Question

2al + 3sn²⁺ → 2al³⁺ + 3sn
which of these represents the line notation of the spontaneous cell?
a. al | al³⁺(1.0 m) || sn²⁺(1.0 m) | sn
b. al³⁺(1.0 m) | al || sn | sn²⁺(1.0 m)
c. sn | sn²⁺(1.0 m) || al³⁺(1.0 m) | al
d. sn²⁺(1.0 m) | sn || al | al³⁺(1.0 m)
enter the answer choice letter.

Explanation:

To determine the line notation of a spontaneous electrochemical cell, we first identify the oxidation and reduction half - reactions. In the given reaction \(2\text{Al}+3\text{Sn}^{2 +}
ightarrow2\text{Al}^{3+}+3\text{Sn}\), aluminum (Al) is oxidized (loses electrons: \(\text{Al}
ightarrow\text{Al}^{3+}+ 3e^-\)) and tin(II) ion (\(\text{Sn}^{2+}\)) is reduced (gains electrons: \(\text{Sn}^{2+}+2e^-
ightarrow\text{Sn}\)).

In line notation, the anode (where oxidation occurs) is written on the left, and the cathode (where reduction occurs) is written on the right. The anode is the species being oxidized (Al in this case), so we write the metal (Al) first, followed by its oxidized form (\(\text{Al}^{3+}\)) in solution. The cathode is the species being reduced (\(\text{Sn}^{2+}\) in this case), so we write the reduced form's solution (\(\text{Sn}^{2+}\)) first, followed by the metal (Sn) it is reduced to. The double vertical line (\(\parallel\)) represents the salt bridge.

Looking at the options:

• Option A: \(\text{Al}|\text{Al}^{3+}(1.0\ \text{M})\parallel\text{Sn}^{2+}(1.0\ \text{M})|\text{Sn}\) has Al (anode, oxidized) on the left and \(\text{Sn}^{2+}\) (cathode, reduced) and Sn on the right, which matches the requirements.
• Option B: Has \(\text{Al}^{3+}\) on the left instead of Al (the species being oxidized), so it's incorrect.
• Option C: Has Sn on the left (Sn is the reduced species, should be on the cathode side, not the anode side), so it's incorrect.
• Option D: Has \(\text{Sn}^{2+}\) and Sn on the left (should be on the cathode side) and Al on the right (should be on the anode side), so it's incorrect.

Answer:

A