Chapter 19, Problem 19.121SP

a)

Interpretation Introduction

Interpretation:

A ${\text{E}}_{cell}^{}$ value has to be calculated.

Concept introduction:

Standard reduction potential: The voltage associated with a reduction reaction at an electrode when all solutes are 1M and all gases are at 1 atm. The hydrogen electrode is called the standard hydrogen electrode (SHE).

Standard emf: ${\text{E}}_{cell}^0$ is composed of a contribution from the anode and a contribution from the cathode is given by,

$E_{cell}^o=E_{cathode}^o-E_{anode}^o$

Where both $E_{cathode}^o$ and $E_{anode}^o$ are the standard reduction potentials of the electrodes.

Thermodynamics of redox reactions:

The change in free-energy represents the maximum amount of useful work that can be obtained in a reaction: $\Delta {\text{G}}^0\text{=}{\text{-nFE}}_{cell}^0$

Relation between ${\text{E}}_{cell}^0$ and equilibrium constant (K) of a redox reaction:

$\begin{array}{l}{\text{E}}_{\text{cell}}^{\text{0}}\text{=}\frac{\text{RT}}{\text{nF}}\text{lnK}\\ \text{where}\text{R}\text{is}\text{gas}\text{constant}\left(\text{8}\text{.314}\text{J/K}\text{.mol}\right)\\ \text{T}\text{is}\text{Temperature}\text{in}\text{Kelvin}\\ \text{n}\text{is}\text{no}\text{.of}\text{electrons}\text{transferred}\text{in}\text{redox}\text{reaction}\\ \text{F}\text{is}\text{Faraday}\text{constant}\left(\text{96500}\text{J/V}\text{.mol}\right)\\ \text{K}\text{is}\text{equilibrium}\text{constant}\\ \end{array}$

Relation between $\Delta {\text{G}}^0$ and K:$\Delta {\text{G}}^0\text{=}\text{-}RT\ln K$

Effect of concentration on cell Emf:

The mathematical relationship between the emf of galvanic cell and the concentration of reactants and products in a redox reaction under nonstandard-state conditions is,

$\begin{array}{l}\text{ΔG}\text{=}{\text{ΔG}}^{\text{0}}\text{+}\text{RTlnQ}\\ {\text{where, ΔG}}^{\text{0}}\text{is}\text{standard}\text{Gibb's}\text{free}\text{energy}\\ \text{ Q}\text{is}\text{reaction}\text{quotient}\text{.}\end{array}$

As known $\Delta {\text{G}}^0\text{=}{\text{-nFE}}_{cell}^0$ and $\Delta \text{G}\text{=}{\text{-nFE}}_{cell}^{}$, above expression can be written as,

$\begin{array}{l}\text{ΔG}\text{=}{\text{ΔG}}^{\text{0}}\text{+}\text{RTlnQ}\\ {\text{-nFE}}_{cell}\text{=}{\text{-nFE}}_{cell}^0\text{+}\text{RTlnQ}\end{array}$

Dividing by –nF, the above equation becomes,

$\begin{array}{l}\frac{{\text{-nFE}}_{cell}}{-nF}\text{=}\frac{{\text{-nFE}}_{cell}^0}{-nF}\text{+}\frac{\text{RTlnQ}}{-nF}\\ \\ {\text{E}}_{cell}\text{=}{\text{E}}_{cell}^0-\frac{\text{RT}}{nF}\text{lnQ}\to \text{Nernst equation}\end{array}$

Nernst equation: The Nernst equation is used to calculate the cell voltage under nonstandard-state conditions.

Explanation of Solution

        Figure.1

A galvanic concentration cell, each compartment consists of Co electrode in $\text{Co}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ solution. The concentrations in the compartments are 2.0M and 0.10M, respectively. The anode and cathode are labelled as shown above. The electron flows from anode to cathode compartments.

Nernst equation of the concentration cell and Substitute known constant values of R, T and F into Nernst equation becomes as follows,

${\text{E}}_{\text{cell}}\text{=}{\text{E}}_{\text{cell}}^{\text{0}}\text{-}\frac{\left(\text{0}\text{.0257}\text{V}\right)}{\text{n}}\text{ln}\frac{{\left[{\text{Co}}^{2+}\right]}_{dil}}{{\left[{\text{Co}}^{2+}\right]}_{conc}}$

The number of electrons transferred in the given redox reaction is TWO (n=2) and ${\text{E}}_{\text{cell}}^{\text{0}}=\text{0}\text{.0}\text{V}$ as the electrodes on two compartments are same.

$\begin{array}{l}{\text{E}}_{\text{cell}}\text{= }\left(\text{+}\text{0}\text{.0}\text{V}\right)\text{-}\frac{\left(\text{0}\text{.0257}\text{V}\right)}{\text{2}}\text{ln}\frac{\text{0}\text{.10}}{\text{2}\text{.0}}\\ \text{=}\left(\text{+}\text{0}\text{.0}\text{V}\right)\text{-}\left(\text{0}\text{.01285}\right)\text{ln}\left(\text{0}\text{.05}\right)\\ \text{=}\left(\text{+}\text{0}\text{.0}\text{V}\right)\text{-}\left(\text{0}\text{.01285}\right)\left(-2.996\right)\\ \text{=}\left(\text{+}\text{0}\text{.0}\text{V}\right)+\text{0}\text{.023549}\\ \text{=}\text{+}\text{0}\text{.0385 V}\end{array}$

The emf of the given galvanic cell reaction is $\text{+}\text{0}\text{.0385 V}$

b)

Interpretation Introduction

Interpretation:

The concentrations in the compartments when ${\text{E}}_{cell}^{}$ drops to $\text{0}\text{.020 V}$ has to be calculated.

