\(FeO^{2-}_4+\underrightarrow{2.2v}\;Fe^{3+}\underrightarrow{+0.70v}\;Fe^{2+}\underrightarrow{-0.45v}\;Fe^0\)
\(E^0_{FeO_4^{2+}}\; is \;x × 10^{-3} V\). The Value of \(x\) is __
\(E^0_{FeO_4^{2+}}\; is \;x × 10^{-3} V\). The Value of \(x\) is __
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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.
Correct Answer: 1825
Solution and Explanation
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.
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