nd anode half-reactions. Given:

The standard reduction potential for the half-reaction involving aluminum:

$$ \mathrm{Al}^{3+} + 3\mathrm{e}^- \rightarrow \mathrm{Al} \quad E^\circ = -1.66 \, \mathrm{V} $$

The standard hydrogen electrode (SHE) is used as the reference with a potential of 0 V:

$$ \mathrm{H}^+ + \mathrm{e}^- \rightarrow \frac{1}{2}\mathrm{H}_2 \quad E^\circ = 0 \, \mathrm{V} $$

Since the cell notation provided is:

$$ \mathrm{Al}|\mathrm{Al}_2(\mathrm{SO}_4)_{3(1M)}||\mathrm{HCl}_{(1M)}|\mathrm{H}_{2(g)(1atm)}, \mathrm{Pt} $$

The cell is an Al|H$_2$ cell, where aluminum acts as the anode and hydrogen as the cathode.

Half-reactions:

Anode (oxidation):

$$ \mathrm{Al} \rightarrow \mathrm{Al}^{3+} + 3\mathrm{e}^- \quad (E^\circ = +1.66 \, \mathrm{V} \, \text{since oxidation is the reverse process}) $$

Cathode (reduction):

$$ 2\mathrm{H}^+ + 2\mathrm{e}^- \rightarrow \mathrm{H}_2 \quad (E^\circ = 0 \, \mathrm{V}) $$

Calculate $\mathrm{E}_{\text{cell}}^\circ$:

The cell potential is given by the difference in the reduction potentials of the cathode and the anode:

$$\mathrm{E}_{\text{cell}}^\circ = E_{\text{cathode}}^\circ - E_{\text{anode}}^\circ$$

Substituting the values,

$$\mathrm{E}_{\text{cell}}^\circ = 0 \, \mathrm{V} - (-1.66 \, \mathrm{V}) = 1.66 \, \mathrm{V}$$

Thus, the standard cell potential $\mathrm{E}_{\text{cell}}^\circ$ is:

Option A: 1.66 V

Electrochemistry

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