The Nernst equation E=`E^(@)`-RT/nF in Q indicates that the Q will be equal to equilibrium constant `K_c` when:

Options

A. E=`E^(@)`
B. RT/nF=1
C. E=zero
D. `E^(@)=1`

Answer

To solve the question regarding the Nernst equation and the condition under which the reaction quotient \( Q \) equals the equilibrium constant \( K_c \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Nernst Equation**:
The Nernst equation is given by:
\[ E = E^0 - \frac{RT}{nF} \ln Q \]
where:
- \( E \) is the cell potential,
- \( E^0 \) is the standard cell potential,
- \( R \) is the universal gas constant,
- \( T \) is the temperature in Kelvin,
- \( n \) is the number of moles of electrons transferred,
- \( F \) is Faraday's constant,
- \( Q \) is the reaction quotient.

2. **Identify the Condition for Equilibrium**:
At equilibrium, the cell potential \( E \) becomes zero. This is because there is no net change in the concentrations of reactants and products, and the system is at a state of balance.

3. **Set \( E \) to Zero**:
To find the condition under which \( Q \) equals \( K_c \), we set \( E = 0 \) in the Nernst equation:
\[ 0 = E^0 - \frac{RT}{nF} \ln Q \]

4. **Rearrange the Equation**:
Rearranging the equation gives:
\[ E^0 = \frac{RT}{nF} \ln Q \]

5. **Substitute \( Q \) with \( K_c \)**:
At equilibrium, \( Q \) is equal to the equilibrium constant \( K_c \). Therefore, we can write:
\[ E^0 = \frac{RT}{nF} \ln K_c \]

6. **Conclusion**:
Thus, we conclude that \( Q \) will be equal to \( K_c \) when the cell potential \( E \) is zero, indicating that the system is at equilibrium.

### Final Answer:
The reaction quotient \( Q \) will be equal to the equilibrium constant \( K_c \) when the cell potential \( E \) is zero.

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