Question

using the activity series, predict if the following reaction will occur, and the chemical formulas of the products that will form.
li(s) + znbr₂(aq) →
○ spontaneous, li₂zn and br
○ spontaneous, libr₂ and zn
○ spontaneous, libr and zn
○ spontaneous, li₂br and zn
○ not spontaneous (no reaction)
activity series of metals
li(s) → li⁺(aq) + e⁻
zn(s) → zn²⁺(aq) + 2e⁻
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question 16
0/4 pts 5 19 details
in the following reaction, how many electrons are exchanged between the reactants?
2al(s) + 3cu²⁺(aq) → 2al³⁺(aq) + 3cu(s)
6
5
3
2
zero, this isnt a redox reaction.

Explanation:

First Question (Reaction Prediction)

Step1: Analyze Activity Series

Li loses 1 e⁻: \( \text{Li}(s)
ightarrow \text{Li}^+(aq) + e^- \) (oxidation). Zn²⁺ gains 2 e⁻: \( \text{Zn}^{2+}(aq) + 2e^-
ightarrow \text{Zn}(s) \) (reduction). Li is more reactive (higher in activity series) than Zn, so Li can displace Zn²⁺ from \( \text{ZnBr}_2 \).

Step2: Balance the Reaction

Oxidation (Li): Multiply by 2 to balance electrons: \( 2\text{Li}(s)
ightarrow 2\text{Li}^+(aq) + 2e^- \).
Reduction (Zn²⁺): \( \text{Zn}^{2+}(aq) + 2e^-
ightarrow \text{Zn}(s) \).
Combine: \( 2\text{Li}(s) + \text{ZnBr}_2(aq)
ightarrow 2\text{LiBr}(aq) + \text{Zn}(s) \). Products are LiBr and Zn, reaction is spontaneous.

Step1: Identify Oxidation/Reduction

Al: \( \text{Al}(s)
ightarrow \text{Al}^{3+}(aq) + 3e^- \) (oxidation, loses 3 e⁻ per Al).
Cu²⁺: \( \text{Cu}^{2+}(aq) + 2e^-
ightarrow \text{Cu}(s) \) (reduction, gains 2 e⁻ per Cu²⁺).

Step2: Balance Electrons

For 2 Al atoms: \( 2\text{Al}(s)
ightarrow 2\text{Al}^{3+}(aq) + 6e^- \) (loses \( 2 \times 3 = 6 \) e⁻).
For 3 Cu²⁺ ions: \( 3\text{Cu}^{2+}(aq) + 6e^-
ightarrow 3\text{Cu}(s) \) (gains \( 3 \times 2 = 6 \) e⁻).
Total electrons exchanged: 6.

Answer:

Spontaneous, \( \text{LiBr} \) and \( \text{Zn} \) (the third option)

Second Question (Electron Exchange)