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DedPeter [7]
2 years ago
14

Determine the reducing agent in the following reaction. Explain your answer. 2 Li(s) + Fe(C2H3O2)2(aq) → 2 LiC2H3O2(aq) + Fe(s)

Chemistry
1 answer:
Brilliant_brown [7]2 years ago
3 0

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:

  • Oxidation reaction

Li⁰(s) → Li⁺(aq) + e⁻   (2)

  • Reduction reaction

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).  

 

Learn more here:

  • brainly.com/question/10547418?referrer=searchResults
  • brainly.com/question/14096111?referrer=searchResults

I hope it helps you!

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Explanation:

Step 1: Given data

  • Initial temperature (T₁): 30 °C
  • Initial volume (V₁): 3.25 L
  • Final temperature (T₂): -10 °C
  • Final volume (V₂): ?

Step 2: Convert the temperatures to Kelvin

We will use the following expression.

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T₁: K = 30°C + 273.15 = 303 K

T₂: K = -10°C + 273.15 = 263 K

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V₁/T₁ = V₂/T₂

V₂ = V₁ × T₂/T₁

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