Question:
For a cell reaction involving a two-electron change, the standard emf of the cell is found to be 0.295 V at 25°C. The equilibrium constant of the reaction at 25°C will be:
For a cell reaction involving a two-electron change, the standard emf of the cell is found to be 0.295 V at 25°C. The equilibrium constant of the reaction at 25°C will be:
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The equilibrium constant \( K \) can be found from the standard electrode potential using the Nernst equation.
Updated On: Apr 22, 2026
- 10\(^1\)
- 1 \(\times\) 10\(^1\)
- 1 \(\times\) 10\(^10\)
- 10 \(\times\) 10\(^2\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Nernst equation.
The Nernst equation relates the standard electrode potential (E°) to the equilibrium constant (K) for a reaction: \[ \Delta G = -nFE = -RT \ln K \] For two-electron change reactions: \[ E = \frac{RT}{nF} \ln K \]
Step 2: Substituting values.
At 25°C, \( E = 0.295 \, \text{V} \), \( n = 2 \), and using the known constants: \[ 0.295 = \frac{0.0592}{2} \log K \] Solving for \( K \), we get: \[ K = 10^{10} \]
Step 3: Conclusion.
The correct answer is (2) \( 1 \times 10^{10} \).
The Nernst equation relates the standard electrode potential (E°) to the equilibrium constant (K) for a reaction: \[ \Delta G = -nFE = -RT \ln K \] For two-electron change reactions: \[ E = \frac{RT}{nF} \ln K \]
Step 2: Substituting values.
At 25°C, \( E = 0.295 \, \text{V} \), \( n = 2 \), and using the known constants: \[ 0.295 = \frac{0.0592}{2} \log K \] Solving for \( K \), we get: \[ K = 10^{10} \]
Step 3: Conclusion.
The correct answer is (2) \( 1 \times 10^{10} \).
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