FORLAD 4 Consider the redox reaction where Ag+ (aq) is oxidized to Ago (s) (E° 0x = -1.77 v) and Helo₂ (aq) is reduced to Helo (aq) (E°red = + 1.67v). a) write the Oxidation half reaction and reduction half reaction b) Calculate the equilibrium constant at 25°c hint (log k = (n. E°) - 0.0592)

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### Redox Reaction Analysis

In this exercise, we will consider the redox reaction where Hg (in its oxidation state) and \( \text{HClO}_2 \) (in its reduction state) participate. Here are the given standard electrode potentials for each species:

- \( E^\circ_{\text{ox}} = -1.67 \text{ V} \) for \( \text{Hg} \)
- \( E^\circ_{\text{red}} = +1.67 \text{ V} \) for \( \text{HClO}_2 \)

To complete this analysis, we are asked to do the following:

#### (a) Write the Oxidation and Reduction Half-Reactions

First, identify the oxidation and reduction half-reactions separately.

- The oxidation half-reaction involves \( \text{Hg} \) being oxidized:
  \[
  \text{Hg} \rightarrow \text{Hg}^{2+} + 2e^{-} \quad (E^\circ_{\text{ox}} = -1.67 \text{ V})
  \]

- The reduction half-reaction involves \( \text{HClO}_2 \) being reduced:
  \[
  \text{HClO}_2 + 2e^{-} \rightarrow \text{HCl} + \text{{other products}} \quad (E^\circ_{\text{red}} = +1.67 \text{ V})
  \]

#### (b) Calculate the Equilibrium Constant at 25°C

Using the Nernst equation and the relationship between the standard electrode potential and the equilibrium constant, we can determine the equilibrium constant \( K \). The relevant formula is:
\[
\log K = \frac{n \cdot E^\circ_{\text{cell}}}{0.0592}
\]
where \( n \) is the number of moles of electrons transferred in the balanced equation and \( E^\circ_{\text{cell}} \) is the standard cell potential. 

To calculate \( E^\circ_{\text{cell}} \):
\[
E^\circ_{\text{cell}} = E^\circ_{\text{red}} - E^\circ_{\text{ox}} = +1.67 \text{ V} - (-1.67 \text{ V}) = +3.34 \text{ V}
\]

Since \( n = 2
Transcribed Image Text:### Redox Reaction Analysis In this exercise, we will consider the redox reaction where Hg (in its oxidation state) and \( \text{HClO}_2 \) (in its reduction state) participate. Here are the given standard electrode potentials for each species: - \( E^\circ_{\text{ox}} = -1.67 \text{ V} \) for \( \text{Hg} \) - \( E^\circ_{\text{red}} = +1.67 \text{ V} \) for \( \text{HClO}_2 \) To complete this analysis, we are asked to do the following: #### (a) Write the Oxidation and Reduction Half-Reactions First, identify the oxidation and reduction half-reactions separately. - The oxidation half-reaction involves \( \text{Hg} \) being oxidized: \[ \text{Hg} \rightarrow \text{Hg}^{2+} + 2e^{-} \quad (E^\circ_{\text{ox}} = -1.67 \text{ V}) \] - The reduction half-reaction involves \( \text{HClO}_2 \) being reduced: \[ \text{HClO}_2 + 2e^{-} \rightarrow \text{HCl} + \text{{other products}} \quad (E^\circ_{\text{red}} = +1.67 \text{ V}) \] #### (b) Calculate the Equilibrium Constant at 25°C Using the Nernst equation and the relationship between the standard electrode potential and the equilibrium constant, we can determine the equilibrium constant \( K \). The relevant formula is: \[ \log K = \frac{n \cdot E^\circ_{\text{cell}}}{0.0592} \] where \( n \) is the number of moles of electrons transferred in the balanced equation and \( E^\circ_{\text{cell}} \) is the standard cell potential. To calculate \( E^\circ_{\text{cell}} \): \[ E^\circ_{\text{cell}} = E^\circ_{\text{red}} - E^\circ_{\text{ox}} = +1.67 \text{ V} - (-1.67 \text{ V}) = +3.34 \text{ V} \] Since \( n = 2
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