The half cell potential of a half-cell `A^(x+), A^((x+n)+)|Pt` were found to be as follows:
`{:(%"of reduced form",24.4,48.8),("Cell potential"//N,0.101,0.115):}`
Determine the value of `n`.
[take `(2.303RT)/(F) = 0.05, log_(10)24.4 = 1.387, log_(10) 75.6 = 1.878 log_(10) 48.8 = 1.688, log_(10) 51.2 = 1.709]`

Explanation

To determine the value of `n` in the half-cell potential of the given electrochemical cell, we will use the Nernst equation and the provided data.

### Step-by-step Solution:

1. **Identify the Given Data:**
- Reduced form concentrations:
- Case 1: \( A^{x+} \) concentration = 24.4%
- Case 2: \( A^{x+} \) concentration = 48.8%
...