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 cell, we first identify the oxidation and reduction half - reactions. In the reaction \(2\text{Al}+3\text{Sn}^{2+}
ightarrow2\text{Al}^{3+}+3\text{Sn}\), Al is oxidized (loses electrons: \(\text{Al}
ightarrow\text{Al}^{3 +}+3e^-\)) and \(\text{Sn}^{2+}\) is reduced (gains electrons: \(\text{Sn}^{2+}+2e^-
ightarrow\text{Sn}\)). In line notation, the anode (where oxidation occurs) is on the left and the cathode (where reduction occurs) is on the right. The anode is represented as the metal | metal ion, and the cathode is metal ion | metal (or metal | metal ion, but in the case of reduction, the order is such that the reduced species is on the right side of the ||). For the given reaction, Al is oxidized (anode: \(\text{Al}|\text{Al}^{3+}\)) and \(\text{Sn}^{2+}\) is reduced (cathode: \(\text{Sn}^{2+}|\text{Sn}\)). So the line notation should have the anode on the left and cathode on the right, separated by ||. Option A is \(\text{Al}|\text{Al}^{3+}(1.0\ \text{M})||\text{Sn}^{2+}(1.0\ \text{M})|\text{Sn}\), which follows the anode (Al oxidation) - cathode (Sn²⁺ reduction) order. Option B has the anode and cathode reversed in terms of ion - metal order. Option C has Sn as the anode (but Sn is reduced, not oxidized) and Al as the cathode (but Al is oxidized, not reduced). Option D also has the anode and cathode reversed.

Answer:

A