Chapter 17, Problem 100QAP

Consider the following standard reduction potentials:
$\begin{array}{l}{\text{Tl}}^{+}\left(aq\right)+e^{-}\to \text{Tl}\left(s\right)E_{\text{red}}^{°}=-0.34\text{V}\\ {\text{Tl}}^{3+}\left(aq\right)+3e^{-}\to \text{Tl}\left(s\right)E_{\text{red}}^{°}=0.74\text{V}\\ {\text{Tl}}^{3+}\left(aq\right)+2e^{-}\to {\text{Tl}}^{+}\left(aq\right)E_{\text{red}}^{°}=1.28\text{V}\end{array}$
and the following abbreviated cell notations:
(1) $\text{Tl}|{\text{Tl}}^{+}\Vert {\text{Tl}}^{3+},{\text{Tl}}^{+}|\text{Pt}$
(2) $\text{Tl}|{\text{Tl}}^{3+}\Vert {\text{Tl}}^{3+},{\text{Tl}}^{+}|\text{Pt}$
(3) $\text{Tl}|{\text{Tl}}^{+}\Vert {\text{Tl}}^{3+}|\text{Tl}$
(a) Write the overall equation for each cell.
(b) Calculate E° for each cell.
(c) Calculate ΔG° for each overall equation.
(d) Comment on whether ΔG° and/or E° are state properties. (Hint: A state property is path-independent.)

Interpretation Introduction

(a)

Interpretation:

The overall equation for each of the cells represented by the following notation needs to be deduced

$\begin{array}{l}{\text{(1) Tl | Tl}}^{\text{+}}{\text{|| Tl}}^{\text{3+}}{\text{, Tl}}^{\text{+}}\text{ | Pt }\\ {\text{(2) Tl | Tl}}^{\text{3+}}{\text{|| Tl}}^{\text{3+}}{\text{, Tl}}^{\text{+}}\text{ | Pt}\\ {\text{(3) Tl | Tl}}^{\text{+}}{\text{|| Tl}}^{\text{3+}}\text{, Tl }\end{array}$

Concept introduction:

Electrochemical cells are represented in a shorthand format via the cell notation method which depicts the anode half-cell on the left separated from the cathode half-cell on the right via a salt bridge (||). The chemical reactions that take place in an electrochemical cell are redox reactions i.e. oxidation (loss of electrons) at the anode and reduction (gain of electrons) at the cathode

Answer to Problem 100QAP

${\text{Overall equation for each cell: 2Tl(s) + Tl}}^{\text{3+}}\text{(aq) }\to {\text{ 3Tl}}^{\text{+}}\text{(aq)}$ .

Explanation of Solution

1. The given cell notation is:

${\text{Tl | Tl}}^{\text{+}}{\text{|| Tl}}^{\text{3+}}{\text{, Tl}}^{\text{+}}\text{ | Pt }$

The half reactions are:

$\begin{array}{l}\text{Anode (oxidation): 2Tl(s) }\to {\text{ 2Tl}}^{\text{+}}{\text{(aq) + 2e}}^{\text{-}}{\text{ E}}^0=-0.34\text{ V}\\ \text{Cathode (reduction)}:{\text{ Tl}}^{\text{3+}}{\text{(aq) + 2e}}^{\text{-}}\to {\text{Tl}}^{\text{+}}{\text{(aq) E}}^0\text{ = 1}\text{.28 V }\\ \text{--------------------------------------------------------------- }\\ {\text{Overall Reaction: 2Tl(s) + Tl}}^{\text{3+}}\text{(aq) }\to {\text{ 3Tl}}^{\text{+}}\text{(aq) }\\ \text{------------------------------------------------------------------}\end{array}$

1. The given cell notation is:

${\text{Tl | Tl}}^{\text{3+}}{\text{|| Tl}}^{\text{3+}}{\text{, Tl}}^{\text{+}}\text{ | Pt}$

The half reactions are:

