Chapter 21, Problem 62QAP

Interpretation Introduction

Interpretation:

The value of pH needs to be calculated if voltage is $1.000\text{V}$.

Concept introduction:

The pH is the negative logarithm of hydrogen ion concentration. Therefore,

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

Nernst equation is written as follows:

$E_{\text{cell}}=E_{{}_{\text{cell}}}^0-\frac{RT}{nF}\ln Q$

Here, R represents gas constant, T represents temperature, n represents number of electron, F represents Faraday’s constant, $Q$ represents reaction quotient, $E_{\text{cell}}$ represents cell potential and $E^0{}_{\text{cell}}$ represents standard electrode potential.

Answer to Problem 62QAP

At voltage $1.000\text{V}$, pH is $4.6$.

Explanation of Solution

Balance the reaction as follows:

Consider oxidation and reduction reaction separately as given below.

Reduction: ${\text{NO}}^{-}{}_3\left(\text{aq}\right)\to \text{NO}\left(\text{g}\right)$

Oxidation: ${\text{SO}}_{\text{2}}\left(\text{g}\right)\to {\text{SO}}_4^{2-}\left(\text{aq}\right)$

Atom balance as follows:

One nitrogen each side.

Hence, no adjustment is required.

Total oxidation number- $\text{N:}\text{5}\to \text{2}$

Add electrons: Oxidation number for N goes from +5 to +2. It is reduced by 3. Add three electrons to reactant side.

${\text{NO}}^{-}{}_3\left(\text{aq}\right)+3{\text{e}}^{-}\to \text{NO}\left(\text{g}\right)$

Balance charge: reactants $-1+3\left(-1\right)=-4$

Products: 0, in acidic medium add ${\text{H}}^{\text{+}}$. To balance add four ${\text{H}}^{\text{+}}$.

${\text{NO}}^{-}{}_3\left(\text{aq}\right)+3{\text{e}}^{-}+4{\text{H}}^{+}\to \text{NO}\left(\text{g}\right)$

Balance hydrogen atom:

Reactant contains 4 H and product contains 0 H. To balance add $2{\text{H}}_{\text{2}}\text{O}$ on right.

${\text{NO}}^{-}{}_3\left(\text{aq}\right)+3{\text{e}}^{-}+4{\text{H}}^{+}\to \text{NO}\left(\text{g}\right)+{\text{2H}}_{\text{2}}\text{O}$

Balance reaction is as follows:

${\text{NO}}_3^{-}\left(\text{aq}\right)+4{\text{H}}^{\text{+}}\left(\text{aq}\right)+3{\text{e}}^{-}\to \text{NO}\left(\text{g}\right){\text{+2H}}_{\text{2}}\text{O}$     ......(1)

Atom balance as follows:

One sulphur each side. Hence, no adjustment is required.

Total oxidation number- $\text{S:}+4\to +6$

Add electrons: Oxidation number for S increases from +4 to +6. It is increased by 2. Add two electrons to product side.

${\text{SO}}_{\text{2}}\left(\text{g}\right)\to {\text{SO}}_4^{2-}\left(\text{aq}\right)+{\text{2e}}^{-}$

Balance charge: Products $-2+2\left(-1\right)=-4$

In acidic medium add ${\text{H}}^{\text{+}}$. To balance add 4 ${\text{H}}^{\text{+}}$.

${\text{SO}}_{\text{2}}\left(\text{g}\right)\to {\text{SO}}_4^{2-}\left(\text{aq}\right)+{\text{2e}}^{-}+4{\text{H}}^{+}$

Balance hydrogen atom:

Reactant contains 0 H and product contains 4 H. To balance, add $2{\text{H}}_{\text{2}}\text{O}$ on right.

${\text{SO}}_{\text{2}}\left(\text{g}\right)+2{\text{H}}_{\text{2}}\text{O}\to {\text{SO}}_4^{2-}\left(\text{aq}\right)+{\text{2e}}^{-}+4{\text{H}}^{+}$

Balance reaction is as follows:

${\text{SO}}_{\text{2}}\left(\text{g}\right)+2{\text{H}}_{\text{2}}\text{O}\to {\text{SO}}_4^{2-}\left(\text{aq}\right)+{\text{2e}}^{-}+4{\text{H}}^{+}$     ......(2)

Multiply equation (1) by 2 and equation (2) by 3 and add the resulting equation.

