Question

In the given reaction, \[ 2Fe^{2+} + 3H_2O \rightarrow Fe_2O_3 + 6H^+ + 2e^- \] consider ideal condition, take concentration of \( Fe^{2+} \) as \( 10^{-5} \) molal, \( E^0 = 0.98 \, \text{V} \) and pH = 6. The value of \( (2.303 \times R \times T / F) = 0.059 \) (where \( F \) is the Faraday constant). The value of \( E_h \) on the \( Fe^{2+} / \) hematite boundary at 25°C is \(\underline{\hspace{1cm}}\) V (round off to 2 decimal places).

Hint:

Use the Nernst equation to calculate the electrode potential, taking into account the concentration of ions and the number of electrons involved in the reaction.

Answer 1

Correct Answer: 0.21

To calculate the value of \( E_h \), we use the Nernst equation: \[ E_h = E^0 - \frac{0.059}{n} \log_{10} \left( \frac{[Fe^{2+}]^2}{[H^+]^6} \right) \] Given:
- \( E^0 = 0.98 \, \text{V} \),
- \( n = 2 \) (number of electrons transferred),
- \( [Fe^{2+}] = 10^{-5} \, \text{molal} \),
- \( [H^+] = 10^{-6} \) (since pH = 6),
- \( (2.303 \times R \times T / F) = 0.059 \).
Substituting these values into the equation: \[ E_h = 0.98 - \frac{0.059}{2} \log_{10} \left( \frac{(10^{-5})^2}{(10^{-6})^6} \right) \] \[ E_h = 0.98 - 0.0295 \log_{10} \left( \frac{10^{-10}}{10^{-36}} \right) \] \[ E_h = 0.98 - 0.0295 \log_{10} (10^{26}) \] \[ E_h = 0.98 - 0.0295 \times 26 = 0.98 - 0.767 = 0.213 \, \text{V} \] Thus, the value of \( E_h \) is approximately \( 0.21 \, \text{V} \).