$\begin{array}{l}\text{Anode (oxidation): 2Tl(s) }\to {\text{ 2Tl}}^{\text{3+}}{\text{(aq) + 6e}}^{\text{-}}{\text{ E}}^0=0.74\text{ V}\\ \text{Cathode (reduction)}:{\text{ 3Tl}}^{\text{3+}}{\text{(aq) + 6e}}^{\text{-}}\to 3{\text{Tl}}^{\text{+}}{\text{(aq) E}}^0\text{ = 1}\text{.28 V }\\ \text{--------------------------------------------------------------- }\\ {\text{Overall Reaction: 2Tl(s) + Tl}}^{\text{3+}}\text{(aq) }\to {\text{ 3Tl}}^{\text{+}}\text{(aq) }\\ \text{------------------------------------------------------------------}\end{array}$

1. The given cell notation is:

${\text{Tl | Tl}}^{\text{+}}{\text{|| Tl}}^{\text{3+}}\text{, Tl}$

The half reactions are:

$\begin{array}{l}\text{Anode (oxidation): 3Tl(s) }\to {\text{ 3Tl}}^{\text{+}}{\text{(aq) + 3e}}^{\text{-}}{\text{ E}}^0=-0.34\text{ V}\\ \text{Cathode (reduction)}:{\text{ Tl}}^{\text{3+}}{\text{(aq) + 3e}}^{\text{-}}\to {\text{Tl(s) E}}^0\text{ = +0}\text{.74 V }\\ \text{--------------------------------------------------------------- }\\ {\text{Overall Reaction: 2Tl(s) + Tl}}^{\text{3+}}\text{(aq) }\to {\text{ 3Tl}}^{\text{+}}\text{(aq) }\\ \text{------------------------------------------------------------------}\end{array}$

Interpretation Introduction

(b)

Interpretation:

The standard voltage (E°) for each of the three cells needs to be deduced

Concept introduction:

The standard voltage (E°) for the cell is the sum of the standard voltages of the reduction $({\text{E}}_{\text{red}}^{\text{0}})$ and oxidation $({\text{E}}_{\text{oxd}}^{\text{0}})$ potentials $\begin{array}{l}{\text{E}}^{\text{0}}{\text{ = E}}_{\text{red}}^{\text{0}}{\text{- E}}_{\text{oxd}}^{\text{0}}\\ {\text{(or) E}}^{\text{0}}{\text{ = E}}_{\text{cathode}}^{\text{0}}{\text{- E}}_{\text{anode}}^{\text{0}}\text{ ----------(1)}\end{array}$

Answer to Problem 100QAP

$\begin{array}{l}{\text{(1) Tl | Tl}}^{\text{+}}{\text{|| Tl}}^{\text{3+}}{\text{, Tl}}^{\text{+}}{\text{ | Pt ----------E}}^0=1.62\text{ V}\\ {\text{(2) Tl | Tl}}^{\text{3+}}{\text{|| Tl}}^{\text{3+}}{\text{, Tl}}^{\text{+}}{\text{ | Pt ----------E}}^0=\text{0}\text{.54 V}\\ {\text{(3) Tl | Tl}}^{\text{+}}{\text{|| Tl}}^{\text{3+}}{\text{, Tl ----------E}}^0=\text{1}\text{.08 V}\end{array}$

Explanation of Solution

1. The given cell notation is:

${\text{Tl | Tl}}^{\text{+}}{\text{|| Tl}}^{\text{3+}}{\text{, Tl}}^{\text{+}}\text{ | Pt }$

The half reactions are:

$\begin{array}{l}\text{Anode (oxidation): 2Tl(s) }\to {\text{ 2Tl}}^{\text{+}}{\text{(aq) + 2e}}^{\text{-}}{\text{ E}}^0=-0.34\text{ V}\\ \text{Cathode (reduction)}:{\text{ Tl}}^{\text{3+}}{\text{(aq) + 2e}}^{\text{-}}\to {\text{Tl}}^{\text{+}}{\text{(aq) E}}^0\text{ = 1}\text{.28 V }\\ \text{--------------------------------------------------------------- }\\ {\text{Overall Reaction: 2Tl(s) + Tl}}^{\text{3+}}\text{(aq) }\to {\text{ 3Tl}}^{\text{+}}\text{(aq) }\\ \text{------------------------------------------------------------------}\\ {\text{E}}^{\text{0}}{\text{ = E}}_{\text{cathode}}^{\text{0}}{\text{- E}}_{\text{anode}}^{\text{0}}=\text{ 1}\text{.28 - (-0}\text{.34) = 1}\text{.62 V}\end{array}$

