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andrew11 [14]
2 years ago
4

Which species in each pair is a better oxidizing agent under standard-state conditions? (a) br2 au3+ (b) h2 ag+ (c) cd2+ cr3+ (d

) o2 in acidic media o2 in basic media?

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

Answer:

(a) Au³⁺;  (b) Ag⁺; (c) Cd²⁺; (d) O₂ in acidic media

Explanation:

You've had Trends in the Periodic Table. Now, here are Trends in Standard Reduction Potentials.

  • The strength of oxidizing agents increases from bottom to top on the left-hand side.
  • The strength of reducing agents increases from top to bottom on the right-hand side.

Thus, for each pair of half reactions, we need to look only at which one has the more positive standard reduction potential.

(a) Br₂/Au³⁺  

Au³⁺(aq) + 3e⁻ ⟶ Au(s)       1.498 V

Br₂(ℓ) + 2e⁻ ⟶ 2Br⁻(aq)      1.066 V

Au³⁺ is the stronger oxidizing agent.

(b) H₂/Ag⁺  

Ag⁺(aq) + e⁻ ⟶ Au(s)          0.7996 V

H₂(g) + 2e⁻ ⟶ 2H⁻(aq)      -2.33      V

Ag⁺ is the stronger oxidizing agent.

(c) Cd²⁺/Cr³⁺  

Cd²⁺(aq) + e⁻ ⟶ Cd(s)      -0.4030 V

Cr³⁺(aq) + 3e⁻ ⟶ Cr(s)      -0.744   V

Cd²⁺ is the stronger oxidizing agent.

(d) O₂, H⁺/O₂, OH⁻  

O₂(g) + 4H⁺(aq) + 4e⁻ ⟶ 2H₂O(ℓ)         1.224 V

O₂(g) + H₂O(ℓ) + 4e⁻ ⟶ 4OH⁻(aq)       0.401  V

O₂ in acid is the stronger oxidizing agent.

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Answer:

<span>23.6 g carbon dioxide comes from 8.6 g of CH4 or 10.7 g carbon dioxide comes from 15.6 g O that means the 15.6 g of oxygen is still the limiting reactant because it gets used up and only makes 10.7 g of CO2. </span>

Explanation:

1) Balanced chemical equation:

CH₄ + 2O₂ → CO₂ + 2H₂O

2) mole ratios:
1 mol CH₄ : 2mol O₂ : 1 mol CO₂ : 2 mol H₂O

3) molar masses
CH₄: 16.04 g/mol
O₂: 32.0 g/mol
CO₂: 44.01 g/mol

4) Convert the reactant masses to number of moles, using the formula 

number of moles = mass in grams / molar mass


CH₄: 8.6g / 16.04 g/mol = 0.5362 moles
<span />

O₂: 15.6 g / 32.0 g/mol = 0.4875 moles

5) If the whole 0.5632 moles of CH₄ reacted that yields to the same number of moles of CO₂ and that is a mass of:
mass of CO₂ = number of moles x molar mass = 23.60 g of CO₂

Which is what the first part of the answer says.

6) If the whole 0.4875 moles of O₂ reacted that would yield 0.4875 / 2 = 0.24375 moles of CO₂, and that is a mass of:
mass of CO₂ = 0.4875 grams x 44.01 g/mol = 10.7 grams of CO₂.

Which is what the second part of the answer says.

7) From the mole ratio you know infere that 0.5362 moles of CH₄ needs more twice number of moles of O₂, that is 1.0724 moles of O₂, and since there are only 0.4875 moles of O₂, this is the limiting reactant.

Which is what the chosen answer says.

8) From the mole ratios 0.4875 moles of O₂ produce 0.4875 / 2 moles of CO₂, and that is:
0.4875 / 2 mols x 44.01 g/mol = 10.7 g of CO₂, which is the last part of the answer.

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Problem One

You are using Nitrogen as your base example. The first thing you should do is fill in the table. Then you should try and make some rules. You need the rules in case the exam you are preparing for picks a different element to talk about these bond tendencies. In any event, it's handy to think this way.

<em><u>Table</u></em>

Bond               Energy Kj/Mol               Bond Length pico meters

N - N                 167                                                145

N=N                  418                                                125

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<em><u>Rules</u></em>

As the number of bonds INCREASES, the energy contained in the bond goes UP

As the number of bonds INCREASES, the length of the bond goes DOWN.

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