Question

The logarithm of equilibrium constant for the reaction $Pd ^{2+}+4 Cl ^{-} \rightleftharpoons PdCl _4^{2-}$ is (Nearest integer) Given : $\frac{230RT }{ F }=006 V$ $Pd _{\text {(aq) }}^{2+}+2 e ^{-} \rightleftharpoons Pd ( s ) \quad E ^{\ominus}=083 V$ $PdCl _4^{2-} \text { (aq) }+2 e ^{-} \rightleftharpoons Pd ( s )+4 Cl ^{-} \text {(aq) } E ^{\ominus}=065 V$

Answer 1

The correct answer is 6
ΔG0=−RTℓnK
−nFEcell o​=−RT×2.303(log10​K)
0.06ECell 0​​×n=logK....(1)
Pd+2 (aq.) +e−⇌Pd(s),Ecat,red n​=0.83
Pd(s)+4Cl−(aq.)⇌PdCl42−​,(aq)+2e−,EAnode, oxiad n​=0.65
Net Reaction →Pd2+ (aq.) +4Cl−(aq.) ⇌PdCl42−​ (aq.)
Ecell o​=Ecat,red n​−EAnode,0xid n​
Ecell o​=0.83−0.65
Ecell o​=0.18...(2)
Also n=2.....(3)
Using equation (1), (2) & (3) logK=6