Chapter 5, Problem 5.77P

(a)

Interpretation Introduction

Interpretation:

Standard potential of ${\text{HCO}}_3^{-}/{\text{CO}}_3^{2-},H_2$ couple has to be determined using the given data.

Concept Introduction:

The value of standard free energy change ${\text{ΔG}}^{\text{ο}}$ of the given reaction is calculated by the formula,

  ${\text{Δ}}_{\text{r}}{\text{G}}^{\text{Θ}}{\text{ = Σν}}_p{\text{Δ}}_{\text{f}}{\text{G}}^{\text{Θ}}{}_{\left(\text{products}\right)}-{\text{ Σν}}_r{\text{Δ}}_{\text{f}}{\text{G}}^{\text{Θ}}{}_{\left(\text{reactants}\right)}$

Where,

  $\begin{array}{c}{\text{Δ}}_{\text{f}}{\text{G}}^{\text{Θ}}{}_{\left(\text{reactants}\right)}\text{=}\text{the standard free energy of formation for the reactants}\\ {\text{Δ}}_{\text{f}}{\text{G}}^{\text{Θ}}{}_{\left(\text{products}\right)}\text{=}\text{the standard free energy of formation for the products}\\ {\text{ν}}_p\text{=}\text{number of products molecule}\\ {\text{ν}}_r\text{=}\text{number of reactants molecule}\end{array}$

Standard Gibbs energy is calculated as below,

  ${\text{Δ}}_r{\text{G}}^{\Theta }\text{= }\text{}-\nu {\text{FE}}_{cell}^{\Theta }\boxed{\begin{array}{c}\text{F = Faraday's constant}\\ \nu \text{ = No}\text{. of electrons transferred}\\ {\text{E}}_{cell}^{\Theta }\text{ = Standard cell potential at temperature, T}\end{array}}$

Explanation of Solution

The given information and half-cell reaction is shown below,

  $\begin{array}{l}{\text{Δ}}_{\text{f}}{\text{G}}^{\text{Θ}}{}_{\left(C{O}_3^{2-}\right)}=-527.81\text{ kJ}{\text{.mol}}^{-1}\\ \\ {\text{Δ}}_{\text{f}}{\text{G}}^{\text{Θ}}{}_{\left(HC{O}_3^{-}\right)}=-586.77\text{ kJ}{\text{.mol}}^{-1}\\ \\ HC{O}_3^{-}\left(aq\right){\text{ + e}}^{-}\stackrel{}{\to }C{O}_3^{2-}\left(aq\right)\text{ + }\frac{1}{2}{H}_2\left(g\right)\end{array}$

Standard Gibbs energy is calculated as follows,

  $\begin{array}{c}{\text{Δ}}_{\text{r}}{\text{G}}^{\text{Θ}}{\text{ = Σν}}_p{\text{Δ}}_{\text{f}}{\text{G}}^{\text{Θ}}{}_{\left(\text{products}\right)}-{\text{ Σν}}_r{\text{Δ}}_{\text{f}}{\text{G}}^{\text{Θ}}{}_{\left(\text{reactants}\right)}\\ \\ =\left[{\text{ν}}_{\left(C{O}_3^{2-}\right)}{\text{Δ}}_{\text{f}}{\text{G}}^{\text{Θ}}{}_{\left(C{O}_3^{2-}\right)}+{\text{ ν}}_{\left(H_2\right)}{\text{Δ}}_{\text{f}}{\text{G}}^{\text{Θ}}{}_{\left(H_2\right)}\right]-\left[{\text{ν}}_{\left(HC{O}_3^{-}\right)}{\text{Δ}}_{\text{f}}{\text{G}}^{\text{Θ}}{}_{\left(HC{O}_3^{-}\right)}\right]\\ \\ =\left[\left(\text{1 mol}\times \left(-527.81\text{ kJ}{\text{.mol}}^{-1}\right)\right)+\left(\frac{1}{2}\text{ mol}\times 0\text{ kJ}{\text{.mol}}^{-1}\right)\right]\\ -\left[\text{1 mol}\times \left(-586.77\text{ kJ}{\text{.mol}}^{-1}\right)\right]\\ \\ =\text{ 58}\text{.96 kJ}\end{array}$

Standard potential of ${\text{HCO}}_3^{-}/{\text{CO}}_3^{2-},H_2$ couple is determined using the given formula,

