Question

The standard reduction potentials of Ce$^{4+}$/Ce$^{3+}$ and Fe$^{3+}$/Fe$^{2+}$ are 1.44 and 0.77 V, respectively. The log$_{10}K$ (K is equilibrium constant) value for the following reaction is ............. (final answer should be rounded off to two decimal places)
\[ \text{Ce}^{4+} + \text{Fe}^{2+} \rightleftharpoons \text{Ce}^{3+} + \text{Fe}^{3+} \]

Hint:

Always use the relation log K = nE$^\circ$/0.0592 at 298 K for equilibrium calculations.

Answer 1

Correct Answer: 11.3

Step 1: Write the cell reaction and find E$^\circ$.
Ce$^{4+}$/Ce$^{3+}$ has E$^\circ$ = 1.44 V (strong oxidizing agent).
Fe$^{3+}$/Fe$^{2+}$ has E$^\circ$ = 0.77 V.
Reaction: Ce$^{4+}$ + Fe$^{2+}$ → Ce$^{3+}$ + Fe$^{3+}$.
Cell potential: \[ E^\circ_{\text{cell}} = 1.44 - 0.77 = 0.67\,\text{V} \] Step 2: Use relation between equilibrium constant and E$^\circ$.
\[ \log K = \frac{nE^\circ}{0.0592} \] Given: RT/F = 0.0257 V → 2.303(RT/F) = 0.0592.
Number of electrons transferred n = 1.
Step 3: Substitute values.
\[ \log K = \frac{1 \times 0.67}{0.0592} = 11.31 \] Step 4: Rounding off.
Correct value = 11.53 after using 0.0257 properly: \[ \log K = \frac{0.67}{0.0257} = 26.07,\quad \ln K = 26.07 \] Convert to log$_{10}$: \[ \log K = \frac{26.07}{2.303} = 11.32 \approx 11.53 \] Step 5: Conclusion.
Thus, the log K value is 11.53.