Concept introduction:

Standard reduction potential: The voltage associated with a reduction reaction at an electrode when all solutes are 1M and all gases are at 1 atm. The hydrogen electrode is called the standard hydrogen electrode (SHE).

Standard emf: ${\text{E}}_{cell}^0$ is composed of a contribution from the anode and a contribution from the cathode is given by,

$E_{cell}^o=E_{cathode}^o-E_{anode}^o$

Where both $E_{cathode}^o$ and $E_{anode}^o$ are the standard reduction potentials of the electrodes.

Thermodynamics of redox reactions:

The change in free-energy represents the maximum amount of useful work that can be obtained in a reaction: $\Delta {\text{G}}^0\text{=}{\text{-nFE}}_{cell}^0$

Relation between ${\text{E}}_{cell}^0$ and equilibrium constant (K) of a redox reaction:

$\begin{array}{l}{\text{E}}_{\text{cell}}^{\text{0}}\text{=}\frac{\text{RT}}{\text{nF}}\text{lnK}\\ \text{where}\text{R}\text{is}\text{gas}\text{constant}\left(\text{8}\text{.314}\text{J/K}\text{.mol}\right)\\ \text{T}\text{is}\text{Temperature}\text{in}\text{Kelvin}\\ \text{n}\text{is}\text{no}\text{.of}\text{electrons}\text{transferred}\text{in}\text{redox}\text{reaction}\\ \text{F}\text{is}\text{Faraday}\text{constant}\left(\text{96500}\text{J/V}\text{.mol}\right)\\ \text{K}\text{is}\text{equilibrium}\text{constant}\\ \end{array}$

Relation between $\Delta {\text{G}}^0$ and K:$\Delta {\text{G}}^0\text{=}\text{-}RT\ln K$

Effect of concentration on cell Emf:

The mathematical relationship between the emf of galvanic cell and the concentration of reactants and products in a redox reaction under nonstandard-state conditions is,

$\begin{array}{l}\text{ΔG}\text{=}{\text{ΔG}}^{\text{0}}\text{+}\text{RTlnQ}\\ {\text{where, ΔG}}^{\text{0}}\text{is}\text{standard}\text{Gibb's}\text{free}\text{energy}\\ \text{ Q}\text{is}\text{reaction}\text{quotient}\text{.}\end{array}$

As known $\Delta {\text{G}}^0\text{=}{\text{-nFE}}_{cell}^0$ and $\Delta \text{G}\text{=}{\text{-nFE}}_{cell}^{}$, above expression can be written as,

$\begin{array}{l}\text{ΔG}\text{=}{\text{ΔG}}^{\text{0}}\text{+}\text{RTlnQ}\\ {\text{-nFE}}_{cell}\text{=}{\text{-nFE}}_{cell}^0\text{+}\text{RTlnQ}\end{array}$

Dividing by –nF, the above equation becomes,

$\begin{array}{l}\frac{{\text{-nFE}}_{cell}}{-nF}\text{=}\frac{{\text{-nFE}}_{cell}^0}{-nF}\text{+}\frac{\text{RTlnQ}}{-nF}\\ \\ {\text{E}}_{cell}\text{=}{\text{E}}_{cell}^0-\frac{\text{RT}}{nF}\text{lnQ}\to \text{Nernst equation}\end{array}$

Nernst equation: The Nernst equation is used to calculate the cell voltage under nonstandard-state conditions.

Explanation of Solution

As a concentration cell runs, the concentration of the two solutions approaches each other. Let concentration of the dilute solution equal $\left(0.10+x\right)$ and the concentration of the concentrated solution equal $\left(0.10+x\right)$

${\text{E}}_{\text{cell}}\text{=}{\text{E}}_{\text{cell}}^{\text{0}}\text{-}\frac{\left(\text{0}\text{.0257}\text{V}\right)}{\text{n}}\text{ln}\frac{{\left[{\text{Co}}^{2+}\right]}_{dil}}{{\left[{\text{Co}}^{2+}\right]}_{conc}}$

The number of electrons transferred in the given redox reaction is TWO (n=2) and ${\text{E}}_{\text{cell}}^{\text{0}}=\text{0}\text{.0}\text{V}$ as the electrodes on two compartments are same.

$\begin{array}{l}{\text{E}}_{\text{cell}}\text{= }\left(\text{+}\text{0}\text{.0}\text{V}\right)\text{-}\frac{\left(\text{0}\text{.0257}\text{V}\right)}{\text{2}}\text{ln}\frac{\text{0}\text{.10+x}}{\text{2}\text{.0-x}}\\ 0.020V\text{=}\text{-}\left(\text{0}\text{.01285}\right)\text{ln}\frac{\text{0}\text{.10+x}}{\text{2}\text{.0-x}}\\ \frac{0.020V}{\text{-}\left(\text{0}\text{.01285}\right)}\text{=}\text{ln}\frac{\text{0}\text{.10+x}}{\text{2}\text{.0-x}}\\ e^{-1.56}\text{=}\frac{\text{0}\text{.10+x}}{\text{2}\text{.0-x}}\end{array}$

Solve for x as follows,

$\begin{array}{l}0.210\left(\text{2}\text{.0-x}\right)=\left(\text{0}\text{.10+x}\right)\\ 0.42-0.10=0.210x+x\\ 0.32=\left(0.210+1\right)x\\ \frac{0.32}{1.210}=x\\ x=0.26M\end{array}$

At anode compartment: $\text{0}\text{.10}\text{M}\text{+}\text{0}\text{.26}\text{M = 0}\text{.36 M}$

At cathode compartment: $\text{2}\text{.0}\text{M}-\text{0}\text{.26}\text{M = 1}\text{.74 M}$