The logarithm of equilibrium constant for the reaction {\Pd}^{2+}+4 {\Cl}^{-} ⇌ {\PdCl}_4^{2-} is _______ (Nearest integer) \text{ Given }: {2.303 {\RT}}/{ {\F}}=0.06 {\V} {\Pd}_{( {\aq})}^{2+}+2 {\e}^{-} ⇌ {\Pd}( { s}) {\E}^{\circ}=0.83 {\V} {\PdCl}_4^{2-}( {\aq})+2 {\e}^{-} ⇌ {\Pd}( { s})+4 {\Cl}^{-}( {\aq}) {\E}^{\circ}=0.65 {\V} [31-Jan-2023 Shift 1]

JEE Mains 31-Jan-2023 Shift 1 Solved Paper

Section: Chemistry

Question : 51 of 90

Marks: +1, -0

The logarithm of equilibrium constant for the reaction Pd2++4Cl−⇌PdCl42− is _______ (Nearest integer)
‌‌ Given ‌:‌

2.303‌RT

=0.06V
‌Pd(aq)2++2e−⇌Pd( s)‌‌E∘=0.83V
‌PdCl42−(aq)+2e−⇌Pd( s)+4Cl−(aq)
‌E∘=0.65V

[31-Jan-2023 Shift 1]

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