Question

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

Short answer

Answer: Yes, both \(E^\circ\) and \(\Delta G^\circ\) are state properties. They are path-independent and depend only on the initial and final states, not the process that changes the state.

Step-by-step solution

Write the overall equation for each cell

For each cell notation, we write the corresponding half-cell reactions, and then sum up the reactions to get the overall cell reactions. Cell (1): Half-cell reactions: $$\mathrm{Tl}^{+}(a q)+e^{-} \to \mathrm{Tl}(s)$$ $$\mathrm{Tl}^{3+}(a q)+2 e^{-} \to \mathrm{Tl}^{+}(a q)$$ Overall reaction: $$\mathrm{Tl}^{3+}(a q)+\mathrm{Tl}(s) \to 2\mathrm{Tl}^{+}(a q)$$ Cell (2): Half-cell reactions: $$\mathrm{Tl}^{3+}(a q)+3 e^{-} \to \mathrm{Tl}(s)$$ $$\mathrm{Tl}^{3+}(a q)+2 e^{-} \to \mathrm{Tl}^{+}(a q)$$ Overall reaction: $$2\mathrm{Tl}^{3+}(a q)+\mathrm{Tl}(s) \to 3\mathrm{Tl}^{+}(a q)$$ Cell (3): Half-cell reactions: $$\mathrm{Tl}^{+}(a q)+e^{-} \to \mathrm{Tl}(s)$$ $$\mathrm{Tl}^{3+}(a q)+3 e^{-} \to \mathrm{Tl}(s)$$ Overall reaction: $$\mathrm{Tl}^{3+}(a q)+\mathrm{Tl}^{+}(a q) \to 2\mathrm{Tl}(s)$$

Calculate \(E^\circ\) for each cell

To calculate the standard cell potential, we subtract the standard reduction potential for the half-cell taking place at the left electrode (oxidation half-reaction) from the standard reduction potential for the half-cell taking place at the right electrode (reduction half-reaction). Cell (1): $$E^\circ_{\text{cell}} = E^\circ_{\text{reduction}}-E^\circ_{\text{oxidation}}$$ $$ E^\circ_{\text{cell}} = 1.28 V - (-0.34 V) = 1.62 V$$ Cell (2): $$E^\circ_{\text{cell}} = E^\circ_{\text{reduction}}-E^\circ_{\text{oxidation}}$$ $$ E^\circ_{\text{cell}} = 1.28 V - 0.74 V = 0.54 V$$ Cell (3): $$E^\circ_{\text{cell}} = E^\circ_{\text{reduction}}-E^\circ_{\text{oxidation}}$$ $$ E^\circ_{\text{cell}} = 0.74 V - (-0.34 V) = 1.08 V$$

Calculate \(\Delta G^\circ\) for each overall equation

We will use the formula \(\Delta G^\circ = -nFE^\circ\), where \(n\) is the number of moles of electrons transferred and \(F=96,485 \, \mathrm{C/mol}\) is the Faraday constant. Cell (1): $$\Delta G^\circ = -2 \times 96,485\, \mathrm{C/mol} \times 1.62\, \mathrm{V} = -313,739\, \mathrm{J/mol}$$ Cell (2): $$\Delta G^\circ = -3 \times 96,485\, \mathrm{C/mol} \times 0.54\, \mathrm{V} = -156,533\, \mathrm{J/mol}$$ Cell (3): $$\Delta G^\circ = -3 \times 96,485\, \mathrm{C/mol} \times 1.08\, \mathrm{V} = -313,739\, \mathrm{J/mol}$$

Comment on \(\Delta G^\circ\) and \(E^\circ\) as state properties

A state property is path-independent, meaning its value only depends on the initial and final states, not the process that changes the state. \(E^\circ\) is a state property because it considers only the difference in the standard reduction potentials of the half-cells. The value of \(E^\circ\) depends only on the specific half-cells and their potentials, and it doesn't matter what path the electrons take between the half-cells. \(\Delta G^\circ\) is also a state property, as it is related to the free energy change between the initial and final states of a process. The relationship between \(\Delta G^\circ\) and \(E^\circ\), i.e., \(\Delta G^\circ = -nFE^\circ\), confirms this property, as both quantities are path-independent.

Key concepts

Standard Reduction Potential

The standard reduction potential, denoted as \(E^\circ_{\text{red}}\), is a measure of the tendency of a chemical species to gain electrons and thereby be reduced. In electrochemistry, it serves as a quantitative indication of the ease with which a species can be reduced, with higher positive values indicating a greater affinity for electrons. For instance, in the solution you read about thallium \( \mathrm{Tl}^{3+} \) demonstrating a higher \(E^\circ_{\text{red}}\) than \( \mathrm{Tl}^{+} \), meaning \( \mathrm{Tl}^{3+} \) is more readily reduced.

Standard reduction potentials are measured under specific conditions: a temperature of 298 K, a 1 M concentration for each aqueous species involved, and a pressure of 1 bar for any gases that may be involved. Importantly, these potentials are conventionally referenced against the standard hydrogen electrode, which is assigned a potential of 0 V.

Cell Notation

Cell notation provides a shorthand representation of the setup of an electrochemical cell, which displays the oxidation and reduction half-reactions at the anode and cathode, respectively. In a typical cell notation, the anode (where oxidation occurs) is written on the left, and the cathode (where reduction occurs) is on the right. The phases of the matter are indicated by vertical lines (|), and salt bridges are often represented by a double line (||).

In the context of the exercise, you encountered three abbreviated cell notations for cells involving thallium. The cell notation expresses the anode and cathode materials, electrolytes, and direction of electron flow in a concise manner, crucial for understanding the electrochemical reactions taking place.

Gibbs Free Energy

Gibbs free energy, represented by \(\Delta G\), is defined as the maximum amount of work a thermodynamic system can perform at constant temperature and pressure. In the realm of electrochemistry, Gibbs free energy is directly associated with the spontaneity of a reaction; a negative \(\Delta G\) indicates a spontaneous process. The relationship between Gibbs free energy and standard cell potential \(E^\circ\) is given by the equation \(\Delta G^\circ = -nFE^\circ\), where \(n\) is the number of moles of electrons involved in the reaction, and \(F\) is the Faraday constant, approximately 96,485 C/mol.

In your study, the calculation of \(\Delta G^\circ\) for the given cells will help determine the feasibility and direction of the electrochemical reactions, which are fundamental for harnessing electrical energy from chemical processes.

State Property

A state property, in thermodynamics, is a characteristic of a system that depends only on the current state of the system, and not on the path used to reach that state. Both Gibbs free energy (\(\Delta G\)) and cell potential (\(E^\circ\)) are state properties. They do not concern themselves with the 'route' taken by the system during a reaction, but only with the initial and final states.

Understanding that these quantities are state properties is vital when analyzing chemical reactions, predicting spontaneity or cell voltage, and for calculations in electrochemistry; it ensures consistency and predictability of these values across a range of conditions and processes. From the solution provided, by establishing that both \(E^\circ\) and \(\Delta G^\circ\) are indeed state properties, it underscores their reliability as indicators of electrochemical behavior.