`K_(sp)` of `Al(OH)_(3)=10^(-36)`
and `E_(Al^(3+)//Al)^(@)=-1.66V`
Reduction potential of `Al^(3+)//Al` couple at `pH=12` and 298K is

Options

A. 1.07V
B. 2.25V
C. `-1.07V`
D. `-2.25V`

Answer

To find the reduction potential of the Al³⁺/Al couple at pH 12, we will follow these steps:

### Step 1: Determine the concentration of OH⁻ ions
Given that the pH is 12, we can calculate the pOH:
\[ \text{pOH} = 14 - \text{pH} = 14 - 12 = 2 \]
Now, we can find the concentration of OH⁻ ions:
\[ [\text{OH}^-] = 10^{-\text{pOH}} = 10^{-2} \, \text{M} \]

### Step 2: Use the Ksp to find the concentration of Al³⁺ ions
The dissociation of Al(OH)₃ can be represented as:
\[ \text{Al(OH)}_3 \rightleftharpoons \text{Al}^{3+} + 3\text{OH}^- \]
The solubility product constant (Ksp) is given by:
\[ K_{sp} = [\text{Al}^{3+}][\text{OH}^-]^3 \]
Substituting the known values:
\[ 10^{-36} = [\text{Al}^{3+}](10^{-2})^3 \]
This simplifies to:
\[ 10^{-36} = [\text{Al}^{3+}](10^{-6}) \]
Now, solving for \([\text{Al}^{3+}]\):
\[ [\text{Al}^{3+}] = \frac{10^{-36}}{10^{-6}} = 10^{-30} \, \text{M} \]

### Step 3: Calculate the reduction potential using the Nernst equation
The Nernst equation for the reduction potential is given by:
\[ E = E^\circ - \frac{0.059}{n} \log \left( \frac{1}{[\text{Al}^{3+}]} \right) \]
Where:
- \(E^\circ\) (standard reduction potential) for Al³⁺/Al is -1.66 V.
- \(n\) (number of electrons transferred) is 3 for the reaction:
\[ \text{Al}^{3+} + 3e^- \rightarrow \text{Al} \]

Substituting the values into the Nernst equation:
\[ E = -1.66 - \frac{0.059}{3} \log \left( \frac{1}{10^{-30}} \right) \]
This simplifies to:
\[ E = -1.66 - \frac{0.059}{3} \cdot 30 \]
Calculating the logarithm:
\[ \log \left( \frac{1}{10^{-30}} \right) = \log(10^{30}) = 30 \]
Now substituting back into the equation:
\[ E = -1.66 - \frac{0.059 \cdot 30}{3} \]
Calculating:
\[ E = -1.66 - 0.59 \]
Thus:
\[ E = -2.25 \, \text{V} \]

### Final Answer
The reduction potential of the Al³⁺/Al couple at pH 12 is:
\[ \boxed{-2.25 \, \text{V}} \]