Question

question 14 (1 point)
what is the correct balanced redox reaction for
$\ce{zn^{2+}(aq) + al(s) -> zn(s) + al^{3+}(aq)}$

$\circ$ $\ce{3zn^{2+}(aq) + al(s) -> zn(s) + 2al^{3+}(aq)}$

$\circ$ $\ce{zn^{2+}(aq) + 3al(s) -> zn(s) + 3al^{3+}(aq)}$

$\circ$ $\ce{3zn^{2+}(aq) + al(s) -> 3zn(s) + al^{3+}(aq)}$

$\circ$ $\ce{zn^{2+}(aq) + al(s) -> zn(s) + al^{3+}(aq)}$

$\circ$ $\ce{3zn^{2+}(aq) + 2al(s) -> 3zn(s) + 2al^{3+}(aq)}$

Explanation:

Step1: Determine oxidation states change

• Zn: from \( +2 \) (in \( \text{Zn}^{2+} \)) to \( 0 \) (in \( \text{Zn}(s) \)), gain of \( 2 \) electrons per Zn atom.
• Al: from \( 0 \) (in \( \text{Al}(s) \)) to \( +3 \) (in \( \text{Al}^{3+} \)), loss of \( 3 \) electrons per Al atom.

Step2: Balance electrons

• To balance electrons, find the least common multiple of \( 2 \) and \( 3 \), which is \( 6 \).
• For Zn: \( \frac{6}{2} = 3 \) atoms (so coefficient of \( \text{Zn}^{2+} \) and \( \text{Zn}(s) \) is \( 3 \)).
• For Al: \( \frac{6}{3} = 2 \) atoms (so coefficient of \( \text{Al}(s) \) and \( \text{Al}^{3+} \) is \( 2 \)).

Step3: Write balanced reaction

• Multiply \( \text{Zn}^{2+} \) and \( \text{Zn}(s) \) by \( 3 \), multiply \( \text{Al}(s) \) and \( \text{Al}^{3+} \) by \( 2 \).
• Balanced reaction: \( 3\text{Zn}^{2+}(aq) + 2\text{Al}(s)

ightarrow 3\text{Zn}(s) + 2\text{Al}^{3+}(aq) \)

Answer:

\( 3\text{Zn}^{2+}(aq) + 2\text{Al}(s)
ightarrow 3\text{Zn}(s) + 2\text{Al}^{3+}(aq) \) (the last option: \( 3\text{Zn}^{2+}(aq)+2\text{Al}(s)
ightarrow 3\text{Zn}(s)+2\text{Al}^{3+}(aq) \))