Question

For a cell reaction, $ A(s) + B^{2+}(aq) \rightarrow A^{2+}(aq) + B(s) $; the standard emf of the cell is 0.295 V at 25°C. The equilibrium constant at 25°C will be

• A. \( 1 \times 10^{10} \)
• B. \( 10 \)
• C. \( 2.95 \times 10^{-2} \)
• D. \( 2.95 \times 10^{-10} \)

Hint:

The Nernst equation allows you to calculate the equilibrium constant from the standard emf, and vice versa. It's useful for electrochemical reactions.

Answer 1

The Correct Option is A

To find the equilibrium constant (\( K_{eq} \)) for the given cell reaction at 25°C, we use the relationship given by the Nernst equation and the expression for the equilibrium constant:
The equation that relates the standard electromotive force (\( E^\circ \)) of the cell and the equilibrium constant is given by:

\( E^\circ = \frac{RT}{nF} \ln K_{eq} \)

Where:

• \( R \) is the universal gas constant, which is 8.314 J/mol·K.
• \( T \) is the temperature in Kelvin, which is 298 K at 25°C.
• \( n \) is the number of moles of electrons exchanged in the reaction. For this reaction, \( n = 2 \) as each A and B undergoes a change of +2 and -2 oxidation states, respectively.
• \( F \) is the Faraday constant, which is 96485 C/mol.
• \( E^\circ = 0.295 \) V (given).

Rearranging gives:

\( \ln K_{eq} = \frac{nFE^\circ}{RT} \)

Substituting the known values:

\( \ln K_{eq} = \frac{2 \times 96485 \times 0.295}{8.314 \times 298} \)

Calculating:

\( \ln K_{eq} \approx 23.026 \)

Thus, \( K_{eq} \) is:

\( K_{eq} = e^{23.026} \)

\( K_{eq} \approx 1 \times 10^{10} \)

The correct equilibrium constant at 25°C is \( 1 \times 10^{10} \).

Answer 2

To determine the equilibrium constant (\( K \)) from the standard electromotive force (\( E^\circ \)) of the cell reaction \( A(s) + B^{2+}(aq) \rightarrow A^{2+}(aq) + B(s) \), we can use the Nernst equation at standard conditions, which relates these quantities:
\[ E^\circ = \frac{RT}{nF} \ln K \]
Where:

• \( R \) is the universal gas constant, \( 8.314 \, \text{J mol}^{-1} \text{K}^{-1} \)
• \( T \) is the temperature in Kelvin, \( 298 \, \text{K} \)
• \( n \) is the number of moles of electrons transferred in the reaction
• \( F \) is Faraday's constant, \( 96485 \, \text{C mol}^{-1} \)
• \( E^\circ = 0.295 \, \text{V} \)

For this reaction, \( n = 2 \) since there is a transfer of 2 electrons. Thus, substituting the values in:
\[ 0.295 = \frac{(8.314 \times 298)}{(2 \times 96485)} \ln K \]
We solve for \( \ln K \) as follows:
\[ \ln K = \frac{0.295 \times 2 \times 96485}{8.314 \times 298} \]
This simplifies to:
\[ \ln K \approx 23.026 \]
Taking the exponential of both sides, we find:
\[ K = e^{23.026} \approx 1 \times 10^{10} \]
Thus, the equilibrium constant \( K \) is \( 1 \times 10^{10} \).