$Pd ^{2+}+4 Cl ^{-} \rightleftharpoons PdCl _4^{2-}$ પ્રક્રિયા માટે સંતુલન અચળાંકનો લધુગણક $...........$ (નજીકનો પૂર્ણાંક)

આપેલ $: \frac{2.303 RT }{ F }=0.06 V$

$Pd _{( aq )}^{2+}+2 e ^{-} \rightleftharpoons Pd ( s ) \quad E ^{\circ}=0.83\,V$

$PdCl _4^{2-}( aq )+2 e ^{-} \rightleftharpoons Pd ( s )+4 Cl ^{-}( aq )$

$E ^{\circ}=0.65\,V$

Question

$Pd ^{2+}+4 Cl ^{-} \rightleftharpoons PdCl _4^{2-}$ પ્રક્રિયા માટે સંતુલન અચળાંકનો લધુગણક $...........$ (નજીકનો પૂર્ણાંક)

આપેલ $: \frac{2.303 RT }{ F }=0.06 V$

$Pd _{( aq )}^{2+}+2 e ^{-} \rightleftharpoons Pd ( s ) \quad E ^{\circ}=0.83\,V$

$PdCl _4^{2-}( aq )+2 e ^{-} \rightleftharpoons Pd ( s )+4 Cl ^{-}( aq )$

$E ^{\circ}=0.65\,V$

JEE MAIN 2023, Medium

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Solution

d
$\Delta G ^{\circ}=- RT \ell nK$

$- nFE$

$\text { cell }=- RT \times 2.303\left(\log _{10} K \right)$

$\frac{ E _{ Cell }^{ o }}{0.06} \times n =\log K$

$Pd ^{+2}( aq .)+ 2 e ^{-} \rightleftharpoons Pd ( s ), E _{\text {cat,red }}^{ o }=0.83$

$\left. Pd ( s )+4 Cl ^{-} \text {(aq. }\right) \rightleftharpoons PdCl _4^{2-},( aq )+2 e ^{-}, E _{\text {Anode,Oxid }}^0=0.65$

Net Reaction $\rightarrow Pd ^{2+}$ (aq.) $+4 Cl ^{-}$(aq.) $\rightleftharpoons PdCl _4^{2-}$ (aq.)

$E _{\text {cell }}^{\circ}= E _{\text {cat,reed }}^0- E _{\text {Andode,oxid }}^0$

$E _{\text {cell }}^{\circ}=0.83-0.65$

$E _{\text {cell }}^{\circ}=0.18$

Also $n =2$

Using equation $(1),(2)$ and $(3)$ $\log K=6$

સૂચિ - $I$	સૂચિ - $II$
$(A)$ $Cd ( s )+2 Ni ( OH )_{3}( s ) \rightarrow CdO ( s )+2 Ni ( OH )_{2}( s )+ H _{2} O (l)$	$(I)$ પ્રાથમિક બેટરી
$(B)$ $Zn ( Hg )+ HgO ( s ) \rightarrow ZnO ( s )+ Hg (l)$	$(II)$ દ્વિતિયક બેટરી (કોષ) નું ડિસચાર્જિંગ
$(C)$ $2 PbSO _{4}( s )+2 H _{2} O (l) \rightarrow Pb ( s )+ PbO _{2}( s )+ 2 H _{2} SO _{4}( aq )$	$(III)$ બળતણા (ઈંઘણ) કોષ
$(D)$ $2 H _{2}( g )+ O _{2}( g ) \quad \rightarrow 2 H _{2} O (l)$	$(IV)$ દ્વિતિયક બેટરીનું ચાર્જિંગ

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