In Nernst equation the constant 0.0591 at 298K represents the value of

Options

A. `(RT)/(nF)`
B. `(RT)/(F)`
C. `(2.303RT)/(nF)`
D. `(2.303RT)/(F)`

Answer

To solve the question regarding the constant 0.0591 in the Nernst equation at 298K, we will follow these steps:

### Step 1: Understand the Nernst Equation
The Nernst equation relates the reduction potential of a half-cell at non-standard conditions to the standard electrode potential, temperature, and the reaction quotient. The equation is given by:

\[ E = E^\circ - \frac{RT}{nF} \ln Q \]

Where:
- \( E \) = cell potential under non-standard conditions
- \( E^\circ \) = standard cell potential
- \( R \) = universal gas constant (8.314 J/(mol·K))
- \( T \) = temperature in Kelvin
- \( n \) = number of moles of electrons transferred in the reaction
- \( F \) = Faraday's constant (approximately 96500 C/mol)
- \( Q \) = reaction quotient

### Step 2: Substitute Known Values
At 298 K (25°C), we can substitute the known values into the equation. The constant 0.0591 is derived from the expression:

\[ \frac{RT}{nF} \]

### Step 3: Calculate \( \frac{RT}{F} \)
First, we need to calculate \( \frac{RT}{F} \) at 298 K.

1. **Calculate \( R \cdot T \)**:
\[ R = 8.314 \, \text{J/(mol·K)} \]
\[ T = 298 \, \text{K} \]
\[ R \cdot T = 8.314 \times 298 = 2478.572 \, \text{J/mol} \]

2. **Divide by Faraday's constant \( F \)**:
\[ F = 96500 \, \text{C/mol} \]
\[ \frac{RT}{F} = \frac{2478.572}{96500} \approx 0.0257 \, \text{V} \]

### Step 4: Incorporate the Logarithmic Factor
The Nernst equation uses the logarithm base 10, which means we need to multiply by \( 2.303 \) to convert natural logarithm to log base 10:

\[ \frac{RT}{nF} = \frac{2.303 \cdot RT}{F} \]

### Step 5: Calculate \( 2.303 \cdot \frac{RT}{F} \)
Now we calculate:

\[ 2.303 \cdot 0.0257 \approx 0.0591 \, \text{V} \]

### Conclusion
Thus, the constant 0.0591 at 298K in the Nernst equation represents the value of \( \frac{2.303 \cdot RT}{F} \).

### Final Answer
The constant 0.0591 in the Nernst equation at 298K represents \( \frac{2.303 \cdot RT}{F} \).

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