Question #44301

A heavy Group 14 metal $\ce{M}$ exhibits two common oxidation states, $+2$ and $+4$. The successive ionization energies of $\ce{M}$ are $IE_1 = 715$, $IE_2 = 1450$, $IE_3 = 3081$, and $IE_4 = 4083 \text{ kJ mol}^{-1}$. Due to the relativistic contraction of the valence s-orbital, one of these oxidation states is thermodynamically more stable at room temperature.

Two $0.1 \text{ M}$ aqueous solutions are prepared: Solution A: Contains the chloride salt of $\ce{M}$ in its most stable oxidation state. Solution B: Contains the chloride salt of $\ce{M}$ in its less stable, higher oxidation state.

The metal ions undergo hydrolysis as follows: $\ce{M^{4+}(aq) + 2H2O(l) <=> M(OH)2^{2+}(aq) + 2H+(aq)} \quad K_{h1} = 1.0 \times 10^{-2}$ $\ce{M^{2+}(aq) + H2O(l) <=> M(OH)+(aq) + H+(aq)} \quad K_{h2} = 1.0 \times 10^{-7}$

A concentration cell is constructed at 298 K: $\ce{Pt(s) | H2(g, 1 bar) | Solution A || Solution B | H2(g, 1 bar) | Pt(s)}$

Calculate the EMF of the cell ($E_{cell}$) at 298 K. (Given: $\frac{2.303 RT}{F} = 0.059 \text{ V}$)