Determine the standard cell potential, E^ext{°}_{ext{cell}}, for the reaction: Al(s) + 3 Ag^+(aq) → 3 Ag(s) + Al^{3+}(aq). Given: E^ext{°}_{ext{Al}^{3+}/ext{Al}} = -1.66 ext{ V}, E^ext{°}_{ext{Ag}^+/ext{Ag}} = +0.80 ext{ V}.

• A. 0.86 V
• B. -2.46 V
• C. -0.86 V
• D. 2.46 V

Answer: (D)

Verified step by step guidance

1. Identify the half-reactions involved in the overall redox reaction. The oxidation half-reaction is aluminum metal being oxidized to aluminum ion: \(\mathrm{Al(s) \rightarrow Al^{3+}(aq) + 3e^-}\), and the reduction half-reaction is silver ion being reduced to silver metal: \(\mathrm{Ag^+(aq) + e^- \rightarrow Ag(s)}\).
2. Write down the standard reduction potentials for each half-reaction as given: \(E^\circ_{\mathrm{Al^{3+}/Al}} = -1.66\ \mathrm{V}\) and \(E^\circ_{\mathrm{Ag^+/Ag}} = +0.80\ \mathrm{V}\). Note that the aluminum half-reaction is given as a reduction potential, but in the overall reaction, aluminum is oxidized, so its sign will be reversed when calculating the cell potential.
3. Calculate the standard cell potential \(E^\circ_{\mathrm{cell}}\) using the formula: \(E^\circ_{\mathrm{cell}} = E^\circ_{\mathrm{cathode}} - E^\circ_{\mathrm{anode}}\) where the cathode is the reduction half-reaction (silver) and the anode is the oxidation half-reaction (aluminum).
4. Substitute the values into the formula: \(E^\circ_{\mathrm{cell}} = (+0.80\ \mathrm{V}) - (-1.66\ \mathrm{V})\) This accounts for the fact that aluminum is oxidized (reverse of the reduction potential).
5. Interpret the result: a positive \(E^\circ_{\mathrm{cell}}\) indicates a spontaneous reaction under standard conditions. The magnitude of the cell potential reflects the driving force of the redox reaction.