The half cell potentials of a halfcell `A^((x+n)+) , A^(x+)|pt` were found to be as follows : `{:(% "of reduced form",24.4,48.8),("Half cell potential (V)",0.101,0.115):}`
Determinwe the value of `n`.

Answer

To determine the value of \( n \) in the given half-cell potentials, we can follow these steps:

### Step 1: Understand the Reduction Reactions
The reduction reactions for the half-cells can be represented as follows:
- For \( A^{(x+n)+} \):
\[ A^{(x+n)+} + n \text{e}^- \rightarrow A^x \]
- For \( A^{(x)+} \):
\[ A^{(x)+} + \text{e}^- \rightarrow A^x \]

### Step 2: Identify the Given Data
From the problem statement, we have:
- Percentage of reduced form:
- \( A^{(x+n)+} \): 24.4%
- \( A^{(x)+} \): 48.8%
- Half-cell potentials:
- \( E(A^{(x+n)+}) = 0.101 \, V \)
- \( E(A^{(x)+}) = 0.115 \, V \)

### Step 3: Calculate the Concentrations
Using the percentage of reduced forms, we can determine the concentrations of the oxidized and reduced forms:
- For \( A^{(x+n)+} \):
- Reduced form concentration = 24.4
- Oxidized form concentration = \( 100 - 24.4 = 75.6 \)

- For \( A^{(x)+} \):
- Reduced form concentration = 48.8
- Oxidized form concentration = \( 100 - 48.8 = 51.2 \)

### Step 4: Apply the Nernst Equation
The Nernst equation is given by:
\[ E = E^0 + \frac{0.059}{n} \log \left( \frac{[\text{Products}]}{[\text{Reactants}]} \right) \]

For the first half-cell:
\[ 0.101 = E^0 + \frac{0.059}{n} \log \left( \frac{75.6}{24.4} \right) \tag{1} \]

For the second half-cell:
\[ 0.115 = E^0 + \frac{0.059}{n} \log \left( \frac{51.2}{48.8} \right) \tag{2} \]

### Step 5: Subtract Equation (1) from Equation (2)
Subtracting equation (1) from equation (2):
\[ 0.115 - 0.101 = \frac{0.059}{n} \left( \log \left( \frac{51.2}{48.8} \right) - \log \left( \frac{75.6}{24.4} \right) \right) \]
\[ 0.014 = \frac{0.059}{n} \left( \log \left( \frac{51.2}{48.8} \right) - \log \left( \frac{75.6}{24.4} \right) \right) \]

### Step 6: Calculate the Logarithmic Values
Calculate the logarithmic values:
- \( \log \left( \frac{51.2}{48.8} \right) \)
- \( \log \left( \frac{75.6}{24.4} \right) \)

Let's compute these:
1. \( \frac{51.2}{48.8} \approx 1.048 \) → \( \log(1.048) \approx 0.020 \)
2. \( \frac{75.6}{24.4} \approx 3.1 \) → \( \log(3.1) \approx 0.491 \)

### Step 7: Substitute Back into the Equation
Substituting the logarithmic values back into the equation:
\[ 0.014 = \frac{0.059}{n} \left( 0.020 - 0.491 \right) \]
\[ 0.014 = \frac{0.059}{n} \left( -0.471 \right) \]

### Step 8: Solve for \( n \)
Rearranging gives:
\[ n = \frac{0.059 \cdot (-0.471)}{0.014} \]
Calculating this yields:
\[ n \approx 2 \]

### Final Answer
The value of \( n \) is approximately \( 2 \).

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