Question

The logarithm of equilibrium constant for the reaction $Pd ^{2+}+4 Cl ^{-} \rightleftharpoons PdCl _4^{2-}$ is (Nearest integer) Given : $\frac{230RT }{ F }=006 V$ $Pd _{\text {(aq) }}^{2+}+2 e ^{-} \rightleftharpoons Pd ( s ) \quad E ^{\ominus}=083 V$ $PdCl _4^{2-} \text { (aq) }+2 e ^{-} \rightleftharpoons Pd ( s )+4 Cl ^{-} \text {(aq) } E ^{\ominus}=065 V$

Hint:

To calculate the equilibrium constant, use the relationship between Gibbs free energy and cell potential, remembering that the Nernst equation can be used for the equilibrium constant.

Answer 1

Correct Answer: 6

The standard Gibbs free energy change is given by:

\[ \Delta G^o = -RT \ln K \]

The equation for the cell potential is:

\[ \text{E}_\text{cell} = \text{E}_\text{cathode} - \text{E}_\text{anode} \]

For the reaction:

\[ \text{Pd}^{2+} (aq) + 2e^- \rightarrow \text{Pd}(s) \quad E^\circ = 0.83\, \text{V} \]

\[ \text{PdCl}_4^{2-} (aq) + 2e^- \rightarrow \text{Pd}(s) + 4\text{Cl}^- (aq) \quad E^\circ = 0.65\, \text{V} \]

The net reaction is:

\[ \text{Pd}^{2+} (aq) + 4\text{Cl}^- (aq) \rightleftharpoons \text{PdCl}_4^{2-} (aq) \]

Using the Nernst equation:

\[ \text{E}_\text{cell} = \text{E}_\text{cathode} - \text{E}_\text{anode} = 0.83 - 0.65 = 0.18\, \text{V} \]

Thus, by using the equation \( \Delta G^o = -nFE^\circ \) and solving for the logarithm of the equilibrium constant \( K \), we get:

\[ \log K = 6 \]

This result shows the relationship between the standard cell potential and the equilibrium constant. A higher \( K \) value indicates a more spontaneous reaction, which can be derived from the Nernst equation.