Question

We've said that the \(+1\) oxidation state is uncommon for indium but is the most stable state for thallium. Verify this statement by calculating \(E^{\circ}\) and \(\Delta G^{\circ}\) (in kilojoules) for the disproportionation reaction $$ 3 \mathrm{M}^{+}(a q) \longrightarrow \mathrm{M}^{3+}(a q)+2 \mathrm{M}(s) \quad \mathrm{M}=\mathrm{In} \text { or } \mathrm{Tl} $$ Is disproportionation a spontaneous reaction for \(\mathrm{In}^{+}\) and \(/\) or \(\mathrm{TI}^{+} ?\) Standard reduction potentials for the relevant half- reactions are $$ \begin{array}{ll} \mathrm{In}^{3+}(a q)+2 \mathrm{e}^{-} \longrightarrow \operatorname{In}^{+}(a q) & E^{\circ}=-0.44 \mathrm{~V} \\ \mathrm{In}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \operatorname{In}(s) & E^{\circ}=-0.14 \mathrm{~V} \\ \mathrm{Tl}^{3+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{T}^{+}(a q) & E^{\circ}=+1.25 \mathrm{~V} \\ \mathrm{TI}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Tl}(s) & E^{\circ}=-0.34 \mathrm{~V} \end{array} $$

Short answer

Disproportionation is spontaneous for In but not for Tl.

Step-by-step solution

Write Disproportionation Reactions

The general disproportionation reaction is given as \[ 3 \mathrm{M}^{+}(a q) \rightarrow \mathrm{M}^{3+}(a q)+2 \mathrm{M}(s) \]For M = In or Tl. Both ions will follow this stoichiometry, where one M becomes M³⁺ and two reduce to M.

Calculate Cell Potential for Indium

To find the cell potential (E^{\circ}) for indium disproportionation, calculate using the standard reduction potentials:\[E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} \]For In:- Anode (oxidation of In^{+1} to In^{3+}): E^{\circ} = -0.44 \, \text{V}- Cathode (reduction of In^{+1} to In(s)): E^{\circ} = -0.14 \, \text{V}\[ E^{\circ}_{\text{cell(In)}} = (-0.14) - (-0.44) = 0.30 \, \text{V} \]

Calculate Cell Potential for Thallium

Similar steps for thallium:- Anode (oxidation ofTl^{+1} to Tl^{3+}): E^{\circ} = +1.25 \, \text{V}- Cathode (reduction of Tl^{+1} to Tl(s)): E^{\circ} = -0.34 \, \text{V}\[ E^{\circ}_{\text{cell(Tl)}} = (-0.34) - (1.25) = -1.59 \, \text{V} \]

Calculate \(\Delta G^{\circ}\) for Indium

The relation between the Gibbs free energy change and cell potential is given by:\[\Delta G^{\circ} = -nFE^{\circ}_{\text{cell}}\]For In:- \( n = 2 \)- \( F = 96,485 \text{ C/mol e}^- \)\[\Delta G^{\circ}_{\text{In}} = -2 \times 96,485 \times 0.30 = -57.89 \text{ kJ/mol}\]

Calculate \(\Delta G^{\circ}\) for Thallium

For Tl:- \( n = 2 \)\[\Delta G^{\circ}_{\text{Tl}} = -2 \times 96,485 \times (-1.59) = 306.26 \text{ kJ/mol}\]

Conclusion: Spontaneity of Reactions

A negative \(\Delta G^{\circ}\) indicates a spontaneous reaction. For indium, \(\Delta G^{\circ} \approx -57.89\, \text{kJ/mol}\) indicates spontaneity, whereas for thallium, \(\Delta G^{\circ} \approx 306.26\, \text{kJ/mol}\) suggests non-spontaneity.

Key concepts

Disproportionation Reaction

A disproportionation reaction is a fascinating type of redox reaction where a single substance is simultaneously oxidized and reduced, resulting in the production of two different products. This type of reaction is unique because it involves a single element undergoing both a decrease and an increase in oxidation state.
In the provided exercise, the reaction involves the conversion of a metal ion in the +1 oxidation state to metallic form (0 oxidation state) and +3 oxidation state, showcasing both reduction and oxidation.

• For Indium (In extsuperscript{+}), the metal in a +1 state is partially reduced to the metallic indium (In), which has an oxidation state of 0.
• Simultaneously, the remaining Indium (In extsuperscript{+}) is oxidized to In extsuperscript{3+}.

This dual change underscores the nature of disproportionation reactions, where single species act both as reducing and oxidizing agents.

Standard Reduction Potentials

Standard reduction potentials ( E^ ext{°} ) are crucial in determining the direction of redox reactions.
These values describe the tendency of a chemical species to gain electrons, serving as a measure of its oxidizing power. A positive E^ ext{°} indicates stronger oxidizing abilities, while a negative value suggests a preference for oxidation.

• For Indium, the standard potentials are -0.44 V (In extsuperscript{3+} to In extsuperscript{+}) and -0.14 V (In extsuperscript{+} to In).
• For Thallium, they are +1.25 V (Tl extsuperscript{3+} to Tl extsuperscript{1+}) and -0.34 V (Tl extsuperscript{+} to Tl).

Understanding these values helps predict the likelihood and direction of electron flow in reactions, affecting the calculated cell potential.

Gibbs Free Energy

Gibbs free energy (\Delta G^° ) connects thermodynamics with electrochemistry, helping predict spontaneity in reactions. The formula linking \Delta G^° with cell potential (E^ ext{°}_ ext{cell} ) is:\[ \Delta G^° = -nFE^ ext{°}_\text{cell} \]where:

• \( n \) is the number of moles of electrons
• \( F \) is Faraday's constant (96,485 C/mol)

For the exercise:
Indium's \Delta G^° of -57.89 kJ/mol indicates the reaction is spontaneous, as negative \Delta G^° denotes spontaneity.
Thallium’s positive \Delta G^° of 306.26 kJ/mol denotes non-spontaneity, meaning energy is required for the reaction to occur.

Cell Potential

The cell potential (E^ ext{°}_ ext{cell} ) is a measure of the electromotive force in electrochemical reactions. It determines if and how a reaction might proceed.
This potential is calculated by taking the difference between the reduction potential of the cathode and the oxidation potential of the anode:\[ E^ ext{°}_\text{cell} = E^ ext{°}_\text{cathode} - E^ ext{°}_\text{anode} \]

• For Indium, the calculated cell potential is 0.30 V, substantiating a spontaneous process.
• For Thallium, a cell potential of -1.59 V indicates the reaction is not naturally favored.

The sign and magnitude of E^ ext{°}_ ext{cell} offer insights into reaction feasibility and energy conversion efficiency.