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: Understand the Reaction
The given half-cell reaction involves the reduction of aluminum ions (\( \text{Al}^{3+} \)) to solid aluminum (\( \text{Al} \)). This process involves the gain of three electrons.

### Step 2: Write the Nernst Equation
The Nernst equation relates the electrode potential of a half-cell to the concentrations of the reactants and products. The general form of the Nernst equation is:

\[ E = E^\circ - \frac{0.0591}{n} \log \left( \frac{[\text{products}]}{[\text{reactants}]} \right) \]

For our half-cell reaction, the Nernst equation becomes:

\[ E = E^\circ - \frac{0.0591}{3} \log \left( \frac{[\text{Al}]}{[\text{Al}^{3+}]} \right) \]

### Step 3: Consider the Activity of Solid Aluminum
Since aluminum is in solid form, its activity is considered to be 1. Therefore, we can simplify the equation:

\[ E = E^\circ - \frac{0.0591}{3} \log \left( \frac{1}{[\text{Al}^{3+}]} \right) \]

This can be rewritten as:

\[ E = E^\circ + \frac{0.0591}{3} \log \left( [\text{Al}^{3+}] \right) \]

### Step 4: Analyze the Effect of Increasing \([\text{Al}^{3+}]\)
From the modified Nernst equation, we can see that the electrode potential \( E \) is directly proportional to the logarithm of the concentration of \( \text{Al}^{3+} \). Therefore, if we increase the concentration of \( \text{Al}^{3+} \), the term \( \log \left( [\text{Al}^{3+}] \right) \) will increase, leading to an increase in the electrode potential \( E \).

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