What is the equilibrium constant for the
reaction taking place at room temperature
(T = 25◦C) in the battery
Zn(s)| Zn2+(aq)|| Ce4+(aq)| Ce3+(aq) ?
Assume that the number of electrons transferred
in the reaction is n = 2.
Zn2+ + 2 e
− → Zn E
◦
red = −0.76 V
Ce4+ + e
− → Ce3+ E
◦
red = +1.61 V
2Ce4+ + 2e ==> 2Ce3+ 1.61 v.
Zn --> Zn2+ + 2e 0.76v
-----------------------
2Ce^4+ + 2e + Zn ==> Zn2+ +2e + 2Ce3+
Eocell = 1.61 + 0.76 = ? v.
Then nFEo = RTlnK
Substitute and solve for K.
Why did the battery go to therapy? Because it couldn't find its equilibrium constant! But don't worry, I'm here to help. To find the equilibrium constant, we use the Nernst equation:
E = E° - (RT/nF) * ln(Q)
Where:
- E is the cell potential
- E° is the standard cell potential
- R is the gas constant (8.314 J/(mol·K))
- T is the temperature in Kelvin (25°C = 298 K)
- n is the number of electrons transferred (2 in this case)
- F is Faraday's constant (96485 C/mol)
- Q is the reaction quotient
So plugging in the values we have:
E = 1.61 V - (8.314 J/(mol·K) * 298 K / (2 * 96485 C/mol)) * ln(Q)
Now, we need the value for Q to calculate the equilibrium constant. If you provide the concentrations of Zn2+ and Ce3+ in the reaction, I'd be happy to help you crunch the numbers!
To calculate the equilibrium constant for the given reaction, we can use the Nernst equation. The Nernst equation relates the standard cell potential to the equilibrium constant.
The Nernst equation 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
F = Faraday's constant (96485 C/mol)
ln = Natural logarithm
Q = Reaction quotient
In this case, the cell potential (E) can be determined by subtracting the reduction potential of the anode from the reduction potential of the cathode:
E = E°(Ce4+/Ce3+) - E°(Zn2+/Zn)
E°(Ce4+/Ce3+) = +1.61 V (Given)
E°(Zn2+/Zn) = -0.76 V (Given)
Substituting these values, we get:
E = +1.61 V - (-0.76 V)
E = +2.37 V
Next, we need to convert the temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T = 25 °C + 273.15
T = 298.15 K
Now we can substitute the values into the Nernst equation and solve for ln(Q):
2.37 = E° - (8.314 * 298.15) / (2 * 96485) * ln(Q)
Simplifying the equation, we get:
ln(Q) = [(2 * 2.37) / (8.314 * 298.15)] * 96485
ln(Q) = 22135.13
Now, we can calculate Q by taking the inverse natural logarithm (e^x) of ln(Q):
Q = e^(22135.13)
Finally, we can calculate the equilibrium constant (K) using the relation K = Q:
K = e^(22135.13)
This is the value of the equilibrium constant for the given reaction.
To calculate the equilibrium constant (K) for the reaction, we need to use the Nernst equation. The Nernst equation relates the equilibrium constant (K) to the standard reduction potentials (E°red) of the half-reactions involved.
The Nernst equation is given by:
Ecell = E°cell - (RT / nF) * ln(K)
Where:
- Ecell is the cell potential at equilibrium
- E°cell is the standard cell potential
- R is the gas constant (8.314 J/(mol·K))
- T is the temperature (in Kelvin)
- n is the number of electrons transferred in the reaction
- F is the Faraday constant (96,485 C/mol)
- K is the equilibrium constant
In this case, the cell potential at equilibrium (Ecell) is zero because it's at equilibrium. Therefore, we can rearrange the equation to solve for K:
K = exp((nF / RT) * E°cell)
Given:
- E°red(Zn2+ / Zn) = -0.76 V
- E°red(Ce4+ / Ce3+) = +1.61 V
We can use the reduction potentials to calculate the standard cell potential (E°cell):
E°cell = E°red(Ce3+/Ce4+) - E°red(Zn2+/Zn)
E°cell = (+1.61 V) - (-0.76 V) = +2.37 V
Now, substituting the values of n, F, R, and T into the equation, we can calculate K:
K = exp((2 * 96485 C/mol) / (8.314 J/(mol·K) * 298 K) * 2.37 V)
Calculating this expression will give you the equilibrium constant (K) for the reaction taking place at room temperature (T = 25°C) in the given battery.