Question:
The standard emf of a cell involving one electron change is found to be 0.591 V at \( 25^\circ C \). The equilibrium constant of the reaction is ________. (\( F = 96,500 \, \text{C/mol} \))
The standard emf of a cell involving one electron change is found to be 0.591 V at \( 25^\circ C \). The equilibrium constant of the reaction is ________. (\( F = 96,500 \, \text{C/mol} \))
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The standard emf of a cell involving one electron change is found to be 0.591 V at $25$)
Updated On: Apr 15, 2026
- $1.0 \times 10^{1}$
- $1.0 \times 10^{5}$
- $1.0 \times 10^{10}$
- $1.0 \times 10^{30}$
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The Correct Option is C
Solution and Explanation
Step 1: Formula
The relationship between standard emf ($E^{\circ}_{cell}$) and equilibrium constant ($K_{eq}$) is given by: $E^{\circ}_{cell} = \frac{0.0591}{n} \log K_{eq}$
Step 2: Substitution
Given $E^{\circ}_{cell} = 0.591 \text{ V}$ and $n = 1$ (for a one electron change). $0.591 = \frac{0.0591}{1} \log K_{eq}$
Step 3: Calculation
$\log K_{eq} = \frac{0.591}{0.0591} = 10$ $K_{eq} = 10^{10} = 1.0 \times 10^{10}$
Step 4: Conclusion
The equilibrium constant is $1.0 \times 10^{10}$.
Final Answer: (C)
The relationship between standard emf ($E^{\circ}_{cell}$) and equilibrium constant ($K_{eq}$) is given by: $E^{\circ}_{cell} = \frac{0.0591}{n} \log K_{eq}$
Step 2: Substitution
Given $E^{\circ}_{cell} = 0.591 \text{ V}$ and $n = 1$ (for a one electron change). $0.591 = \frac{0.0591}{1} \log K_{eq}$
Step 3: Calculation
$\log K_{eq} = \frac{0.591}{0.0591} = 10$ $K_{eq} = 10^{10} = 1.0 \times 10^{10}$
Step 4: Conclusion
The equilibrium constant is $1.0 \times 10^{10}$.
Final Answer: (C)
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