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

 To find the logarithm of the equilibrium constant (K) for the reaction $Pd^{2+} + 4Cl^{-} \rightleftharpoons PdCl_4^{2-}$, we use the Nernst equation and given standard electrode potentials.

1. For the overall reaction: $Pd^{2+} + 4Cl^{-} \rightleftharpoons PdCl_4^{2-}$, combine the half-reactions:
   • $Pd^{2+} + 2e^- \rightleftharpoons Pd(s) \quad E^{\ominus} = 0.83 \mathrm{V}$
   • $PdCl_4^{2-} + 2e^- \rightleftharpoons Pd(s) + 4Cl^- \quad E^{\ominus} = 0.65 \mathrm{V}$
2. Calculate the standard electromotive force (EMF) for the reaction:
3. Use the Nernst equation at equilibrium, where the EMF becomes zero:$$E_{cell} = E^{\ominus}_{cell} - \frac{RT}{nF} \ln K = 0$$
4. Rearrange to find K from standard EMF:
5. Given $\frac{230RT}{F} = 0.06 \mathrm{V}$, and using n = 2 for the electrons transferred:
6. Therefore, the logarithm of the equilibrium constant is the nearest integer:
7. The solution fits within the expected range of 6.