Question

If $\text{E}^o \left( \text{Al}_{(\text{eq})}^{+3} \mid \text{Al}_{(\text{s})} \right) = -1.66 \text{ V}$. What is potential of $\text{Al}_{(\text{s})} \longrightarrow \text{Al}^{+3}(0 \cdot 1\text{M}) + 3\text{e}^-$ at 298 K ?

• A. $+1.540 \text{ V}$
• B. $-1.540 \text{ V}$
• C. $+1.679 \text{ V}$
• D. $-1.679 \text{ V}$

Hint:

Oxidation potential $= - (\text{Reduction potential})$. Use the Nernst equation to adjust for non-standard concentrations.

Answer 1

The Correct Option is C

Step 1: Identify the process
The question asks for the oxidation potential ($E_{ox}$).
Given reduction potential $E^\circ_{red} = -1.66 \text{ V}$, so $E^\circ_{ox} = +1.66 \text{ V}$.
Step 2: Nernst Equation
For oxidation $\text{Al} \to \text{Al}^{3+} + 3e^-$:
$E_{ox} = E^\circ_{ox} - \frac{0.059}{n} \log [\text{Al}^{3+}]$
$E_{ox} = 1.66 - \frac{0.059}{3} \log (0.1)$
Step 3: Calculation
$E_{ox} = 1.66 - (0.0197 \times -1)$
$E_{ox} = 1.66 + 0.0197 = 1.6797 \text{ V}$
Final Answer: (C)