JEE Mains · Chemistry · STD 12 - 2. Electrochemistry
The logarithm of equilibrium constant for the reaction \(Pd ^{2+}+4 Cl ^{-} \rightleftharpoons PdCl _4^{2-}\) is (Nearest integer) Given \(: \frac{2.303 RT }{ F }=0.06 V\) \(Pd _{( aq )}^{2+}+2 e ^{-} \rightleftharpoons Pd ( s ) \quad E ^{\circ}=0.83\,V\) \(PdCl _4^{2-}( aq )+2 e ^{-} \rightleftharpoons Pd ( s )+4 Cl ^{-}( aq )\) \(E ^{\circ}=0.65\,V\)
- A \(3\)
- B \(4\)
- C \(12\)
- D \(6\)
Answer & Solution
Correct Answer
(D) \(6\)
Step-by-step Solution
Detailed explanation
\(\Delta G ^{\circ}=- RT \ell nK\) \(- nFE\) \(\text { cell }=- RT \times 2.303\left(\log _{10} K \right)\) \(\frac{ E _{ Cell }^{ o }}{0.06} \times n =\log K\) \(Pd ^{+2}( aq .)+ 2 e ^{-} \rightleftharpoons Pd ( s ), E _{\text {cat,red }}^{ o }=0.83\)…
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