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djverab [1.8K]
3 years ago
10

CHEGG Consider the following reaction: LaTeX: Fe\left(CN\right)_6^{3-}+e^-\longrightarrow Fe\left(CN\right)_6^{4-}F e ( C N ) 6

3 − + e − ⟶ F e ( C N ) 6 4 − From the cyclic voltammetry data, a cathodic peak potential (Epa) of 0.271 V and an anodic peak potential (Epc) of 0.330 V was obtained. Is the reaction reversible?
Chemistry
1 answer:
sladkih [1.3K]3 years ago
4 0

Answer:

E°= E°cathode- E° anode= 0.271-0.330= -0.59V

Explanation:

NB: the stoichiometry does not affect E°values,

And the more positive the E° values , the greater it's tendency to become spontaneous and hence irreversible, and the more negative the E° values the more likely to become less spontaneous and reversible, hence the above reaction is reversible

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What is the theoretical yield of fluorenone if you oxidize 175 mg of fluorene?
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Answer:

  • 602 mg of CO₂ and 94.8 mg of H₂O

Explanation:

The<em> yield</em> is measured by the amount of each product produced by the reaction.

The chemical formula of <em>fluorene</em> is C₁₃H₁₀, and its molar mass is 166.223 g/mol.

The <em>oxidation</em>, also know as combustion, of this hydrocarbon is represented by the following balanced chemical equation:

        2C_{13}H_{10}+31O_2\rightarrow 26CO_2+10H_2O

To calculate the yield follow these steps:

<u>1. Mole ratio</u>

          2molC_{13}H_{10}:31molO_2:26molCO_2:10molH_2O

<u />

<u>2. Convert 175mg of fluorene to number of moles</u>

  • 175mg/times 1g/1,000mg=0.175g

  • Number of moles = mass in grams / molar mass

  • \text{number of moles}=0.175g/166.223g/mol=0.0010528mol

<u>3. Set a proportion for each product of the reaction</u>

a) <u>For CO₂</u>

i) number of moles

         2molC_{10}H_{13}/26molCO_2=0.0010528molC_{10}H{13}/x

x=0.0010528molC_{10}H_{13}\times 26molCO_2/2molC_{10}H_{13}=0.013686molCO_2

ii) mass in grams

The molar mass of CO₂ is 44.01g/mol

  • mass = number of moles × molar mass
  • mass = 0.013686 moles × 44.01 g/mol = 0.602 g = 602mg

b) <u>For H₂O</u>

i) number of moles

0.0010528molC_{10}H_{13}\times10molH_2O/2molC_{10}H_{13}=0.00526molH_2O

ii) mass in grams

The molar mass of H₂O is 18.015g/mol

  • mass = number of moles × molar mass
  • mass = 0.00526 moles × 18.015 g/mol = 0.0948mg = 94.8 mg
4 0
3 years ago
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