$\begin{array}{l}{\text{2NO}}_3^{-}\left(\text{aq}\right)+8{\text{H}}^{\text{+}}\left(\text{aq}\right)+6{\text{e}}^{-}\to 2\text{NO}\left(\text{g}\right){\text{+4H}}_{\text{2}}\text{O}\\ {\text{3SO}}_{\text{2}}\left(\text{g}\right)+6{\text{H}}_{\text{2}}\text{O}\to 12{\text{H}}^{\text{+}}\left(\text{aq}\right)+3{\text{SO}}_4^{2-}\left(\text{aq}\right)+6{\text{e}}^{-}\end{array}$

${\text{2NO}}_3^{-}\left(\text{aq}\right)+{\text{3SO}}_{\text{2}}\left(\text{g}\right)+2{\text{H}}_{\text{2}}\text{O}\to 4{\text{H}}^{\text{+}}\left(\text{aq}\right)+2\text{NO}\left(\text{g}\right)+3{\text{SO}}_4^{2-}\left(\text{aq}\right)$ …… (3)

Calculate $E°$ as follows:

$\begin{array}{c}E°=E{°}_{\text{red}}+E{°}_{\text{ox}}\\ =0.964\text{V}+\left(-0.155\text{V}\right)\\ =+0.809\text{V}\end{array}$

Use Nernst equation to calculate reaction quotient ( Q) as follows:

Nernst equation is given below.

$E_{\text{cell}}=E_{{}_{\text{cell}}}^0-\frac{RT}{nF}\ln Q$

Here, R represents gas constant, T represents temperature, n represents number of electron, F represents Faraday’s constant, $Q$ represents reaction quotient, $E_{\text{cell}}$ represents cell potential and $E^0{}_{\text{cell}}$ represents standard electrode potential.

Here,

$\begin{array}{c}R=8.314\text{}\text{J/}\text{K}\text{mol}\\ T=298\text{K}\\ F=96480\\ E_{\text{cell}}=1.000\text{V}\end{array}$

$\begin{array}{c}{E}_{\text{cell}}^0=+0.809\text{V}\\ n=6\end{array}$

$\begin{array}{c}1.000=0.809-\frac{RT}{nF}\ln Q\\ =0.809-\frac{8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\times 298\text{K}}{n\times 96480}\ln Q\\ \frac{0.0257}{n}\ln Q=0.809-1.000\\ \frac{0.0257}{6}\ln Q=-0.191\end{array}$

$\begin{array}{c}0.0043\ln Q=-0.191\\ \ln Q=\frac{-0.191}{0.0043}\\ \ln Q=-44.41\\ Q=e^{-44.41}\end{array}$

$Q=5.11\times {10}^{-20}$

Calculate concentration of ${\text{H}}^{\text{+}}$ for balance reaction (equation 3) as follows:

Given, pressure of all gases is $1.00\text{atm}$ and pressure of ionic species is $0.100\text{M}$.

$\begin{array}{c}Q=\frac{{\left(P_{\text{NO}}\right)}^2{\left[{\text{SO}}_{\text{4}}{}^{2-}\right]}^3{\left[{\text{H}}^{\text{+}}\right]}^4}{\left[{\text{NO}}_{\text{3}}{}^{-}\right]{\left[P_{{\text{SO}}_{\text{2}}}\right]}^3}\\ 5.11\times {10}^{-20}=\frac{{\left(1.00\right)}^2{\left(0.100\right)}^3{\left[{\text{H}}^{\text{+}}\right]}^4}{{\left(0.100\right)}^2{\left(1.00\right)}^3}\\ \left(5.11\times {10}^{-20}\right){\left(0.100\right)}^2{\left(1.00\right)}^3={\left(1.00\right)}^2{\left(0.100\right)}^3{\left[{\text{H}}^{\text{+}}\right]}^4\end{array}$

$\begin{array}{c}{\left[{\text{H}}^{\text{+}}\right]}^4=5.11\times {10}^{-19}\\ \left[{\text{H}}^{\text{+}}\right]={\left(5.11\times {10}^{-19}\right)}^{\frac{1}{4}}\\ =2.674\times {10}^{-5}\end{array}$

Calculate pH as follows:

$\begin{array}{c}\text{pH}=-\log \left[{\text{H}}^{\text{+}}\right]\\ =-\log \left(2.674\times {10}^{-5}\right)\\ =4.6\end{array}$

Conclusion

Therefore, at voltage $1.000\text{V}$, pH is $4.6$.