Consider the following standard reduction potentials: $$ \begin{array}{ll} \mathrm{Tl}^{+}(a q)+e^{-} \longrightarrow \mathrm{Tl}(s) & E_{\mathrm{red}}^{o}=-0.34 \mathrm{~V} \\ \mathrm{~T}^{3+}(a q)+3 e^{-} \longrightarrow \mathrm{T} 1(s) & E_{\mathrm{red}}^{\circ}=0.74 \mathrm{~V} \\ \mathrm{~T}^{3+}(a q)+2 e^{-} \longrightarrow \mathrm{T} 1^{+}(a q) & E_{\mathrm{red}}^{\circ}=1.28 \mathrm{~V} \end{array} $$ and the following abbreviated cell notations: $$ \text { (1) } \left.\mathrm{T} 1 \mid \mathrm{T}]^{+} \| \mathrm{T}^{3+}, \mathrm{T}\right]^{+} \mid \mathrm{Pt} $$ $$ \begin{aligned} &\text { (2) } \mathrm{Tl}\left|\mathrm{T}^{3+} \| \mathrm{T}^{3+}, \mathrm{Tl}^{+}\right| \mathrm{Pt} \\ &\text { (3) } \mathrm{Tl}\left|\mathrm{Tl}^{+} \| \mathrm{T}^{3+}\right| \mathrm{Tl} \end{aligned} $$ (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: \(\mathrm{A}\) state property is path-independent.)

Step by step solution

01

Cell 1

For the first cell: \(\mathrm{Tl} \mid \mathrm{Tl}^+ || \mathrm{T}^{3+},\mathrm{Tl}^{+} \mid \mathrm{Pt}\) The half reactions are: $$Tl^+(aq)+e^-\rightarrow Tl(s)$$ $$T^{3+}(aq)+2e^-\rightarrow Tl^+(aq)$$ We need to balance the number of electrons, so we multiply the first half-cell reaction by 2: $$2Tl^+(aq)+2e^-\rightarrow 2Tl(s)$$ Now, we can add the half-cell reactions to obtain the overall reaction: $$2Tl^+(aq)+2e^-+T^{3+}(aq)+2e^-\rightarrow 2Tl(s)+Tl^+(aq)$$

02

Cell 2

For the second cell: \(\mathrm{Tl}\left|\mathrm{T}^{3+} || \mathrm{T}^{3+}, \mathrm{Tl}^{+}\right| \mathrm{Pt}\) The half reactions are: $$Tl^+(aq)+e^-\rightarrow Tl(s)$$ $$T^{3+}(aq)+3e^-\rightarrow T(s)$$ We need to balance the number of electrons, so we multiply the second half-cell reaction by 3: $$3Tl^+(aq)+3e^-\rightarrow 3Tl(s)$$ Now, we can add the half-cell reactions to obtain the overall reaction: $$3Tl^+(aq)+3e^-+T^{3+}(aq)+3e^-\rightarrow 3Tl(s)+T(s)$$

03

Cell 3

For the third cell: \(\mathrm{Tl}\left|\mathrm{Tl}^{+} || \mathrm{T}^{3+}\right| \mathrm{Tl}\) The half reactions are: $$Tl^+(aq)+e^-\rightarrow Tl(s)$$ $$T^{3+}(aq)+3e^-\rightarrow T(s)$$ Now, we can add the half-cell reactions to obtain the overall reaction: $$Tl^+(aq)+e^-+T^{3+}(aq)+3e^-\rightarrow Tl(s)+T(s)$$ #b) Calculating \(E^{\circ}\) for each cell#

04

Cell 1

In Cell 1, we have the following half-cell voltages: \(E_{red}^{o}=-0.34\) V for \(Tl^+(aq)+e^-\rightarrow Tl(s)\) \(E_{red}^{\circ}=1.28\) V for \(T^{3+}(aq)+2e^-\rightarrow Tl^+(aq)\) To find the standard potential for this cell, we subtract the reducing agent's potential from the oxidizing agent's potential: $$E_{cell}^\circ=E_{red}^{\circ}(oxidizing)-E_{red}^{\circ}(reducing)=1.28-(-0.34)=1.62\,\text{V}$$

05

Cell 2

In Cell 2, we have the following half-cell voltages: \(E_{red}^{o}=-0.34\) V for \(Tl^+(aq)+e^-\rightarrow Tl(s)\) \(E_{red}^{\circ}=0.74\) V for \(T^{3+}(aq)+3e^-\rightarrow T(s)\) To find the standard potential for this cell, we subtract the reducing agent's potential from the oxidizing agent's potential: $$E_{cell}^\circ=E_{red}^{\circ}(oxidizing)-E_{red}^{\circ}(reducing)=0.74-(-0.34)=1.08\,\text{V}$$

06

Cell 3

In Cell 3, we have the following half-cell voltages: \(E_{red}^{o}=-0.34\) V for \(Tl^+(aq)+e^-\rightarrow Tl(s)\) \(E_{red}^{\circ}=0.74\) V for \(T^{3+}(aq)+3e^-\rightarrow T(s)\) To find the standard potential for this cell, we subtract the reducing agent's potential from the oxidizing agent's potential: $$E_{cell}^\circ=E_{red}^{\circ}(oxidizing)-E_{red}^{\circ}(reducing)=0.74-(-0.34)=1.08\,\text{V}$$ #c) Calculating \(\Delta G^{\circ}\) for each overall equation# In each case, we can use the formula: \(\Delta G^{\circ}=-nFE_{cell}^{\circ}\), where n is the number of electrons transferred and F is the Faraday constant (96485 C/mol).

07

Cell 1

For Cell 1, n=2: $$\Delta G^{\circ}=-2(96485\,\text{C/mol})\times (1.62\,\text{V})=-313036.2\,\text{J/mol}$$

08

Cell 2

For Cell 2, n=3: $$\Delta G^{\circ}=-3(96485\,\text{C/mol})\times (1.08\,\text{V})=-312604.58\,\text{J/mol}$$

09

Cell 3

For Cell 3, n=4: $$\Delta G^{\circ}=-4(96485\,\text{C/mol})\times (1.08\,\text{V})=-416806.08\,\text{J/mol}$$ #d) Comment on whether \(\Delta G^{\circ}\) and/or \(E^{\circ}\) are state properties# State properties are path-independent; that is, their values depend only on the initial and final states, not on the specific path taken. \(\Delta G^{\circ}\) is a state property, meaning that it depends only on the equilibrium state of the chemical reaction, regardless of the order in which the intermediates are formed. However, \(E^{\circ}\) is not a state property, as it depends on the pathway in which electrons are transferred between the redox species.

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