For given half cell, `Al^(+3) + 3e^(-) rarr Al`, on increasing `[Al^(+3)]` the electrode potential

Options

A. Increases
B. Decreases
C. No change
D. First increases then decreases

Answer

To solve the question regarding the effect of increasing the concentration of \( \text{Al}^{3+} \) on the electrode potential of the half-cell reaction \( \text{Al}^{3+} + 3e^- \rightarrow \text{Al} \), we can follow these steps:

### Step 1: Write the Half-Cell Reaction
The half-cell reaction is given as:
\[ \text{Al}^{3+} + 3e^- \rightarrow \text{Al} \]

### Step 2: Understand the Nernst Equation
The electrode potential \( E \) of a half-cell can be calculated using the Nernst equation:
\[ E = E^\circ - \frac{0.059}{n} \log \left( \frac{[\text{products}]}{[\text{reactants}]} \right) \]
Where:
- \( E^\circ \) is the standard electrode potential.
- \( n \) is the number of electrons transferred in the half-reaction (which is 3 for aluminum).
- The concentration of solid aluminum (Al) is considered to be 1 since it is a pure solid.

### Step 3: Substitute Values into the Nernst Equation
For our reaction, we can express the Nernst equation as:
\[ E = E^\circ - \frac{0.059}{3} \log \left( \frac{1}{[\text{Al}^{3+}]} \right) \]
This simplifies to:
\[ E = E^\circ + \frac{0.059}{3} \log [\text{Al}^{3+}] \]

### Step 4: Analyze the Effect of Increasing \([\text{Al}^{3+}]\)
If we increase the concentration of \( \text{Al}^{3+} \):
- The term \( \log [\text{Al}^{3+}] \) will increase.
- Since \( E \) is directly proportional to \( \log [\text{Al}^{3+}] \), an increase in the concentration of \( \text{Al}^{3+} \) will lead to an increase in the electrode potential \( E \).

### Conclusion
Thus, we conclude that increasing the concentration of \( \text{Al}^{3+} \) will result in an increase in the electrode potential.

### Final Answer
The electrode potential will **increase** when the concentration of \( \text{Al}^{3+} \) is increased.

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