}{\text{O + 5e}}^{-}$

Step 2: Calculation of standard electrode potential $\left(E^o\right)$ by the formula,

$E^{\text{ο}}\text{ = }{E}^o{}_{\text{cathode}}-E^o{}_{\text{anode}}$   .... $\left(\text{i}\right)$

Here,

$\begin{array}{c}{E}^o{}_{\text{anode}}\text{ = 1}\text{.077 V}\\ E^o{}_{\text{cathode}}\text{ = 1}\text{.512 V}\end{array}$

From equation $\left(\text{i}\right)$ ,

$\begin{array}{c}{E}^{\text{ο}}\text{ = }{E}^o{}_{\text{cathode}}-E^o{}_{\text{anode}}\\ \text{= }\left(\text{1}\text{.512 }-\text{ 1}\text{.077}\right)\text{V}\\ \text{= +0}\text{.435 V }\end{array}$

Thus, the reaction is spontaneous $\left(E^o\text{ is positive}\right)$.

Interpretation Introduction

(b)

Interpretation:

The spontaneity of the reaction at a $\text{pH}$ of $\text{2}\text{.00}$ when all other ionic species are at $\text{0}\text{.100 M}$ needs to be determined.

Concept introduction:

The formula of Nernst equation is used to calculate the spontaneity of the reaction, given by,

$E_{\text{cell}}\text{ = }{E}^o-\frac{\text{0}\text{.0592}}{n}\log Q$

where,

$\begin{array}{c}{E}^o\text{ = standard electrode potential}\\ Q\text{ = reaction quotient}\\ n\text{ = replacable electrons }\end{array}$

Answer to Problem 73QAP

The reaction is spontaneous $\left(E_{cell}\text{ is positive}\right)$ at a $\text{pH}$ of $\text{2}\text{.00}$.

Explanation of Solution

Calculation of reaction quotient $\left(Q\right)$

$\begin{array}{c}Q\text{ = }\frac{{\left[{\text{Mn}}^{\text{2+}}\left(\text{aq}\right)\right]}^{\text{2}}}{{\left[{\text{MnO}}_{\text{4}}{}^{-}\left(\text{aq}\right)\right]}^{\text{2}}{\left[{\text{H}}^{\text{+}}\left(\text{aq}\right)\right]}^{\text{16}}{\left[{\text{Br}}^{-}\left(\text{aq}\right)\right]}^{10}}\\ \text{= }\frac{{\left[\text{0}\text{.1}\right]}^{\text{2}}}{{\left[0.1\right]}^{\text{2}}{\left[\text{0}\text{.01}\right]}^{\text{16}}{\left[0.1\right]}^{10}}\\ {\text{= 10}}^{42}\end{array}$

From Nernst equation,

$E_{\text{cell}}\text{ = }{E}^o-\frac{\text{0}\text{.0592}}{n}\log Q$

Put the values of $n\text{ = 10, }{E}^{\text{ο}}\text{ = 0}\text{.435 V and }Q{\text{ = 10}}^{42}$ in above equation,

$\begin{array}{c}{E}_{cell}\text{ = 0}\text{.435 V}-\frac{\text{0}\text{.0592}}{\text{10}}\text{log}\left({\text{10}}^{42}\right)\\ =\text{0}\text{.435 V}-0\cdot 00592\left[42\log 10\right]\\ \text{= 0}\text{.435 V}-0\cdot 00592\times 42\\ =\text{0}\text{.435 V}-0\cdot 248\end{array}$

$\text{= 0}\cdot \text{187 V}$

Thus, the reaction is spontaneous $\left(E_{cell}\text{ is positive}\right)$ at a $\text{pH}$ of $\text{2}\text{.00}$.

Interpretation Introduction

(c)

Interpretation:

The spontaneity of the reaction at a $\text{pH}$ of $\text{5}\text{.00}$ when all other ionic species are at $\text{0}\text{.100 M}$ needs to be determined.

Concept introduction:

The formula of Nernst equation is used to calculate the spontaneity of the reaction, given by,

$E_{\text{cell}}\text{ = }{E}^o-\frac{\text{0}\text{.0592}}{n}\log Q$

where,

$\begin{array}{c}{E}^o\text{ = standard electrode potential}\\ Q\text{ = reaction quotient}\\ n\text{ = replacable electrons }\end{array}$

Answer to Problem 73QAP

The reaction is non-spontaneous $\left(E_{cell}\text{ is negative}\right)$ at a $\text{pH}$ of $\text{5}\text{.00}$.

