Ionic Equilibrium Question 9
Question 9 - 31 January - Shift 1
The logarithm of equilibrium constant for the reaction $Pd^{2+}+4 Cl^{-} \rightarrow PdCl_4^{2-} \quad$ is
(Nearest integer)
Given: $\frac{2.303 RT}{F}=0.06 V$
$Pd _{(aq)}^{2+}+2 e^{-} \rightarrow Pd(s) \quad E^{\circ}=0.83 V$
$PdCl_4^{2-}(aq)+2 e^{-} \rightarrow Pd(s)+4 Cl^{-}(aq)$
Show Answer
Answer: (6)
Solution:
Formula: Relation between $\Delta G^{o}$ and $lnK_{eq}$
$E^{o}=0.65 V$
$\Delta G^{o}=-RT \ell nK$
$-nFE _{\text{cell }}^{o}=-RT \times 2.303(\log _{10} K)$
$\frac{E _{\text{Cell }}^{0}}{0.06} \times n=\log K$
$Pd^{+2}$ (aq.) $+\not 2 e^{-} \rightarrow Pd(s), E _{\text{cat }, \text{ red }^{n}}^{o}=0.83$
$Pd(s)+4 Cl^{-}($aq. $) \rightarrow PdCl_4^{2-},(aq)+2 e^{-}, E _{\text{Andot.,Odid }}^{O}=0.65$
Net Reaction ->
$ Pd^{2+} \text{ (aq.) }+4 Cl^{-} \text{(aq.) } \rightarrow PdCl_4^{2-} \text{ (aq.) } $
$E _{\text{cell }}^{0}=E _{\text{cat }, \text{ red }^{n}}^{o}-E _{\text{Anode,0xid }} C^{n}$
$E _{\text{cell }}^{o}=0.83-0.65=0.18$
Also $n=2$
Using equation (1), (2) & (3)
