Question

A standard galvanic cell is constructed so that the overall cell reaction is $$2 \mathrm{Al}^{3+}(a q)+3 \mathrm{M}(s) \longrightarrow 3 \mathrm{M}^{2+}(a q)+2 \mathrm{Al}(s)$$ where \(\mathrm{M}\) is an unknown metal. If \(\Delta G^{\circ}=-411 \mathrm{~kJ}\) for the overall cell reaction, identify the metal used to construct the standard cell.

Short answer

The unknown metal M used to construct the standard cell is Cadmium (Cd). This is determined by calculating the number of moles of electrons transferred in the given reaction, finding the standard cell potential (E°), and then using the relationship between Gibbs free energy and cell potential to identify the metal from the Standard Reduction Potentials table. The calculated E° value was 0.35 V, and adding it to the E° of the Al³⁺/Al half-cell (about -1.66 V) gave an E° for the M²⁺/M half-cell of 2.01 V. Comparing this value to the standard reduction potentials, the Cd²⁺/Cd half-cell has a standard reduction potential of 2.02 V, which is the closest match to our calculated value.

Step-by-step solution

Write down the formula connecting ∆G°, E°, and n

The relationship between the standard change in Gibbs free energy (∆G°), the standard cell potential (E°), and the number of moles of electrons transferred (n) in a redox reaction is given by the formula: \[\Delta G^\circ = -nFE^\circ\] Where F is the Faraday's constant, approximately equal to \(96485\,\text{C/mol}\).

Calculate the number of moles of electrons transferred (n)

In the given cell reaction, each Al³⁺ ion gains 3 electrons to form Al, and each M ion loses 2 electrons to form M²⁺. There are two Al³⁺ ions and three M atoms in the balanced reaction, so the total number of electrons transferred is: \(n = 2 \times 3\text{ electrons for Al³⁺} + 3 \times 2\text{ electrons for M} = 6 + 6 = 12\)

Calculate the standard cell potential (E°)

Now, we will use the given value of ∆G° and the calculated value of n to find the standard cell potential (E°) using the formula from Step 1. Given ∆G° = -411 kJ, convert it to J by multiplying by 1000, which gives us -411000 J. Then, plug in the values for ∆G°, n, and F: \[-411000\,\text{J} = -12\,\text{mol} \times 96485 \,\text{C/mol} \times E^\circ\] Now solve for E°: \[E^\circ = \frac{-411000}{(-12)(96485)} \approx 0.35\,\text{V}\]

Identify metal M using the standard reduction potentials table

Refer to the Standard Reduction Potentials table and find an element whose half-cell reaction E° value, when added to the E° value for the reduction of Al³⁺ to Al, equals the calculated E° value in Step 3. The standard reduction potential for the Al³⁺/Al half-cell is about -1.66 V. Since E° for the overall cell reaction is positive, the metal M must have a higher standard reduction potential than Al³⁺/Al. We can find the E° for M²⁺/M half-cell by adding the E° of Al³⁺/Al half-cell to the overall cell potential: \[E^\circ_{M^{2+}/M} = 0.35\,\text{V} - (-1.66\,\text{V}) = 2.01\,\text{V}\] Now, go through the Standard Reduction Potentials table and look for a half-reaction with a standard reduction potential close to 2.01 V. Cd²⁺/Cd half-cell has a standard reduction potential of 2.02 V, which is very close to the calculated value of 2.01 V. Therefore, the unknown metal M used to construct the standard cell is Cadmium (Cd).

Key concepts

Standard Cell Potential

The concept of Standard Cell Potential, represented as \(E^\circ\), refers to the electromotive force or voltage of a galvanic cell under standard conditions, which include 1 M concentration for solutions, a pressure of 1 atm for gases, and a temperature of 25°C (298 K). This potential essentially measures the tendency of the cell reaction to occur spontaneously.

When analyzing a galvanic cell, the standard cell potential is determined by the combination of the potentials of the two half-reactions involved in the cell. One half-reaction focuses on the reduction event at the cathode, while the other focuses on the oxidation event at the anode.

The cell potential is calculated by the difference between the cathode and anode potentials (i.e., the reduction half-cell potential minus the oxidation half-cell potential). When the overall cell reaction has a positive \(E^\circ\), it indicates a spontaneous reaction under standard conditions.

In our specific case, the \(E^\circ\) value helps us to deduce the potential because it tells us that the combination of the known aluminum half-cell and the unknown metal half-cell leads to a potential of 0.35 V. This information aids in identifying the unknown metal based on its standard reduction potential.

Standard Reduction Potential

The Standard Reduction Potential is a crucial part of understanding the electrochemical reactions that occur in a galvanic cell. It represents the tendency of a chemical species to gain electrons and be reduced, measured under standard conditions.

It is denoted by \(E^\circ_{red}\) and is compared to the standard hydrogen electrode, which is arbitrarily assigned a potential of 0 V. The table of standard reduction potentials lists these values for various half-reactions, allowing us to predict the direction of electron flow in electrochemical cells.

In the given exercise, we use the principle that the overall cell potential \(E^\circ\) is derived from balancing the standard reduction potentials of two half-reactions. Specifically, it is crucial to identify the half-reaction of the unknown metal alloy \(M\), which made it necessary to refer to the standard reduction potentials table.

The real power of using these standard potential tables lies in predicting which species is more likely to be reduced or oxidized. Metals with higher \(E^\circ_{red}\) values are more likely to be reduced and act as a cathode, whereas metals with lower values tend to become anodes in electrochemical reactions.

Gibbs Free Energy

Gibbs Free Energy, represented as \(\Delta G^\circ\), is a thermodynamic quantity that provides critical insights on the spontaneity of a reaction under standard conditions. A negative value of \(\Delta G^\circ\) suggests that a reaction is spontaneous and energetically favorable.

The relationship between \(\Delta G^\circ\) and the standard cell potential \(E^\circ\) is derived from the equation: \[\Delta G^\circ = -nFE^\circ\]In this equation, \(n\) is the number of moles of electrons transferred in the reaction, and \(F\) is the Faraday constant (approximately 96485 C/mol).

For the problem in question, the given value of \(\Delta G^\circ\) was -411 kJ, which needed conversion to Joules for calculations (i.e., -411000 J). Understanding the interplay between Gibbs Free Energy and the standard cell potential helps us identify the feasibility and direction of the redox processes occurring in the cell.

Moreover, calculating \(E^\circ\) from \(\Delta G^\circ\) provides essential information about the cell's voltage, which is crucial for determining the reactivity and identifying unknown metals in electrochemical reactions.