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: Understanding the Concept:
The given reaction is an oxidation half-reaction. Its standard oxidation potential (\(\text{E}^\circ_{\text{ox}}\)) is equal in magnitude but opposite in sign to the standard reduction potential (\(\text{E}^\circ_{\text{red}}\)).
Step 2: Key Formula or Approach:
Nernst Equation for the oxidation half-reaction:
\[ \text{E}_{\text{ox}} = \text{E}^\circ_{\text{ox}} - \frac{0.0592}{n} \log [\text{Al}^{3+}] \] where \(\text{E}^\circ_{\text{ox}} = -\text{E}^\circ_{\text{red}} = -(-1.66 \text{ V}) = +1.66 \text{ V}\).
Step 3: Detailed Explanation:
1. \(n = 3\) (number of electrons involved).
2. \([\text{Al}^{3+}] = 0.1 \text{ M} = 10^{-1} \text{ M}\).
3. Plug values into the equation:
\[ \text{E}_{\text{ox}} = 1.66 - \frac{0.0592}{3} \log(10^{-1}) \] \[ \text{E}_{\text{ox}} = 1.66 - 0.01973 \times (-1) \] \[ \text{E}_{\text{ox}} = 1.66 + 0.0197 = 1.6797 \text{ V} \] Step 4: Final Answer:
The potential is \(+1.679 \text{ V}\).