The logarithm of equilibrium constant for the reaction Pd2+ + 4Cl–⇌ 4 PdCl is ____________(Nearest integer) Given : V . F RT . ) aq ( Pd + 2e– ⇌ Pd(s) E– = 0.83V 4 PdCl (aq)+ 2e– ⇌ Pd(s) +4Cl–(aq) E–

Question

The logarithm of equilibrium constant for the reaction Pd2+ + 4Cl–⇌ 4 PdCl is ____________(Nearest integer) Given : V . F RT . ) aq ( Pd + 2e– ⇌ Pd(s) E– = 0.83V 4 PdCl (aq)+ 2e– ⇌ Pd(s) +4Cl–(aq) E– = 0.65V NTA (6) RESO. (6)

Prashant Kotian · Accepted Answer

Given reaction Pd2+ + 4Cl– ⇌ [PdCl4]2– Eºcell = o Pd | Pd2 E  – o Pd | ) PdCl ( 4 E  = 0.83 – 0.65 = 0.18 Eºcell = 06 . log Keq 0.18 = 06 . log Keq log Keq = 6

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