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For a cell reaction involving a two-electron change the standard e.m.f. of the cell is found to be 0.295 V at 25ºC. The equilibrium constant of the reaction at 25º C will be – [AIEEE-2003]
- a)10
- b)1 × 1010
- c)1 × 10-10
- d)29.5 × 10-2
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
For a cell reaction involving a two-electron change the standard e.m.f...
The correct answer is Option B.
Eocell = (0.0591/n) log Kc
0.259 = (0.0591/2) log Kc
0.259 = 0.0295 log Kc
Kc = antilog 10
Kc = 1 x 1010
Eocell = (0.0591/n) log Kc
0.259 = (0.0591/2) log Kc
0.259 = 0.0295 log Kc
Kc = antilog 10
Kc = 1 x 1010
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For a cell reaction involving a two-electron change the standard e.m.f...
°C. Calculate the equilibrium constant, K, for the reaction.
The Nernst equation relates the standard electrode potential (E°), the actual electrode potential (E), the gas constant (R), the temperature (T), the Faraday constant (F), and the reaction quotient (Q):
E = E° - (RT/F) ln(Q)
At equilibrium, the reaction quotient is equal to the equilibrium constant:
Q = K
Substituting the given values, we get:
0.295 V = E° - (8.314 J/K/mol) x (298 K) / (2 x 96,485 C/mol) x ln(K)
Simplifying:
ln(K) = (0.295 V - E°) x (2 x 96,485 C/mol) / (8.314 J/K/mol x 298 K)
ln(K) = 2.303 x (0.295 V - E°) / (8.314 J/K/mol)
Assuming E° is positive (as it is for most spontaneous reactions), we can rearrange the equation to solve for K:
K = exp[(0.295 V - E°) x (2 x 96,485 C/mol) / (8.314 J/K/mol x 298 K)]
We need to know the specific reaction to determine E°, but the general formula for a two-electron change is:
aA + 2e- ⇌ bB
where A and B are the reactant and product, respectively, and a and b are the stoichiometric coefficients. The standard electrode potential for this reaction is given by:
E° = (RT/F) ln(K/a^2)
Substituting the given values, we get:
K = a^2 x exp(E° x (2 x 96,485 C/mol) / (8.314 J/K/mol x 298 K))
We can use this equation to solve for K once we know the value of E°.
The Nernst equation relates the standard electrode potential (E°), the actual electrode potential (E), the gas constant (R), the temperature (T), the Faraday constant (F), and the reaction quotient (Q):
E = E° - (RT/F) ln(Q)
At equilibrium, the reaction quotient is equal to the equilibrium constant:
Q = K
Substituting the given values, we get:
0.295 V = E° - (8.314 J/K/mol) x (298 K) / (2 x 96,485 C/mol) x ln(K)
Simplifying:
ln(K) = (0.295 V - E°) x (2 x 96,485 C/mol) / (8.314 J/K/mol x 298 K)
ln(K) = 2.303 x (0.295 V - E°) / (8.314 J/K/mol)
Assuming E° is positive (as it is for most spontaneous reactions), we can rearrange the equation to solve for K:
K = exp[(0.295 V - E°) x (2 x 96,485 C/mol) / (8.314 J/K/mol x 298 K)]
We need to know the specific reaction to determine E°, but the general formula for a two-electron change is:
aA + 2e- ⇌ bB
where A and B are the reactant and product, respectively, and a and b are the stoichiometric coefficients. The standard electrode potential for this reaction is given by:
E° = (RT/F) ln(K/a^2)
Substituting the given values, we get:
K = a^2 x exp(E° x (2 x 96,485 C/mol) / (8.314 J/K/mol x 298 K))
We can use this equation to solve for K once we know the value of E°.
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For a cell reaction involving a two-electron change the standard e.m.f. of the cell is found to be 0.295 V at 25ºC. The equilibrium constant of the reaction at 25º C will be – [AIEEE-2003]a)10b)1 × 1010c)1 × 10-10d)29.5 × 10-2Correct answer is option 'B'. Can you explain this answer?
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