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.

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 \(96485\,\text{C/mol}\).

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: \(n = 2 \times 3\text{ electrons for Al³⁺} + 3 \times 2\text{ electrons for M} = 6 + 6 = 12\)

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°: \[E^\circ = \frac{-411000}{(-12)(96485)} \approx 0.35\,\text{V}\]

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).