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QJEEM23Ch79c3: JEE Main 2023 Chemistry - Class 12 JEEM23Ch79c3 Solution: The logarithm of equilibrium constant for the reaction is…

The logarithm of equilibrium constant for the reaction Pd2++4Cl−⇌PdCl42−\mathrm{Pd}^{2+}+4 \mathrm{Cl}^{-} \rightleftharpoons \mathrm{PdCl}_{4}^{2-} is ___________ (Nearest integer)

Given : 2.303R T F=0.06 V\frac{2.303 R \mathrm{~T}}{\mathrm{~F}}=0.06 \mathrm{~V}

Pd(aq)2++2e−⇌Pd(s)E⊖=0.83 V\mathrm{Pd}_{(\mathrm{aq})}^{2+}+2 \mathrm{e}^{-} \rightleftharpoons \mathrm{Pd}(\mathrm{s}) \quad \mathrm{E}^{\ominus}=0.83 \mathrm{~V}

amp;PdCl42−(aq)+2e−⇌Pd(s)+4Cl−(aq)E⊖=0.65 V\begin{aligned}& \mathrm{PdCl}_{4}^{2-}(\mathrm{aq})+2 \mathrm{e}^{-} \rightleftharpoons \mathrm{Pd}(\mathrm{s})+4 \mathrm{Cl}^{-}(\mathrm{aq}) \mathrm{E}^{\ominus}=0.65 \mathrm{~V}\end{aligned}

Step-by-Step Verified Solution

Sol. ΔG∘=−RTℓnK\Delta G^{\circ}=-R T \ell n K

amp;−nFEcell o=−RT×2.303(log⁡10 K)amp;ECell o0.06×n=log⁡K.......(1)amp;Pd+2 (aq.) +2e−⇌Pd(s),Ecat, reductiono=0.83amp;Pd(s)+4Cl−(aq.) ⇌PdCl42−, (aq) +2e−,EAnode, Oxidation 0=0.65\begin{aligned}& -\mathrm{nFE}_{\text {cell }}^{\mathrm{o}}=-\mathrm{RT} \times 2.303\left(\log _{10} \mathrm{~K}\right) \\\\& \frac{\mathrm{E}_{\text {Cell }}^{\mathrm{o}}}{0.06} \times \mathrm{n}=\log \mathrm{K} .......(1) \\\\& \mathrm{Pd}^{+2} \text { (aq.) }+ \mathrm{2e}^{-} \rightleftharpoons \mathrm{Pd}(\mathrm{s}), \mathrm{E}_{\text {cat, }{ reduction}}^{\mathrm{o}}=0.83 \\\\& \mathrm{Pd}(\mathrm{s})+4 \mathrm{Cl}^{-} \text {(aq.) } \rightleftharpoons \mathrm{PdCl}_4^{2-}, \text { (aq) }+2 \mathrm{e}^{-}, \mathrm{E}_{\text {Anode, Oxidation }}^0=0.65\end{aligned}

Net Reaction →Pd2+\rightarrow \mathrm{Pd}^{2+} (aq.) +4Cl−+4 \mathrm{Cl}^{-} (aq.) ⇌PdCl42−\rightleftharpoons \mathrm{PdCl}_4^{2-} (aq.)

amp;Ecell 0=Ecat,red o−EAnded, oxidid 0amp;Ecell o=0.83−0.65amp;Ecell 0=0.18.........(2)\begin{aligned}& \mathrm{E}_{\text {cell }}^0=\mathrm{E}_{\text {cat,red }}^{\mathrm{o}}-\mathrm{E}_{\text {Anded, oxidid }}^0 \\\\& \mathrm{E}_{\text {cell }}^{\mathrm{o}}=0.83-0.65 \\\\& \mathrm{E}_{\text {cell }}^0=0.18 .........(2)\end{aligned}

Also n=2\mathrm{n}=2 .......(3)

Using equation (1), (2) and (3)

log⁡K=6\log K=6

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