0.34 V - (8.314 J/(mol·K) * 298.15 K / (2 * 96485 C/mol)) * ln(4.48 / 3.2×10−3)

1. Break Down the Problem

We need to compute the expression given, which involves constants and a natural logarithm function.

2. Relevant Concepts

The formula involves the following:

• $E = E^0 - \frac{RT}{nF} \ln\left(\frac{[oxidized]}{[reduced]}\right)$ where:
• $E =$ cell potential (V)
• $E^0 = 0.34 \, V$
• $R = 8.314 \, \frac{J}{mol \cdot K}$ (ideal gas constant)
• $T = 298.15 \, K$ (temperature)
• $n = 2$ (number of moles of electrons transferred)
• $F = 96485 \, \frac{C}{mol}$ (Faraday's constant)
• $[oxidized] = 4.48$
• $[reduced] = 3.2 \times 10^{-3}$

3. Analysis and Detail

First, let's compute the term: $\frac{RT}{nF} = \frac{8.314 \, J/(mol \cdot K) \times 298.15 \, K}{2 \times 96485 \, C/mol}$

Calculating $R \times T$: $RT = 8.314 \times 298.15 \approx 2478.85 \, J/mol$

Next, calculate $nF$: $nF = 2 \times 96485 \approx 192970 \, C/mol$

Now, compute the ratio: $\frac{RT}{nF} = \frac{2478.85}{192970} \approx 0.01284 \, V$

Next, calculate the natural logarithm term: $\ln\left(\frac{4.48}{3.2 \times 10^{-3}}\right) = \ln\left(1400\right)$ Computing $\ln(1400)$: $\ln(1400) \approx 7.244$

Now we can substitute back into the formula: $E = 0.34 - (0.01284 \times 7.244)$ Calculating the multiplication: $0.01284 \times 7.244 \approx 0.0930$

Thus, the cell potential becomes: $E = 0.34 - 0.0930 \approx 0.247 \, V$

4. Verify and Summarize

The calculations have been verified at each step.

Final Answer

The final result is: $E \approx 0.247 \, V$