The logarithm of equilibrium constant for the reaction Pd ^ 2 + + 4...

Given, $\dfrac{2 . 303RT}{F} = 0 . 06V$

Using gibbs free energy equation,

$\Delta G ° = - RTlnK$

$- {nFE}_{\text{cell}}^{o} = - RT \times 2 . 303\left(\log_{10}K\right)$

${E}_{\text{Cell}}^{0}0 . 06 \times n = logK - - - - - - - - - ( i )$

${Pd}^{2 +} ( aq ) + 2{e}^{-} ⇌ Pd ( s ) , {E}_{\text{cathode, reduction}}^{0} = 0 . 83V$

$Pd ( s ) + 4{Cl}^{-} ( aq . ) ⇌ {PdCl}_{4}^{2 -} ( aq ) + 2{e}^{-}, {E}_{\text{anode, oxidation}}^{o} = 0 . 65V$

Net Reaction $\to {Pd}^{2 +} ( aq . ) + 4{Cl}^{-} ( aq . ) ⇌ {{PdCl}_{4}}^{2 -} ( aq . )$

${E}_{\text{cell}}^{o} = {E}_{\text{cathode,reduction}}^{o} - {E}_{\text{anode, oxidation}}^{o}$

$⇒ {E}_{cell}^{o} = 0 . 83 - 0 . 65 ⇒ {E}_{cell}^{o} = 0 . 18 - - - - - - - - ( ii )$

Also $n = 2 - - - - - - - - ( iii )$

Using equation $( i ) , ( ii )$& $( iii )$, we get

${E}_{\text{Cell}}^{0}0 . 06 \times n = logK$

$⇒ \dfrac{0 . 18}{0 . 06} \times 2 = logK$

$⇒ logK = 6$