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, lower reduction potential) and cathode (reduction, higher reduction potential). Al³⁺ has a lower E° (-1.66 V) than Sn²⁺ (-0.14 V), so Al is oxidized (loses electrons: \( \text{Al}
ightarrow \text{Al}^{3+} + 3e^- \)) and Sn²⁺ is reduced (gains electrons: \( \text{Sn}^{2+} + 2e^-
ightarrow \text{Sn} \)). To balance electrons, multiply oxidation by 2 (total 6 e⁻ lost) and reduction by 3 (total 6 e⁻ gained). This gives \( 2\text{Al} + 3\text{Sn}^{2+}
ightarrow 2\text{Al}^{3+} + 3\text{Sn} \), which matches option B.

Answer:

B