  $\begin{array}{c}{\text{Δ}}_r{\text{G}}^{\Theta }\text{= }\text{}-\nu {\text{FE}}_{cell}^{\Theta }\\ \\ \left(\text{58}\text{.96 kJ}\right)=-\left(1\text{ mol}\right)\left(\text{96485C/mol}\right){\text{E}}_{cell}^{\Theta }\\ \\ {\text{E}}_{cell}^{\Theta }\text{ = }-\frac{\left(\text{58}\text{.96 kJ}\right)}{\left(1\text{ mol}\right)\left(\text{96485C/mol}\right)}\\ \\ =-0.611\text{ V}\end{array}$

Standard potential of ${\text{HCO}}_3^{-}/{\text{CO}}_3^{2-},H_2$ couple is $-0.611\text{ V}$.

(b)

Interpretation Introduction

Interpretation:

Standard cell potential for the given reaction has to be determined using the given data.

Concept Introduction:

Standard potential ${\text{(E}}_{\text{cell}}^{\text{o}}\text{)}$ can be calculated by the following formula.

  ${\text{E}}_{\text{cell}}^{\Theta }{\text{=E}}_{\text{R}}^{\Theta }-{\text{E}}_{\text{L}}^{\Theta }$

Explanation of Solution

Net ionic equation can be given as below,

  $C{O}_3^{2-}\left(aq\right)\text{ + }{H}_{2}O\left(l\right)\stackrel{}{\to }HC{O}_3^{-}\left(aq\right){\text{ + OH}}^{-}\left(aq\right)$

Standard potential of ${\text{HCO}}_3^{-}/{\text{CO}}_3^{2-},H_2$ couple is $-0.61\text{ V}$.

Also,

  $H_{2}O{\text{ + e}}^{-}\underset{}{\to }\frac{1}{2}{H}_2\left(g\right){\text{ + OH}}^{-}\left(aq\right){\text{E}}^o=-0.83\text{ V}$

${\text{HCO}}_3^{-}/{\text{CO}}_3^{2-},H_2$ couple is right hand half-cell whereas ${\text{H}}_{\text{2}}{\text{O/H}}_{\text{2}}{\text{, OH}}^{-}$ couple is left hand half-cell. Therefore, standard cell potential for the given reaction is calculated as given below,

  $\begin{array}{c}{\text{E}}_{\text{cell}}^{\Theta }{\text{=E}}_{\text{R}}^{\Theta }-{\text{E}}_{\text{L}}^{\Theta }\\ \\ ={\text{ E}}_{\left({\text{CO}}_{\text{3}}^{\text{2}-}{\text{/HCO}}_{\text{3}}^{-}\right)}^{\text{Θ}}-{\text{E}}_{\left({\text{H}}_{\text{2}}{\text{O/H}}^{\text{+}}{\text{,OH}}^{-}\right)}^{\text{Θ}}\\ \\ =\left(-0.83\text{ V}\right)-\left(-0.61\text{ V}\right)\\ \\ =-0.22\text{ V}\end{array}$

Standard cell potential for the given reaction is $-0.22\text{ V}$.

(c)

Interpretation Introduction

Interpretation:

Nernst equation for the given cell reaction has to be determined.

Concept Introduction:

Cell potential of a cell can be determined using Nernst equation as follows,

  $E_{cell}{\text{ = E}}_{cell}^{\Theta }-\frac{RT}{\nu F}\ln Q\boxed{\begin{array}{c}\text{F = Faraday's constant}\\ \nu \text{ = No}\text{. of electrons transferred}\\ {\text{E}}_{cell}^{\Theta }\text{ = Standard cell potential}\end{array}}$

Explanation of Solution

Net ionic equation can be given as below,

  $C{O}_3^{2-}\left(aq\right)\text{ + }{H}_{2}O\left(l\right)\stackrel{}{\to }HC{O}_3^{-}\left(aq\right){\text{ + OH}}^{-}\left(aq\right)$

Nernst equation for the given cell reaction is shown below,

  $\begin{array}{c}{E}_{cell}{\text{ = E}}_{cell}^{\Theta }-\frac{RT}{\nu F}\ln Q\\ \\ ={\text{E}}_{cell}^{\Theta }-\frac{RT}{1\times F}\ln \frac{a_{HC{O}_3^{-}}{a}_{O{H}^{-}}}{a_{C{O}_3^{2-}}{a}_{H_{2}O}}\left(\text{Since}{\text{ a}}_{{\text{H}}_{\text{2}}\text{O}}\text{= 1}\right)\\ \\ E_{cell}{\text{ = E}}_{cell}^{\Theta }-\frac{RT}{1\times F}\ln \frac{a_{HC{O}_3^{-}}{a}_{O{H}^{-}}}{a_{C{O}_3^{2-}}}\end{array}$

(d)

Interpretation Introduction

Interpretation:

Change in potential when pH of the solution becomes 7 has to be determined using the given data.

