Question

\(FeO^{2-}_4+\underrightarrow{2.2v}\;Fe^{3+}\underrightarrow{+0.70v}\;Fe^{2+}\underrightarrow{-0.45v}\;Fe^0\)
\(E^0_{FeO_4^{2+}}\; is \;x × 10^{-3} V\). The Value of \(x\) is __

Hint:

To calculate the overall standard reduction potential for a series of redox reac tions, use the relationship \(∆G = −nFE^◦\) and combine the contributions from all steps.

Answer 1

Correct Answer: 1825

Step 1: Combine the Half-Cell Reactions

The reactions are as follows:

• \( \text{FeO}_4^{2-} + 3e^- \rightarrow \text{Fe}^{3+} \) (\( \Delta G_1 \))
• \( \text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+} \) (\( \Delta G_2 \))
• \( \text{FeO}_4^{2-} + 4e^- \rightarrow \text{Fe}^{2+} \) (\( \Delta G_3 \))

Step 2: Use Gibbs Free Energy Relationships

The total Gibbs free energy for the combined reaction is:

\[ \Delta G_3 = \Delta G_1 + \Delta G_2 \]

Substitute \( \Delta G = -nFE^\circ \):

\[ -4FE^\circ_3 = -3F(2.2) + (-1F)(0.7) \]

Step 3: Simplify to Calculate \( E^\circ_3 \)

Simplify the equation:

\[ 4E^\circ_3 = 6.6 + 0.7 = 7.3 \]

Divide by 4:

\[ E^\circ_3 = \frac{7.3}{4} = 1.825 \, \text{V} \]

Express in millivolts:

\[ E^\circ_3 = 1.825 \times 10^3 \, \text{mV} \]

Final Answer:

The value of \( x \) is 1825.