Question

The logarithm of equilibrium constant for the reaction $Pd ^{2+}+4 Cl ^{-} \rightleftharpoons PdCl _4^{2-}$ is (Nearest integer)
Given : $\frac{2.30RT }{ F }=0.06 V$
$Pd _{\text {(aq) }}^{2+}+2 e ^{-} \rightleftharpoons Pd ( s ) \quad E ^{\ominus}=0.83 V$
$PdCl _4^{2-} \text { (aq) }+2 e ^{-} \rightleftharpoons Pd ( s )+4 Cl ^{-} \text {(aq) } E ^{\ominus}=0.65 V$

Answer 1

$ \Delta G ^0=- RT \ell nK $
$ - nFE _{\text {cell }}^o=- RT \times 2.303\left(\log _{10} K \right) $
$ \frac{ E _{\text {Cell }}^0}{0.06} \times n =\log K $....(1)
$ Pd ^{+2} \text { (aq.) }+\not \not e ^{-} \rightleftharpoons Pd ( s ), E _{\text {cat,red }}^n=0.83$
$Pd ( s )+4 Cl ^{-}( aq .) \rightleftharpoons PdCl _4^{2-},( aq )+2 e ^{-}, E _{\text {Anode, oxiad }}^n=0.65$
Net Reaction $\rightarrow Pd ^{2+}$ (aq.) $+4 Cl ^{-}$(aq.) $\rightleftharpoons PdCl _4^{2-}$ (aq.)
$ E _{\text {cell }}^{ o }= E _{\text {cat,red}^n }^{ \circ }- E _{\text {Anode,oxid}^n }^{ \circ } $
$ E _{\text {cell }}^{ o }=0.83-0.65$
$ E _{\text {cell }}^{ o }=0.18$...(2)
Also $n =2$.....(3)
Using equation (1), (2) & (3) $\log K =6$