Chapter 18.5, Problem 6PPB

Interpretation Introduction

Interpretation:

Partial pressure of $H_2$ at ${80}^{0}C$ should be calculated

Concept introduction:

The Nernst equation,

$\begin{array}{c}{\text{E}}_{\text{cell}}\text{=}{\text{E}}_{\text{cell}}^{\text{0}}-\frac{\text{2}\text{.303}\text{RT}}{\text{nF}}\log \text{Q}\\ \text{where,}\\ \text{Q}\text{is}\text{the}\text{reactant}\text{quotient}\text{=}\frac{\left[\text{anode}\right]}{\left[\text{cathode}\right]}\end{array}$

• $\begin{array}{c}{\text{E}}_{\text{cell}}^{}\text{=}\text{Electrochemical}\text{cell}\text{potential}\\ {\text{E}}_{\text{cell}}^{\text{0}}=\text{Standard}\text{electrochemical}\text{cell}\text{potential}\\ \text{n}\text{=}\text{number}\text{of}\text{electrons}\text{passed}\text{from}\text{anode}\text{to}\text{cathode}\\ \text{T}\text{=}\text{Temperature}\end{array}$

The $\text{pH}$ of a solution is a figure that expresses the acidity or the alkalinity of a given solution

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

Answer to Problem 6PPB

Partial pressure of $H_2$ is $\text{18}\text{atm}$

Explanation of Solution

The cell reaction involved is,

$H_{2\left(g\right)}+2AgC{l}_{\left(s\right)}\underset{}{\rightleftarrows }2A{g}_{\left(s\right)}+2H^{+}+2C{l}_{\left(aq\right)}^{-}$

Partial pressure of $H_2$ is calculated for the above reaction.

Given,

$\begin{array}{c}{E}_{cell}=0.27V\\ E_{cell}^0=0.18V\\ ConcentrationofC{l}^{-},\left[C{l}^{-}\right]=2.5M\\ pH=1.05\\ T={80}^{0}C\\ ={80}^{0}C+273K\\ =353K\end{array}$

Concentration of $H^{+}$, $\left[H^{+}\right]$ is calculated as,

$\begin{array}{c}\text{pH}=-\log \left[{\text{H}}^{+}\right]\\ 1.05=-\log \left[{\text{H}}^{+}\right]\\ \left[{\text{H}}^{+}\right]={\log }^{-1}\left(-1.05\right)\\ =0.089\text{M}\end{array}$

According to Nernst equation,

${\text{E}}_{\text{cell}}\text{=}{\text{E}}_{\text{cell}}^{\text{0}}-\frac{\text{2}\text{.303}\text{RT}}{\text{nF}}\log \text{Q}$

Since $H_2$ is gaseous, partial pressure is taken as the concentration of $H_2$.

Hence,

$Q=\frac{{\left[H^{+}\right]}^2{\left[C{l}^{-}\right]}^2}{partialpressureof{H}_2}$

Therefore, the equation becomes

$\begin{array}{c}{\text{E}}_{\text{cell}}\text{=}{\text{E}}_{\text{cell}}^{\text{0}}-\frac{\text{2}\text{.303}\text{RT}}{\text{nF}}\log \frac{{\left[H^{+}\right]}^2{\left[C{l}^{-}\right]}^2}{partialpressureof{H}_2}\\ \text{0}\text{.27}\text{V}\text{=}\text{0}\text{.18}\text{V}-\frac{\text{2}\text{.303}\times \text{8}\text{.314}\text{J/K}\text{.mol}\times \text{353}\text{K}}{\text{2}\times \text{96500}\text{J/V}\text{.mol}}\log \frac{{\left[0.089M\right]}^2{\left[2.5M\right]}^2}{partialpressureof{H}_2}\\ \text{0}\text{.09}\text{V}=-\left(0.0350\right)\log \frac{\left(7.921\times {10}^{-3}\right)\left(6.25\right)}{partialpressureof{H}_2}\\ \left(-2.571\right)=\log \frac{\left(0.0495\right)}{partialpressureof{H}_2}\\ \frac{\left(0.0495\right)}{partialpressureof{H}_2}={\log }^{-1}\left(-2.571\right)\\ partialpressureof{H}_2=\frac{\left(0.0495\right)}{\left(2.685\times {10}^{-3}\right)}\\ =18atm\\ \end{array}$

Therefore,

Partial pressure of $H_2$ is $\text{18}\text{atm}$

Conclusion

Partial pressure of $H_2$ is $\text{18}\text{atm}$