The logarithm of equilibrium constant for the reaction Pd 2 + + 4 Cl - ⇌ PdCl 4 2 - is (Nearest integer) Given: 2 . 303 RT F = 0 . 06 V Pd ( aq ) 2 + + 2 e - ⇌ Pd ( s ) E o = 0 . 83 V PdCl 4 2 - ( aq ) + 2 e - ⇌ Pd ( s ) + 4 Cl - ( aq ) E o = 0 . 65 V

Step-by-step solution

Given, 2 . 303 RT F = 0 . 06   V Using gibbs free energy equation, ΔG ° = - RT   lnK - nFE cell  o = - RT × 2 . 303 log 10   K E Cell 0 0 . 06 × n = log   K - - - - - - - - - ( i ) Pd 2 +   ( aq )   +   2 e - ⇌   Pd ( s ) ,   E cathode, reduction  0   =   0 . 83   V Pd   ( s ) + 4 Cl -   ( aq . )   ⇌   PdCl 4 2 -   ( aq )   +   2 e - ,   E anode, oxidation o