Concept Introduction:

Cell potential of a cell can be determined using Nernst equation as follows,

  $E_{cell}{\text{ = E}}_{cell}^{\Theta }-\frac{RT}{\nu F}\ln Q\boxed{\begin{array}{c}\text{F = Faraday's constant}\\ \nu \text{ = No}\text{. of electrons transferred}\\ {\text{E}}_{cell}^{\Theta }\text{ = Standard cell potential}\end{array}}$

Explanation of Solution

The pH of the solution is 7. Hence, $pH=pOH\text{ = 7}$. Therefore, $a_{O{H}^{-}}={\text{ 10}}^{-pOH}={\text{ 10}}^{-7}$. Also $a_{HC{O}_3^{-}}\text{ = }{a}_{C{O}_3^{2-}}=\text{ 1}$.

Therefore, change in potential when pH of the solution becomes 7 can be determined as given below,

  $\begin{array}{c}{E}^{\oplus }{}_{cell}{\text{ = E}}_{cell}^{\Theta }-\frac{RT}{\nu F}\ln Q\\ \\ E^{\oplus }{}_{cell}-{\text{E}}^{\Theta }{}_{cell}=-\frac{RT}{F}\ln \frac{a_{HC{O}_3^{-}}{a}_{O{H}^{-}}}{a_{C{O}_3^{2-}}}\\ \\ =-\frac{\left(1\text{ mol}\right)\left(8.314\text{ J}{\text{.K}}^{-1}mo{l}^{-1}\right)\left(298\text{ K}\right)}{\left(96485\text{ C}{\text{.mol}}^{-1}\right)}\ln {10}^{-7}\\ \\ =\text{ 0}\text{.41 V}\end{array}$

Change in potential when pH of the solution becomes 7 is $\text{0}\text{.41 V}$.

(e)

Interpretation Introduction

Interpretation:

$K_a$ of $HC{O}_3{}^{-}$ has to be determined using the given data.

Concept Introduction:

The relation between standard cell potential and base dissociation constant is as follows,

  ${\text{lnK}}_b\text{ = }\frac{-\nu {\text{FE}}_{cell}^{\Theta }}{RT}\boxed{\begin{array}{c}{K}_b=\text{Base dissociation constant}\\ \text{F = Faraday's constant}\\ \nu \text{ = No}\text{. of electrons transferred}\\ {\text{E}}_{cell}^{\Theta }\text{ = Standard cell potential at temperature, T}\end{array}}$

The relation between acidic and basic constants $p{K}_{a}andp{K}_b$ with $p{K}_w$ (equilibrium constant for water) is,

  $p{K}_w{\text{ = pK}}_a{\text{ + pK}}_b$

Explanation of Solution

Standard cell potential for the given reaction is $-0.22\text{ V}$.

Net ionic equation can be given as below,

  $C{O}_3^{2-}\left(aq\right)\text{ + }{H}_{2}O\left(l\right)\stackrel{}{\to }HC{O}_3^{-}\left(aq\right){\text{ + OH}}^{-}\left(aq\right)$

Equation for calculating the $K_b$ of $C{O}_3{}^{2-}$ is given below,

  ${\text{lnK}}_b\text{ =}\frac{\nu {\text{FE}}_{cell}^{\Theta }}{RT}$

Therefore, $K_a$ of $HC{O}_3{}^{-}$ is determined as follows,

  $\begin{array}{c}{\text{pK}}_a\text{ = }p{K}_w-{\text{ pK}}_b\\ \\ =\text{ 14}\text{.0}-\frac{\left(1\text{ mol}\right)\left(96485\text{ C}{\text{.mol}}^{-1}\right)\left(-0.22\text{ V}\right)}{\left(8.314\text{ J}{\text{.K}}^{-1}mo{l}^{-1}\right)\left(298\text{ K}\right)}\\ \\ =\text{ 5}\text{.43}\end{array}$

$K_a$ of $HC{O}_3{}^{-}$ is $\text{5}\text{.43}$