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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The kinetic energy of an object increases when heat is added.
Answer:
2H⁺(aq) + 2OH⁻(aq) --> 2H2O(l)
Explanation:
2HBr(aq)+Ba(OH)2(aq)⟶2H2O(l)+BaBr2(aq)
We break the compounds into ions. Only compounds in the aqueous form can be turned into ions.
The ionic equation is given as;
2H⁺(aq) + 2Br⁻(aq) + Ba²⁺(aq) + 2OH⁻(aq) --> 2H2O(l) + Ba²⁺(aq) + 2Br⁻(aq)
Upon eliminating the spectator ions; The net equation is given as;
2H⁺(aq) + 2OH⁻(aq) --> 2H2O(l)
1) concentration or partial pressure of species
involved. 2) temperature • 3) presence of catalyst
4) nature of reactants.
Answer:
what are the answer choices?
Explanation: