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

Correct Answer: 6

The correct answer is 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.)
$E_{\text{cell}}^o=E_{\text{cat,red}}^n-E_{\text{Anode,0xid}}^n$
$E_{\text{cell}}^o=0.83-0.65$
$E_{\text{cell}}^o=0.18$...(2)
Also $n=2$.....(3)
Using equation (1), (2) & (3) $\log K=6$

Answer 2

The equilibrium constant, K, for a reaction, can be related to the standard Gibbs free energy change, \(ΔG°\), by the equation:
\(ΔG\degree = -RT\) ln K
where R is the gas constant, T is the temperature in Kelvin, and ln is the natural logarithm.

Using the Nernst equation, we can also relate the standard reduction potentials of the half-reactions to the equilibrium constant:
\(\text{E° cell} = \text{E° reduction (cathode) - E° reduction (anode)}\) \(ΔG° = -nF \text{ }E°cell\)
where n is the number of electrons transferred in the reaction and F is the Faraday constant.

We can use these equations to find the logarithm of the equilibrium constant for the given reaction:
\(Pd^{2+} + 4 Cl^{-} ⇌ PdCl_{4}^{2-}\)

The overall reaction involves the reduction of \(Pd^{2+}\) to \(Pd\), and the oxidation of \(Cl^{-}\) to \(Cl^{2}\). We can write the half-reactions and their standard reduction potentials as follows:
\(Pd^{2+} + 2 e^{-} ⇌ Pd(s) \text{E° reduction} = 0.83 V2 Cl^{-} ⇌ Cl^{2} + 2 e^{-} \text{E° reduction} = -1.36 V\)

The overall cell potential is the sum of these reduction potentials:
\(\text{E° cell} = 0.83 - (-1.36) = 2.19 V\)

We can use the Nernst equation to relate this cell potential to the equilibrium constant:
\(\text{E cell} = \text{E° cell} - \frac{RT}{nF}\)
ln Q where Q is the reaction quotient, which is the concentration of products raised to their stoichiometric coefficients divided by the concentration of reactants raised to their stoichiometric coefficients.

For the given reaction, we have:
\(Q = \frac{[PdCl_{4} ^{2-}]}{([Pd^{2+}][Cl^{-}]^{4})}\)
Substituting the values given in the question, we get:
\(2.19 V = 0.83 V - \frac{2.30RT}{F}\text{ ln K}\)
\(\text{ln K} = \frac{(0.83 - 2.19) V}{\frac{2.30 RT}{F}}\)
\(\text{ ln K} = -0.497\)
Taking the antilogarithm of both sides, we get:
\(K = 0.61\)

Therefore, the nearest integer to the logarithm of the equilibrium constant for the reaction is -1.