2) The given cell notation is:

${\text{Tl | Tl}}^{\text{3+}}{\text{|| Tl}}^{\text{3+}}{\text{, Tl}}^{\text{+}}\text{ | Pt}$

The half reactions are:

$\begin{array}{l}\text{Anode (oxidation): 2Tl(s) }\to {\text{ 2Tl}}^{\text{3+}}{\text{(aq) + 6e}}^{\text{-}}{\text{ E}}^0=0.74\text{ V}\\ \text{Cathode (reduction)}:{\text{ 3Tl}}^{\text{3+}}{\text{(aq) + 6e}}^{\text{-}}\to 3{\text{Tl}}^{\text{+}}{\text{(aq) E}}^0\text{ = 1}\text{.28 V }\\ \text{--------------------------------------------------------------- }\\ {\text{Overall Reaction: 2Tl(s) + Tl}}^{\text{3+}}\text{(aq) }\to {\text{ 3Tl}}^{\text{+}}\text{(aq) }\\ \text{------------------------------------------------------------------}\\ {\text{E}}^{\text{0}}{\text{ = E}}_{\text{cathode}}^{\text{0}}{\text{- E}}_{\text{anode}}^{\text{0}}=\text{ 1}\text{.28 - (0}\text{.74) = 0}\text{.54 V}\end{array}$

3) The given cell notation is:

${\text{Tl | Tl}}^{\text{+}}{\text{|| Tl}}^{\text{3+}}\text{, Tl}$

The half reactions are:

$\begin{array}{l}\text{Anode (oxidation): 3Tl(s) }\to {\text{ 3Tl}}^{\text{+}}{\text{(aq) + 3e}}^{\text{-}}{\text{ E}}^0=-0.34\text{ V}\\ \text{Cathode (reduction)}:{\text{ Tl}}^{\text{3+}}{\text{(aq) + 3e}}^{\text{-}}\to {\text{Tl(s) E}}^0\text{ = +0}\text{.74 V }\\ \text{--------------------------------------------------------------- }\\ {\text{Overall Reaction: 2Tl(s) + Tl}}^{\text{3+}}\text{(aq) }\to {\text{ 3Tl}}^{\text{+}}\text{(aq) }\\ \text{------------------------------------------------------------------}\\ {\text{E}}^{\text{0}}{\text{ = E}}_{\text{cathode}}^{\text{0}}{\text{- E}}_{\text{anode}}^{\text{0}}=\text{ 0}\text{.74 - (-0}\text{.34) = 1}\text{.06 V}\end{array}$

Interpretation Introduction

(c)

Interpretation:

The standard free energy change (ΔG°) for each of the three cells needs to be deduced

Concept introduction:

The change in the Gibbs free energy, ΔG is a thermodynamic function which governs the spontaneity of a chemical reaction. If ΔG is negative the reaction is spontaneous, positive value indicated that the reaction is non-spontaneous and if ΔG = 0, then the reaction is said to be at equilibrium

The standard Gibbs free energy change (ΔG°) is related to the standard voltage (E°) by the equation:

$\begin{array}{l}{\text{ΔG}}^{\text{0}}{\text{ = -nFE}}^0\text{ -------(2)}\\ \text{where n = number of electrons, F = Faraday constant (96500 C)}\end{array}$

Answer to Problem 100QAP

The standard free energy change for all three cells is: ΔG° = -312.66 kJ

Explanation of Solution

1. The given cell notation is:

${\text{Tl | Tl}}^{\text{+}}{\text{|| Tl}}^{\text{3+}}{\text{, Tl}}^{\text{+}}\text{ | Pt }$

The half reactions are:

