Question

The logarithm of equilibrium constant for the reaction $P{d}^{2+}+4C{l}^{-}\rightleftharpoons PdC{l}_4^{2-}$ is (Nearest integer)
Given : $\frac{2.30RT}{F}=0.06V$
$P{d}_{\text{(aq) }}^{2+}+2e^{-}\rightleftharpoons Pd(s)E^{\ominus }=0.83V$
$PdC{l}_4^{2-}\text{ (aq) }+2e^{-}\rightleftharpoons Pd(s)+4C{l}^{-}\text{(aq) }{E}^{\ominus }=0.65V$

Answer 1

Answer: 6

$\Delta G^0=-RT\ell nK$
$-nF{E}_{\text{cell }}^o=-RT\times 2.303\left({\log }_{10}K\right)$
$\frac{E_{\text{Cell }}^0}{0.06}\times n=\log K$....(1)
$P{d}^{+2}\text{ (aq.) }+/e^{-}\rightleftharpoons Pd(s),E_{\text{cat,red }}^n=0.83$
$Pd(s)+4C{l}^{-}(aq.)\rightleftharpoons PdC{l}_4^{2-},(aq)+2e^{-},E_{\text{Anode, oxiad }}^n=0.65$
Net Reaction $\to P{d}^{2+}$ (aq.) $+4C{l}^{-}$(aq.) $\rightleftharpoons PdC{l}_4^{2-}$ (aq.)
$E_{\text{cell }}^o=E_{{\text{cat,red}}^n}^{\circ }-E_{{\text{Anode,oxid}}^n}^{\circ }$
$E_{\text{cell }}^o=0.83-0.65$
$E_{\text{cell }}^o=0.18$...(2)
Also $n=2$.....(3)
Using equation (1), (2) & (3) $\log K=6$