At `25^@C`, Nernst equation is

Options

A. `E_"cell"=E_"cell"^@-"0.0591"/n "log" "[ion]"_"RHS"/"[ion]"_"LHS"`
B. `E_"cell"=E_"cell"^@-"0.0591"/n "log" "[M]"_"RHS"/"[M]"_"LHS"`
C. `E_"cell"=E_"cell"^@+"0.0591"/n "log" "[ion]"_"RHS"/"[ion]"_"LHS"`
D. `E_"cell"=E_"cell"^@-"0.0591"/n "log" "[ion]"_"LHS"/"[ion]"_"RHS"`

Answer

To derive the Nernst equation at 25°C, we will follow these steps:

### Step-by-Step Solution:

1. **Write the Nernst Equation**:
The general form of the Nernst equation is given by:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{RT}{nF} \ln Q \]
where:
- \(E_{\text{cell}}\) = cell potential
- \(E^\circ_{\text{cell}}\) = 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 (96485 C/mol)
- \(Q\) = reaction quotient

2. **Convert Temperature to Kelvin**:
At 25°C, the temperature in Kelvin is:
\[ T = 25 + 273 = 298 \, \text{K} \]

3. **Substitute Known Values**:
Substitute \(R\), \(T\), and \(F\) into the Nernst equation:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{(8.314 \, \text{J/(mol·K)}) \times (298 \, \text{K})}{n \times (96485 \, \text{C/mol})} \ln Q \]

4. **Simplify the Equation**:
Calculate the fraction:
\[ \frac{8.314 \times 298}{96485} \approx 0.008314 \times 298 \approx 0.0591 \]
Thus, the equation simplifies to:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \ln Q \]

5. **Convert Natural Logarithm to Logarithm Base 10**:
Since \( \ln Q = 2.303 \log_{10} Q \):
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \times 2.303 \log_{10} Q \]
This can be simplified to:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log_{10} Q \]

6. **Final Form of the Nernst Equation**:
The final form of the Nernst equation at 25°C is:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log_{10} \left( \frac{[\text{products}]}{[\text{reactants}]} \right) \]