$\begin{array}{l}\text{Anode (oxidation): 2Tl(s) }\to {\text{ 2Tl}}^{\text{+}}{\text{(aq) + 2e}}^{\text{-}}{\text{ E}}^0=-0.34\text{ V}\\ \text{Cathode (reduction)}:{\text{ Tl}}^{\text{3+}}{\text{(aq) + 2e}}^{\text{-}}\to {\text{Tl}}^{\text{+}}{\text{(aq) E}}^0\text{ = 1}\text{.28 V }\\ \text{--------------------------------------------------------------- }\\ {\text{Overall Reaction: 2Tl(s) + Tl}}^{\text{3+}}\text{(aq) }\to {\text{ 3Tl}}^{\text{+}}\text{(aq) }\\ \text{------------------------------------------------------------------}\\ {\text{E}}^{\text{0}}{\text{ = E}}_{\text{cathode}}^{\text{0}}{\text{- E}}_{\text{anode}}^{\text{0}}=\text{ 1}\text{.28 - (-0}\text{.34) = 1}\text{.62 V}\\ {\text{ΔG}}^{\text{0}}{\text{ = -nFE}}^0=-2(96500)(1.62)=-312.66\text{ kJ}\end{array}$

2) The given cell notation is:

${\text{Tl | Tl}}^{\text{3+}}{\text{|| Tl}}^{\text{3+}}{\text{, Tl}}^{\text{+}}\text{ | Pt}$

The half reactions are:

$\begin{array}{l}\text{Anode (oxidation): 2Tl(s) }\to {\text{ 2Tl}}^{\text{3+}}{\text{(aq) + 6e}}^{\text{-}}{\text{ E}}^0=0.74\text{ V}\\ \text{Cathode (reduction)}:{\text{ 3Tl}}^{\text{3+}}{\text{(aq) + 6e}}^{\text{-}}\to 3{\text{Tl}}^{\text{+}}{\text{(aq) E}}^0\text{ = 1}\text{.28 V }\\ \text{--------------------------------------------------------------- }\end{array}$

$\begin{array}{l}{\text{Overall Reaction: 2Tl(s) + Tl}}^{\text{3+}}\text{(aq) }\to {\text{ 3Tl}}^{\text{+}}\text{(aq) }\\ \text{------------------------------------------------------------------}\\ {\text{E}}^{\text{0}}{\text{ = E}}_{\text{cathode}}^{\text{0}}{\text{- E}}_{\text{anode}}^{\text{0}}=\text{ 1}\text{.28 - (0}\text{.74) = 0}\text{.54 V}\\ {\text{ΔG}}^{\text{0}}{\text{ = -nFE}}^0=-6(96500)(0.54)=-312.66\text{ kJ}\end{array}$

3) The given cell notation is:

${\text{Tl | Tl}}^{\text{+}}{\text{|| Tl}}^{\text{3+}}\text{, Tl}$

The half reactions are:

$\begin{array}{l}\text{Anode (oxidation): 3Tl(s) }\to {\text{ 3Tl}}^{\text{+}}{\text{(aq) + 3e}}^{\text{-}}{\text{ E}}^0=-0.34\text{ V}\\ \text{Cathode (reduction)}:{\text{ Tl}}^{\text{3+}}{\text{(aq) + 3e}}^{\text{-}}\to {\text{Tl(s) E}}^0\text{ = +0}\text{.74 V }\\ \text{--------------------------------------------------------------- }\\ {\text{Overall Reaction: 2Tl(s) + Tl}}^{\text{3+}}\text{(aq) }\to {\text{ 3Tl}}^{\text{+}}\text{(aq) }\\ \text{------------------------------------------------------------------}\\ {\text{E}}^{\text{0}}{\text{ = E}}_{\text{cathode}}^{\text{0}}{\text{- E}}_{\text{anode}}^{\text{0}}=\text{ 0}\text{.74 - (-0}\text{.34) = 1}\text{.08 V}\\ {\text{ΔG}}^{\text{0}}{\text{ = -nFE}}^0=-3(96500)(1.08)=-312.66\text{ kJ}\end{array}$

Interpretation Introduction

(d)

Interpretation:

An explanation needs to be provided substantiating whether ΔG° and ?E° are state properties

Concept introduction:

In thermodynamics, a ‘state property/function’ is a quantity which is independent of the path taken in order to reach the final value. In contrast, a ‘path property/function’ depends on the path taken to reach the final state.

Answer to Problem 100QAP

ΔG°: State property

E° : Path property

Explanation of Solution

For the given three cells, the ΔG° remains unchanged at -316.22 kJ. However, the standard voltage (E°) is different for each cell. Hence, ΔG° is a state property whereas; E° is a path property.