Question

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).

Answer 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.