Question

A voltaic cell is made up of a silver electrode in a \(1 \mathrm{M}\) silver nitrate solution and an aluminum electrode in a \(1 \mathrm{M}\) aluminum nitrate solution. The half reactions are 1) \(\mathrm{Ag}^{+}+\mathrm{e}^{-} \rightarrow \mathrm{Ag}\) (s) \(\mathrm{E}^{\circ}=.80\) volt, and (2) \(\mathrm{Al}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Al}(\mathrm{s})\) \(E^{\circ}=-1.66\) volt, calculate the standard free energy change, \(\Delta \mathrm{G}^{\circ}\), in \(\mathrm{Kcal} /\) mole. \(1 \mathrm{cal}=4.184\) joule \(1 \mathrm{~F}=96,500\) coul, \(1 \mathrm{Kcal}=1000 \mathrm{cal}\)

Short answer

The standard free energy change for the voltaic cell, \(\Delta G^\circ\), is \(-340.2 \mathrm{Kcal/mol}\).

Step-by-step solution

Combine the half-reactions

Write down the balanced overall redox reaction by combining the two half-reactions: \[ \mathrm{2 Al + 3 Ag^+ \rightarrow 2 Al^{3+} + 3 Ag} \]

Calculate the standard cell potential

Calculate the overall standard cell potential, \(E^\circ_{cell}\): \[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \] In our case, the reduction of silver ions is the cathode half-reaction, and the oxidation of aluminum is the anode half-reaction. \[ E^\circ_{cell} = (0.80 \, \mathrm{V}) - (-1.66 \, \mathrm{V}) \] \[ E^\circ_{cell} = 2.46 \, \mathrm{V} \]

Calculate the standard free energy change

Use the relationship between \(\Delta G^\circ\) and \(E^\circ_{cell}\) to calculate the standard free energy change: \[ \Delta G^\circ = - nFE^\circ_{cell} \] In the balanced redox reaction, the number of electrons exchanged, \(n\), is 6. \[ \Delta G^\circ = - (6) (96,500 \, \mathrm{C/mol}) (2.46 \, \mathrm{V}) \] \[ \Delta G^\circ = - 1,424,760 \, \mathrm{J/mol} \]

Convert the free energy change to Kcal/mol

Now, we must convert the free energy change from Joules/mol to Kcal/mol: \[ \Delta G^\circ = -\frac{1,424,760 \, \mathrm{J/mol}}{(4.184 \, \mathrm{J/cal} \times 1000 \, \mathrm{cal/Kcal})} \] \[ \Delta G^\circ = -340.2 \, \mathrm{Kcal/mol} \] Thus, the standard free energy change for the voltaic cell is \(-340.2 \mathrm{Kcal/mol}\).

Key concepts

Understanding Standard Cell Potential

The standard cell potential, symbolized as \( E^\circ_{cell} \), is a crucial factor in determining the voltage of a voltaic cell. Voltaic cells consist of two half-cells, each containing a different metal electrode immersed in a solution of its ions. The standard cell potential is calculated using the potential of the cathode and the anode:

• The cathode, where reduction occurs, typically has a positive potential.
• The anode, where oxidation occurs, typically has a negative potential.

In terms of calculation, the formula used is:
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \]
This formula helps us find the maximum potential difference across a cell when operating under standard conditions, like in this case where the standard cell potential calculated was 2.46 V. This reflects the cell's ability to do work and indicates the spontaneity of the redox reaction involved.

The Role of Redox Reactions

A redox reaction, short for reduction-oxidation reaction, is one where electrons are transferred between two substances. In a voltaic cell, these reactions allow for the flow of electricity:

• Oxidation occurs at the anode, where electrons are released.
• Reduction occurs at the cathode, where electrons are accepted.

The overall reaction in a voltaic cell includes both the oxidation and reduction half-reactions. For example, in the case of the silver and aluminum electrodes, silver ions are reduced at the cathode, and aluminum is oxidized at the anode, creating a complete circuit that generates electricity. Balancing these half-reactions is essential, ensuring the overall charge and the number of atoms on each side of the equation are balanced.

Understanding Free Energy Change

The concept of free energy change, symbolized as \( \Delta G^\circ \), connects redox reactions in voltaic cells to thermodynamic principles. It provides insight into the spontaneity of a reaction:

• A negative \( \Delta G^\circ \) indicates a spontaneous reaction.
• The formula relating \( \Delta G^\circ \) to \( E^\circ_{cell} \) is \( \Delta G^\circ = - nFE^\circ_{cell} \).

Here, \( n \) is the number of moles of electrons exchanged, and \( F \) is the Faraday constant. In the context of the silver-aluminum voltaic cell, the calculation revealed a \( \Delta G^\circ \) of -340.2 Kcal/mol, indicating a spontaneous process. Converting Joules to Kcal is essential in providing a more convenient energy unit, which allows easier comparison in thermodynamic calculations.

Understanding Half-Reactions

Half-reactions are a vital concept in electrochemistry, representing the separate oxidation and reduction processes occurring in a redox reaction. Each half-reaction shows either the loss or gain of electrons, distinctively:

• A reduction half-reaction includes a gain of electrons and is represented by a cathode reaction. For example, \( \mathrm{Ag}^+ + \mathrm{e}^- \rightarrow \mathrm{Ag} \).
• An oxidation half-reaction includes a loss of electrons and is represented by an anode reaction. For instance, \( \mathrm{Al}^{3+} + 3\mathrm{e}^- \rightarrow \mathrm{Al(s)} \).

Each half-reaction is associated with a standard electrode potential, which helps determine the direction of electron flow when combined in the overall cell equation. Balancing such reactions ensures accurate stoichiometry and charge neutrality, like in the voltaic cell combining silver and aluminum electrodes.