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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
(AIEEE 2003)
- a)1 x 10-10
- b)29.5 x 10-2
- c)10
- d)1 x 1010
Correct answer is option 'D'. Can you explain this answer?
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For a cell reaction involving a two electron change, the standard emf ...
To determine the equilibrium constant of a cell reaction involving a two-electron change, we can use the Nernst equation, which relates the standard electrode potential (E°) of the cell to the equilibrium constant (K) of the reaction.
The Nernst equation is given by:
E = E° - (RT/nF) * ln(K)
Where:
E = cell potential under non-standard conditions
E° = standard electrode potential
R = gas constant (8.314 J/(mol·K))
T = temperature in Kelvin
n = number of electrons transferred in the reaction
F = Faraday constant (96485 C/mol)
K = equilibrium constant of the reaction
Given that the standard emf of the cell is 0.295 V at 25°C, we can substitute the known values into the Nernst equation and solve for ln(K):
0.295 = E° - (8.314 * (25 + 273.15) / (2 * 96485)) * ln(K)
Simplifying the equation:
0.295 = E° - (2.303 * 8.314 * 298.15 / 96485) * ln(K)
0.295 = E° - 0.0592 * ln(K)
Rearranging the equation:
0.0592 * ln(K) = E° - 0.295
ln(K) = (E° - 0.295) / 0.0592
Now, we can take the exponential of both sides to eliminate the natural logarithm:
K = e^((E° - 0.295) / 0.0592)
Substituting the given value of E° (0.295 V) into the equation:
K = e^((0.295 - 0.295) / 0.0592)
K = e^0
K = 1
Therefore, the equilibrium constant of the reaction at 25°C is 1, which corresponds to option 'D' in the given choices.
The Nernst equation is given by:
E = E° - (RT/nF) * ln(K)
Where:
E = cell potential under non-standard conditions
E° = standard electrode potential
R = gas constant (8.314 J/(mol·K))
T = temperature in Kelvin
n = number of electrons transferred in the reaction
F = Faraday constant (96485 C/mol)
K = equilibrium constant of the reaction
Given that the standard emf of the cell is 0.295 V at 25°C, we can substitute the known values into the Nernst equation and solve for ln(K):
0.295 = E° - (8.314 * (25 + 273.15) / (2 * 96485)) * ln(K)
Simplifying the equation:
0.295 = E° - (2.303 * 8.314 * 298.15 / 96485) * ln(K)
0.295 = E° - 0.0592 * ln(K)
Rearranging the equation:
0.0592 * ln(K) = E° - 0.295
ln(K) = (E° - 0.295) / 0.0592
Now, we can take the exponential of both sides to eliminate the natural logarithm:
K = e^((E° - 0.295) / 0.0592)
Substituting the given value of E° (0.295 V) into the equation:
K = e^((0.295 - 0.295) / 0.0592)
K = e^0
K = 1
Therefore, the equilibrium constant of the reaction at 25°C is 1, which corresponds to option 'D' in the given choices.
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For a cell reaction involving a two electron change, the standard emf ...
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For a cell reaction involving a two electron change, the standard emf of the cell is found to be 0.295 V at 25C. The equilibrium constant of the reaction at 25C will be(AIEEE 2003)a)1 x 10-10b)29.5x 10-2c)10d)1 x 1010Correct answer is option 'D'. Can you explain this answer?
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