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Q2.12(i):
How much charge is required for the following reduction: 1 mol of \(Al^{3+}\) to Al?

Solution :

Step 1: Establishing the Half-Reaction

To determine the charge required for the reduction of the aluminum ion ($Al^{3+}$) to metallic aluminum ($Al$), we must first write the balanced half-cell reduction equation. The reduction of a metal ion involves the gain of electrons to reach a neutral oxidation state.

The reduction half-reaction is defined as:

$Al^{3+}(aq) + 3e^{-} \rightarrow Al(s)$

[Per the definition of reduction as the gain of electrons, the oxidation state decreases from $+3$ to $0$.]

Step 2: Stoichiometric Analysis of Electrons

From the balanced chemical equation, it is evident that for the reduction of $1$ mole of $Al^{3+}$ ions, exactly $3$ moles of electrons are required.

  • Stoichiometric coefficient of $Al^{3+}$ = $1$ mole
  • Stoichiometric coefficient of $e^{-}$ = $3$ moles

Step 3: Calculating Total Charge ($Q$)

The total charge ($Q$) required for a redox reaction is calculated using the relationship between the number of moles of electrons ($n$) and the Faraday constant ($F$).

The Faraday constant ($F$) represents the charge of $1$ mole of electrons, which is approximately $96,487\ C \cdot mol^{-1}$. For standard academic calculations, we use $F \approx 96,500\ C \cdot mol^{-1}$.

The formula for total charge is:

$Q = n \times F$

Substituting the known values:

$n = 3\ mol\ e^{-}$

$F = 96,500\ C \cdot mol^{-1}$

$Q = 3\ mol \times 96,500\ C \cdot mol^{-1}$

$Q = 289,500\ C$

Step 4: Schematic Representation of Electron Transfer

Al³⁺ + 3e⁻ Al Charge Required: 3F = 289,500 C

Final Solution: The charge required for the reduction of $1$ mole of $Al^{3+}$ to $Al$ is $289,500\ C$.


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