Question #44865

Consider the Group 10 elements ($\ce{Ni}$, $\ce{Pd}$, $\ce{Pt}$) at $298\text{ K}$. The standard reduction potentials $E^\circ(\ce{M^{2+}/M})$ are $-0.25\text{ V}$, $+0.92\text{ V}$, and $+1.18\text{ V}$, respectively. The sum of the first two ionization enthalpies ($\Delta_{i}H_1 + \Delta_{i}H_2$) is highest for $\ce{Pt}$ due to the Lanthanoid contraction, whereas the sum of the first four ionization enthalpies ($\sum_{j=1}^{4} \Delta_{i}H_j$) is highest for $\ce{Ni}$.

A solution contains $1.0\text{ M}$ each of $\ce{Ni^{2+}}$, $\ce{Pd^{2+}}$, and $\ce{Pt^{2+}}$ in $1.0\text{ M } \ce{HCl}$. Standard hydrogen gas at $1\text{ bar}$ is bubbled through this solution until thermodynamic equilibrium is reached.

Identify the metal(s) that will precipitate from the solution, and the central metal that forms the most thermodynamically stable hexachloro complex $\ce{[MCl_6]^{2-}}$, respectively.