Question #44314

Consider the standard reduction potentials at $298\text{ K}$ for Group 14 elements \ce{Sn} and \ce{Pb}: $E^\circ(\ce{Sn^{4+}}/\ce{Sn^{2+}}) = +0.15\text{ V}$ $E^\circ(\ce{Pb^{4+}}/\ce{Pb^{2+}}) = +1.69\text{ V}$ $E^\circ(\ce{Sn^{2+}}/\ce{Sn}) = -0.14\text{ V}$ $E^\circ(\ce{Pb^{2+}}/\ce{Pb}) = -0.13\text{ V}$

Two thermodynamic equilibrium constants are defined as follows: $K_1$: Equilibrium constant for the oxidation of \ce{Sn^{2+}(aq)} by \ce{Pb^{4+}(aq)}$.$K_2$: The equilibrium concentration ratio$\frac{[\ce{Sn^{2+}}]}{[\ce{Pb^{2+}}]}$ established when excess finely divided \ce{Sn(s)} and \ce{Pb(s)} metals are added to an aqueous solution containing \ce{Pb^{2+}(aq)} and \ce{Sn^{2+}(aq)} ions.

Assuming $\frac{2.303 RT}{F} = 0.06\text{ V}$, calculate the values of $K_1$ and $K_2$ respectively.