Chapter 18, Problem 18.62QP

(a)

Interpretation Introduction

Interpretation: The value of $E^{\text{ο}}$ for the given reaction and concentration of ${\text{ClO}}_{\text{3}}{}^{-}$ is to be calculated.

Concept introduction: Nernst equation relates the cell potential of a half cell reaction to its standard potential and to the concentrations of its reactants and products.

To determine: The value of $E^{\text{ο}}$ for the given reaction.

Answer to Problem 18.62QP

Solution

The value of $E^{\text{ο}}$ for the given reaction is $\underset{\_}{-0.206V}$.

Explanation of Solution

Explanation

Chlorine dioxide ( ${\text{ClO}}_{\text{2}}$) is produced by the following reaction of chlorate ( ${\text{ClO}}_{\text{3}}{}^{-}$) with ${\text{Cl}}^{-}$ in acid solution:

${\text{2ClO}}_{\text{3}}{}^{-}\left(aq\right)+{\text{2Cl}}^{-}\left(aq\right)+4{\text{H}}^{\text{+}}\left(aq\right)\to 2{\text{ClO}}_{\text{2}}\left(g\right)+{\text{Cl}}_2\left(g\right)+2{\text{H}}_2\text{O}\left(l\right)$

The half cell reaction at cathode is,

${\text{ClO}}_{\text{3}}{}^{-}\left(aq\right)+2{\text{H}}^{+}\left(aq\right){\text{+e}}^{-}\to {\text{ClO}}_2\left(g\right)+{\text{H}}_2\text{O}{E}_{\text{cathode}}^{\text{ο}}=1.152\text{V}$ (1)

The half cell reaction at anode is,

${\text{Cl}}_2\left(g\right)\to 2{\text{Cl}}^{-}\left(aq\right)+{\text{2e}}^{-}{E}_{\text{anode}}^{\text{ο}}=1.3583\text{V}$ (2)

Multiply equation (1) by 2 to balance the electrons.

${\text{2ClO}}_{\text{3}}{}^{-}\left(aq\right)+4{\text{H}}^{+}\left(aq\right){\text{+2e}}^{-}\to 2{\text{ClO}}_2\left(g\right)+2{\text{H}}_2\text{O}{E}_{\text{cathode}}^{\text{ο}}=1.152\text{V}$

The standard electrode potential ( $E_{\text{cell}}^{\text{ο}}$) is calculated by the formula,

$E_{\text{cell}}^{\text{ο}}=E_{\text{cathode}}^{\text{ο}}-E_{\text{anode}}^{\text{ο}}$

Substitute the value of $E_{\text{cathode}}^{\text{ο}}$ and $E_{\text{anode}}^{\text{ο}}$ in the above equation.

$\begin{array}{c}{E}_{\text{cell}}^{\text{ο}}=1.152\text{V}-\left(1.3583\text{V}\right)\\ =\underset{\_}{-0.206V}\end{array}$

(b)

Interpretation Introduction

To determine: The concentration of ${\text{ClO}}_{\text{3}}{}^{-}$.

Answer to Problem 18.62QP

Solution

The concentration of ${\text{ClO}}_{\text{3}}{}^{-}$ is $\underset{\_}{0.66\times 10^{-36}M}$.

Explanation of Solution

Explanation

The ${\text{ClO}}_{\text{3}}{}^{-}$ for the given reaction is calculated by the formula,

$E_{\text{cell}}=E^{\text{ο}}-\frac{0.0592}{n}\log \frac{{\left(p_{{\text{ClO}}_{\text{2}}}\right)}^2\left(p_{{\text{Cl}}_2}\right)}{{\left[{\text{H}}^{\text{+}}\right]}^4{\left[{\text{Cl}}^{-}\right]}^2{\left[{\text{ClO}}_{\text{3}}{}^{-}\right]}^2}$

Where,

• $n$ is number of electrons.
• $E_{\text{cell}}$ is the cell potential.

The value of $n=2$.

The value of $E_{\text{cell}}$ at equilibrium is zero.

The concentration values of reactants and products are as follows:

$\left[{\text{H}}^{\text{+}}\right]=\left[{\text{Cl}}^{-}\right]=10\text{M}$

$p_{{\text{ClO}}_{\text{2}}}=2.0\text{atm}$

$p_{{\text{Cl}}_{\text{2}}}=1.0\text{atm}$

Substitute the value of $n$, $E_{\text{cell}}$ and $E_{}^{\text{ο}}$ and repective concentrations in the above formula.

$\begin{array}{c}0=-0.206\text{V}-\frac{0.0592}{6}\log \frac{{\left(2\right)}^2\times 1}{{\left(10\right)}^4{\left(10\right)}^2{\left[{\text{ClO}}_{\text{3}}{}^{-}\right]}^2}\\ \left[{\text{ClO}}_{\text{3}}{}^{-}\right]=\underset{\_}{0.66\times 10^{-36}M}\end{array}$

Therefore, the concentration of ${\text{ClO}}_{\text{3}}{}^{-}$ is $\underset{\_}{0.66\times 10^{-36}M}$.

Conclusion

The concentration of ${\text{ClO}}_{\text{3}}{}^{-}$ is $\underset{\_}{0.66\times 10^{-36}M}$