Explanation of Solution

Calculation of reaction quotient $\left(Q\right)$

$\begin{array}{c}Q\text{ = }\frac{{\left[{\text{Mn}}^{\text{2+}}\left(\text{aq}\right)\right]}^{\text{2}}}{{\left[{\text{MnO}}_{\text{4}}{}^{-}\left(\text{aq}\right)\right]}^{\text{2}}{\left[{\text{H}}^{\text{+}}\left(\text{aq}\right)\right]}^{\text{16}}{\left[{\text{Br}}^{-}\left(\text{aq}\right)\right]}^{10}}\\ \text{= }\frac{{\left[\text{0}\text{.1}\right]}^{\text{2}}}{{\left[0.1\right]}^{\text{2}}{\left[\text{0}\text{.00001}\right]}^{\text{16}}{\left[0.1\right]}^{10}}\\ {\text{= 10}}^{90}\end{array}$

From Nernst equation,

$E_{\text{cell}}\text{ = }{E}^o-\frac{\text{0}\text{.0592}}{n}\log Q$

Put the values of $n\text{ = 10, }{E}^{\text{ο}}\text{ = 0}\text{.435 V and }Q{\text{ = 10}}^{90}$ in above equation,

$\begin{array}{c}{E}_{cell}\text{ = 0}\text{.435 V}-\frac{\text{0}\text{.0592}}{\text{10}}\text{log}\left({\text{10}}^{90}\right)\\ \text{= 0}\text{.435 V}-\text{ 0}\text{.00592}\times \text{90log10}\\ \text{= 0}\text{.435 V}-\text{0}\text{.00592}\times 90\\ =\text{0}\text{.435 V}-0\cdot 532\text{V}\end{array}$

$\text{=}-0\cdot 097\text{V}$

Thus, the reaction is non-spontaneous $\left(E_{cell}\text{ is negative}\right)$ at a $\text{pH}$ of $\text{5}\text{.00}$.

Interpretation Introduction

(d)

Interpretation:

The $\text{pH}$ should be calculated when the reaction establishes an equilibrium with all other ionic species at $\text{0}\text{.100 M}$.

Concept introduction:

The relation between $pH$ and concentration of, $\left[{\text{H}}^{\text{+}}\right]$ is given by,

$\text{pH = }-\text{log}\left[{\text{H}}^{\text{+}}\left(\text{aq}\right)\right]$

Answer to Problem 73QAP

At equilibrium, $\text{pH = }3.96$

Explanation of Solution

From Nernst equation,

$E_{\text{cell}}\text{ = }{E}^o-\frac{\text{0}\text{.0592}}{n}\log Q$

Put the values of $E_{cell}\text{ = 0, }n\text{ = 10, }{E}^{\text{ο}}\text{ = 0}\text{.435 V and }Q\text{ = }\frac{{\left[\text{0}\text{.1}\right]}^{\text{2}}}{{\left[0.1\right]}^{\text{2}}{\left[{\text{H}}^{\text{+}}\left(\text{aq}\right)\right]}^{\text{16}}{\left[0.1\right]}^{10}}$ in above equation,

$\begin{array}{c}\text{0 = 0}\text{.435 V}-\frac{\text{0}\text{.0592}}{\text{10}}\text{log}\left\{\frac{{\left[\text{0}\text{.1}\right]}^{\text{2}}}{{\left[0.1\right]}^{\text{2}}{\left[{\text{H}}^{\text{+}}\left(\text{aq}\right)\right]}^{\text{16}}{\left[0.1\right]}^{10}}\right\}\\ \cancel{-}0\cdot 435=\cancel{-}0\cdot 00592\log \frac{1}{{\left[{\text{H}}^{+}\left(aq\right)\right]}^{16}{\left(0\cdot 1\right)}^{10}}\\ \frac{0\cdot 435}{0\cdot 00592}=\log \frac{{10}^{10}}{{\left[{\text{H}}^{+}\left(aq\right)\right]}^{16}}\\ 73\cdot 47=\log {10}^{10}-\log {\left[{\text{H}}^{+}\left(aq\right)\right]}^{16}\end{array}$

$\begin{array}{c}\text{73}\cdot \text{47 = 10log10}-16\log \left[{\text{H}}^{\text{+}}\left(\text{aq}\right)\right]\\ \text{73}\cdot \text{47}-10=-16\log \left[{\text{H}}^{\text{+}}\left(\text{aq}\right)\right]\\ \log \left[{\text{H}}^{\text{+}}\left(\text{aq}\right)\right]=-\frac{63\cdot 47}{16}\\ -\log \left[{\text{H}}^{\text{+}}\left(\text{aq}\right)\right]=3\cdot 96\end{array}$

Now,

$\begin{array}{c}\text{pH = }-\text{log}\left[{\text{H}}^{\text{+}}\left(\text{aq}\right)\right]\\ \text{= }3\cdot 96\end{array}$