Pt|h2(2atm)|h+(0.02M)||H+(0.1M)|H2(1atm)|Pt

AT 298k the EMF o cell is

To determine the electromotive force (EMF) of the given electrochemical cell, we can use the Nernst equation. This equation relates the cell potential to the concentrations of the reactants and products involved in the electrochemical reaction. Let's break this down step by step.

Understanding the Cell Components

The cell you've described can be represented as follows:

• Left half-cell: Pt | H⁺ (0.02 M) | H₂ (2 atm)
• Right half-cell: H⁺ (0.1 M) | H₂ (1 atm) | Pt

This setup indicates that we have a hydrogen electrode on both sides, with different concentrations of hydrogen ions and different pressures of hydrogen gas.

Standard Electrode Potential

The standard electrode potential (E_°) for the hydrogen electrode is defined as 0 V under standard conditions (1 M concentration of H⁺ and 1 atm pressure of H₂). However, since we have different conditions in our cell, we will need to apply the Nernst equation to find the actual EMF.

The Nernst Equation

The Nernst equation is given by:

E = E_° - (RT/nF) * ln(Q)

Where:

• E = cell potential
• E_° = standard cell potential (0 V for hydrogen electrodes)
• R = universal gas constant (8.314 J/(mol·K))
• T = temperature in Kelvin (298 K in this case)
• n = number of moles of electrons transferred (2 for the hydrogen reaction)
• F = Faraday's constant (96485 C/mol)
• Q = reaction quotient

Calculating the Reaction Quotient (Q)

For the hydrogen half-reaction, the reaction quotient Q can be expressed as:

Q = (PH2 / PH+2)

For the left half-cell:

Q_left = (2 atm) / (0.02 M)² = (2) / (0.0004) = 5000

For the right half-cell:

Q_right = (1 atm) / (0.1 M)² = (1) / (0.01) = 100

Overall Reaction Quotient

The overall reaction quotient for the cell is given by:

Q = Q_left / Q_right = 5000 / 100 = 50

Substituting into the Nernst Equation

Now we can substitute the values into the Nernst equation:

E = 0 V - (8.314 J/(mol·K) * 298 K / (2 * 96485 C/mol)) * ln(50)

Calculating the constants:

E = 0 V - (0.004129 V) * ln(50)

ln(50) ≈ 3.912

E = 0 V - (0.004129 V * 3.912) ≈ 0 V - 0.01616 V

E ≈ -0.01616 V

Final Result

The EMF of the cell at 298 K is approximately -0.016 V. This negative value indicates that the reaction is not spontaneous under the given conditions, meaning that the cell would not produce a positive voltage without external energy input.