Question

In the above diagram, the standard electrode potentials are given in volts (over the arrow). The value of \( E^\circ_{\text{FeO}_4^{2-}/\text{Fe}^{2+}} \) is:

• A. 1.7 V
• B. 1.2 V
• C. 2.1 V
• D. 1.4 V

Hint:

For problems involving electrode potentials, remember to use the relationship between Gibbs free energy and the electrode potential. The total potential can be found by combining the potentials of the individual half-reactions.

Answer 1

The Correct Option is A

We are given the following standard electrode potentials:
\( E_1^\circ = 2.0 \, \text{V} \) for \( \text{FeO}_4^{2-} \to \text{Fe}^{3+} \)
\( E_2^\circ = 0.8 \, \text{V} \) for \( \text{Fe}^{2+} \to \text{Fe}^{3+} \)
\( E_3^\circ = -0.5 \, \text{V} \) for \( \text{Fe}^{2+} \to \text{Fe} \)
We need to find \( E_4^\circ \) for the reaction:
\[ \text{FeO}_4^{2-} \to \text{Fe}^{2+} + \text{Fe}^{3+} \] The equation for the standard electrode potential is: \[ \Delta G^\circ_4 = \Delta G^\circ_1 + \Delta G^\circ_2 \] Using the relationship \( \Delta G^\circ = -nF E^\circ \), we get: \[ -n_4 E_4^\circ = -n_1 E_1^\circ - n_2 E_2^\circ \] For the number of electrons transferred (\( n_4 = 4 \)), the equation becomes: \[ 4 E_4^\circ = 3 \times 2 + (1 \times 0.8) \] Simplifying the equation: \[ E_4^\circ = \frac{6.8}{4} = 1.7 \, \text{V} \] Thus, the value of \( E_4^\circ \) is \( \boxed{1.7 \, \text{V}} \).