The reducing agent in the reaction 2Li(s) + Fe(CH₃COO)₂(aq) → 2LiCH₃COO(aq) + Fe(s) is lithium (Li).
The general reaction is:
2Li(s) + Fe(CH₃COO)₂(aq) → 2LiCH₃COO(aq) + Fe(s) (1)
We can write the above reaction in <u>two reactions</u>, one for oxidation and the other for reduction:
Li⁰(s) → Li⁺(aq) + e⁻ (2)
Fe²⁺(aq) + 2e⁻ → Fe⁰(s) (3)
We can see that Li⁰ is oxidizing to Li⁺ (by <u>losing</u> one electron) in the lithium acetate (<em>reaction 2</em>) and that Fe²⁺ in iron(II) acetate is reducing to Fe⁰ (by <u>gaining</u> two <em>electrons</em>) (<em>reaction 3</em>).
We must remember that the reducing agent is the one that will be oxidized by <u>reducing another element</u> and that the oxidizing agent is the one that will be reduced by <u>oxidizing another species</u>.
In reaction (1), the<em> reducing agent</em> is <em>Li</em> (it is oxidizing to Li⁺), and the <em>oxidizing agent </em>is<em> Fe(CH₃COO)₂</em> (it is reducing to Fe⁰).
Therefore, the reducing agent in reaction (1) is lithium (Li).
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Answer:
Se detailed explanation.
Explanation:
Hello,
In this case, since both magnesium and calcium ions are in group IIA, we can review the following similar properties:
- Since both calcium and magnesium are in group IIA they have two valence electrons, it means that the both of them have two electrons at their outer shells.
- They are highly soluble in water when forming ionic bonds with nonmetals such as those belonging to halogens and oxygen's family.
- Calcium has 18 electrons and magnesium 10 which are two less than the total protons (20 and 12 respectively) since the both of them have lost two electrons due their ionized form.
- Their electron configurations are:

It means that the both of them are at the
region since it is the last subshell at which their electrons are.
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Answer:
2Ag⁺ (aq) + 2OH⁻ (aq) → Ag₂O (s) + H₂O (l)
Explanation:
Step 1: RxN
2AgNO₃ + 2NaOH → Ag₂O + 2NaNO₃ + H₂O
Step 2: Define states of matter
2AgNO₃ (aq) + 2NaOH (aq) → Ag₂O (s) + 2NaNO₃ (aq) + H₂O (l)
Step 3: Total Ionic Equation
2Ag⁺ (aq) + 2NO₃⁻ (aq) + 2Na⁺ (aq) + 2OH⁻ (aq) → Ag₂O (s) + 2Na⁺ (aq) + 2NO₃⁻ (aq) + H₂O (l)
Step 4: Cancel out spectator ions
2Ag⁺ (aq) + 2OH⁻ (aq) → Ag₂O (s) + H₂O (l)
Answer:It is C i got it correct
Explanation: