The voltaic cell described by the balanced chemical reaction has a standard emf of 4.05 V. Calculate the equilibrium constant (Kc) for the reaction at 25 oC. Round your answer to 3 significant figures.
F2(g) + Mn(s) → 2F-(aq) + Mn2+(aq)
Mn2+/Mn -1.18
F2/F- +2.87
0 °C = 273.15 K
R = 8.314 J/(K x mol)
Faraday's Constant
96485 C/mol
nEF = RTlnK
I tried doing that but the value was so high that I wanted to compare answers with something you would get DrBob222. My answer was something e^136
To calculate the equilibrium constant (Kc) for the given reaction at 25 °C, we will use the equation:
E°cell = (0.0592/n) log(Kc)
Where:
- E°cell is the standard emf of the cell (4.05 V)
- n is the number of electrons transferred in the balanced chemical equation (in this case, n = 2, since 2 electrons are transferred)
- Kc is the equilibrium constant we are trying to find
First, we need to find the value of ln(Kc) using the equation above. Rearranging the equation, we have:
ln(Kc) = (n/E°cell) * log(10)
Substituting the given values into the equation, we get:
ln(Kc) = (2/4.05) * log(10) = 0.488
Next, we can find Kc by taking the exponential of ln(Kc):
Kc = e^(ln(Kc)) = e^(0.488) = 1.629
Rounding to 3 significant figures, the equilibrium constant (Kc) for the reaction at 25 °C is 1.63.