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Calculate the emf of the following cell at 298 K:

Fe₃|Fe²⁺(0.001M)||H⁺(1M)|H₂(1 bar), Pt

(Given: Ecell0 = +0.44V)

Aniket Singh , 5 Months ago
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anser 1 Answers
Askiitians Tutor Team

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
  • = 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.

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