Question #44732

A galvanic cell is constructed at $298\text{ K}$ using Zirconium and Hafnium electrodes immersed in their respective hydroxide suspensions:

$\ce{Zr(s) | Zr(OH)4(s) | Buffer (pH = 2.0) || Buffer (pH = 1.5) | Hf(OH)4(s) | Hf(s)}$

Due to the lanthanoid contraction, $\ce{Zr}$ and $\ce{Hf}$ exhibit nearly identical atomic/ionic radii and chemical properties. Consequently, assume their standard reduction potentials are equal ($E^\circ_{\ce{Zr^{4+}/Zr}} = E^\circ_{\ce{Hf^{4+}/Hf}}$) and their hydroxide solubility products are identical ($K_{sp} = 10^{-52}$). Assuming $\ce{M^{4+}}$ is the only soluble metal species, calculate the cell potential ($E_{cell}$) at $298\text{ K}$.

(Given: $\frac{2.303 RT}{F} = 0.059\text{ V}$)