Question

what is the balanced chemical reaction for the cell with a potential of 1.52 v?
standard reduction potentials
half-reaction | e° (v)
al³⁺ + 3e⁻ → al | -1.66
sn²⁺ + 2e⁻ → sn | -0.14
a. al + sn²⁺ → al³⁺ + sn
b. 2al + 3sn²⁺ → 2al³⁺ + 3sn
c. al³⁺ + sn → al + sn²⁺
d. 2al³⁺ + 3sn → 2al + 3sn²⁺
enter the answer choice letter.

Explanation:

To determine the balanced reaction, we first identify the anode (oxidation) and cathode (reduction) from standard reduction potentials. A more negative \( E^\circ \) means the species is a stronger reducing agent (undergoes oxidation). \( \text{Al}^{3+} + 3e^-
ightarrow \text{Al} \) has \( E^\circ = -1.66 \, \text{V} \) (more negative), so Al is oxidized (anode: \( \text{Al}
ightarrow \text{Al}^{3+} + 3e^- \)). \( \text{Sn}^{2+} + 2e^-
ightarrow \text{Sn} \) has \( E^\circ = -0.14 \, \text{V} \), so \( \text{Sn}^{2+} \) is reduced (cathode: \( \text{Sn}^{2+} + 2e^-
ightarrow \text{Sn} \)). To balance electrons, multiply oxidation by 2 (\( 2\text{Al}
ightarrow 2\text{Al}^{3+} + 6e^- \)) and reduction by 3 (\( 3\text{Sn}^{2+} + 6e^-
ightarrow 3\text{Sn} \)). Add them: \( 2\text{Al} + 3\text{Sn}^{2+}
ightarrow 2\text{Al}^{3+} + 3\text{Sn} \), which matches option B. The cell potential is \( E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = -0.14 - (-1.66) = 1.52 \, \text{V} \), confirming the reaction.

Answer:

B