Consider the following redox reaction:\[\text{MnO}_4^- + \text{H}^+ + \text{H}_2 \text{C}_2 \text{O}_4 \rightleftharpoons \text{Mn}^{2+} + \text{H}_2 \text{O} + \text{CO}_2\]The standard reduction potentials are given as below (E\(_\text{red}\)):\[E^\circ_{\text{MnO}_4^-/\text{Mn}^{2+}} = +1.51 \, \text{V}\]\[E^\circ_{\text{CO}_2/\text{H}_2 \text{C}_2 \text{O}_4} = -0.49 \, \text{V}\]If the equilibrium constant of the above reaction is given as \(K_\text{eq} = 10^x\), then the value of (x = ________ ) (nearest integer).
Correct Answer: 338
Approach Solution - 1
Step-by-step Calculation:
The overall cell potential \( E^\circ_{\text{cell}} \) for the redox reaction is calculated as:
\[E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}\]
Here:
\( E^\circ_{\text{cathode}} = +1.51 \, \text{V} \) (Reduction potential for \( \text{MnO}_4^- \))
\( E^\circ_{\text{anode}} = -0.49 \, \text{V} \) (Reduction potential for \( \text{H}_2\text{C}_2\text{O}_4 \))
Therefore:
\[E^\circ_{\text{cell}} = 1.51 - (-0.49) = 2.00 \, \text{V}\]
The equilibrium constant \( K_\text{eq} \) is related to the cell potential by the Nernst equation:
\[\Delta G^\circ = -nF E^\circ_{\text{cell}} \quad \text{and} \quad \Delta G^\circ = -RT \ln K_\text{eq}\]
Equating the two expressions:
\[nF E^\circ_{\text{cell}} = RT \ln K_\text{eq}\]
Rearranging to find \( K_\text{eq} \):
\[\ln K_\text{eq} = \frac{nF E^\circ_{\text{cell}}}{RT}\]
Given:
\( n = 5 \) (number of electrons transferred)
\( F = 96500 \, \text{C mol}^{-1} \)
\( R = 8.314 \, \text{J K}^{-1} \text{mol}^{-1} \)
\( T = 298 \, \text{K} \)
\( E^\circ_{\text{cell}} = 2.00 \, \text{V} \)
Substituting the values:
\[\ln K_\text{eq} = \frac{5 \times 96500 \times 2.00}{8.314 \times 298}\]
\[\ln K_\text{eq} \approx 778.19\]
Converting to base 10:
\[\log_{10} K_\text{eq} = \frac{\ln K_\text{eq}}{\ln 10} \approx \frac{778.19}{2.303} \approx 337.78\]
Rounding to the nearest integer:
\[x \approx 338\]
Conclusion: The value of \( x \) is approximately 338 or 339.
Approach Solution -2
Given:
The standard reduction potentials for the half-reactions are:
\( \text{MnO}_4^- / \text{Mn}^{2+} \, E^\circ = +1.51 \, \text{V} \)
\( \text{CO}_2 / \text{H}_2\text{C}_2\text{O}_4 \, E^\circ = -0.49 \, \text{V} \)
Step 1: The net standard reduction potential for the reaction is:
\[ E^\circ_{\text{net}} = E^\circ_{\text{MnO}_4^- / \text{Mn}^{2+}} - E^\circ_{\text{CO}_2 / \text{H}_2\text{C}_2\text{O}_4} \] \[ E^\circ_{\text{net}} = 1.51 - (-0.49) = 2.00 \, \text{V} \]
Step 2: Using the Nernst equation to find \( K_{eq} \):
The equation is given by:
\[ K_{eq} = \exp\left(\frac{n F E^\circ_{\text{net}}}{R T}\right) \]
Where:
- \(n = 2\) (number of electrons transferred)
- \( F = 96485 \, \text{C/mol} \) (Faraday constant)
- \( R = 8.314 \, \text{J/molK} \) (gas constant)
- \( T = 298 \, \text{K} \) (temperature)
Substituting the values:
\[ K_{eq} = \exp\left(\frac{2 \times 96485 \times 2.00}{8.314 \times 298}\right) \] \[ K_{eq} = \exp\left(\frac{385940}{2477.572}\right) \] \[ K_{eq} = \exp(155.75) \] \[ K_{eq} \approx 10^{338} \]
Final Answer:
Therefore, \( x = 338 \).
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