Question

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

a) $3mathrm{zn^{2+}(aq) + al(s) \
ightarrow zn(s) + 2al^{3+}(aq)}$

b) $mathrm{zn^{2+}(aq) + al(s) \
ightarrow zn(s) + al^{3+}(aq)}$

c) $mathrm{zn^{2+}(aq) + 3al(s) \
ightarrow zn(s) + 3al^{3+}(aq)}$

d) $3mathrm{zn^{2+}(aq) + al(s) \
ightarrow 3zn(s) + al^{3+}(aq)}$

e) $3mathrm{zn^{2+}(aq) + 2al(s) \
ightarrow 3zn(s) + 2al^{3+}(aq)}$

Explanation:

Step1: Determine oxidation states change

• Zinc: \( \text{Zn}^{2+} \) (oxidation state +2) to \( \text{Zn}(s) \) (oxidation state 0). Gain of 2 electrons per Zn atom.
• Aluminum: \( \text{Al}(s) \) (oxidation state 0) to \( \text{Al}^{3+} \) (oxidation state +3). Loss of 3 electrons per Al atom.

Step2: Balance electrons transferred

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

Step3: Write balanced reaction

• Multiply \( \text{Zn}^{2+} \) and \( \text{Zn}(s) \) by 3, \( \text{Al}(s) \) 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:

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