A certain voltaic cell has a standard cell voltage of 1.22 V, and 3 electrons are transferred in the overall cell reaction.
What is the value of the thermodynamic equilibrium constant for this cell reaction at 298K?
nFEoCell = RTlnK
To find the value of the thermodynamic equilibrium constant for this cell reaction at 298K, you can make use of the Nernst equation.
The Nernst equation relates the cell potential to the equilibrium constant and the standard cell potential. It is given by:
E = E° - (RT/nF) * ln(Q)
Where:
E = cell potential
E° = standard cell potential
R = gas constant (8.314 J/mol K)
T = temperature in Kelvin
n = number of electrons transferred in the cell reaction
F = Faraday's constant (96,485 C/mol)
ln = natural logarithm
Q = reaction quotient
In this case, we are given the standard cell potential (E°) as 1.22 V and the number of electrons transferred (n) as 3. We are asked to find the thermodynamic equilibrium constant, which is related to Q.
At equilibrium, Q = K, where K is the equilibrium constant.
To find K, we can rearrange the Nernst equation and solve for Q:
Q = exp((E° - E) * (nF/RT))
Now, plug in the given values:
E° = 1.22 V
E = 0 (since we are considering the equilibrium condition)
n = 3
R = 8.314 J/mol K
T = 298K
F = 96,485 C/mol
Substituting these values into the equation:
Q = exp((1.22 - 0) * (3 * 96,485) / (8.314 * 298))
Evaluating this expression will give you the value of Q, which is equal to the thermodynamic equilibrium constant (K) for the cell reaction at 298K.