To calculate the electromotive force (emf) of the given electrochemical cell at 298 K, we can use the Nernst equation. This equation allows us to determine the cell potential under non-standard conditions by taking into account the concentrations of the reactants and products involved in the redox reactions.
Understanding the Components of the Cell
The cell notation provided is:
- Fe₃ | Fe²⁺(0.001 M)
- ||
- H⁺(1 M) | H₂(1 bar), Pt
This indicates that we have a half-cell involving iron ions and a hydrogen half-cell. The standard cell potential (E°) is given as +0.44 V.
Applying the Nernst Equation
The Nernst equation is expressed as:
E = E° - (RT/nF) * ln(Q)
Where:
- E = cell potential under non-standard conditions
- E° = standard cell potential (+0.44 V)
- R = universal gas constant (8.314 J/(mol·K))
- T = temperature in Kelvin (298 K)
- n = number of moles of electrons transferred in the reaction
- F = Faraday's constant (96485 C/mol)
- Q = reaction quotient
Determining the Reaction Quotient (Q)
For the given cell, the half-reaction for the reduction of Fe³⁺ to Fe²⁺ can be represented as:
Fe³⁺ + e⁻ ⇌ Fe²⁺
In this case, the number of electrons transferred (n) is 1. The reaction quotient (Q) can be calculated using the concentrations of the species involved:
Q = [Fe²⁺] / [Fe³⁺]
Assuming that the concentration of Fe³⁺ is in a saturated state, we can denote it as 1 M for simplicity. Thus:
Q = 0.001 / 1 = 0.001
Calculating the Cell Potential (E)
Now, we can substitute the values into the Nernst equation:
E = 0.44 V - (8.314 J/(mol·K) * 298 K / (1 * 96485 C/mol)) * ln(0.001)
Calculating the term:
- RT/F = (8.314 * 298) / 96485 ≈ 0.008314 V
- ln(0.001) ≈ -6.907755
Substituting these values back into the equation gives:
E = 0.44 V - (0.008314 * -6.907755)
E = 0.44 V + 0.0575 V
E ≈ 0.4975 V
Final Result
The emf of the cell at 298 K is approximately 0.498 V. This value indicates the potential difference that can drive the electrochemical reaction under the specified conditions.