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Vika [28.1K]
3 years ago
14

Use the standard reduction potentials located in the 'Tables' linked above to calculate the equilibrium constant for the reactio

n:
2H+(aq) + Cu(s) H2(g) + Cu2+(aq)
Hint: Carry at least 5 significant figures during intermediate calculations to avoid round off error when taking the antilogarithm. You may use the OWL references to find the values you may need in this question.
Chemistry
1 answer:
solmaris [256]3 years ago
7 0

Answer:

3.3 * 10^-12

Explanation:

The balanced equation of the reaction is;

2H+(aq) + Cu(s) ---------> H2(g) + Cu2+(aq)

Hence two electrons were transferred so n=2

E°cell = E°cathode - E°anode

E°cell = 0 V - 0.34 V

E°cell = - 0.34 V

Then;

E°cell = 0.0592/n log K

Substituting values;

- 0.34 = 0.0592/2 log K

- 0.34/0.0296 = log K

-11.486 = log K

K = Antilog (-11.486)

K = 3.3 * 10^-12

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Answer:

Partial pressure O₂ → 2.74 atm

Explanation:

Let's analyse the data given:

Volume → 3L

In the bottle there is a mixture of gases that contains, O₂, N₂ and CO₂.

Total pressure is 4.80 atm

Let's apply the Ideal Gases Law to determine the total moles of the mixture

P . V = n .  R. T

4.80 atm . 3L = n . 0.082 . 273K

n = 4.80 atm . 3L / 0.082 . 273K → 0.643 moles

We apply the concept of mole fraction:

Mole fraction of a gas X = moles of gas X / Total moles

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We determine mole fraction of N₂ → 0.230 / 0.643 = 0.357

We determine mole fraction of CO₂ → 0.350 atm / 4.80 atm = 0.0729

1 - mole fraction N₂ - mole fraction CO₂ = mole fraction O₂

1 - 0.357 - 0.0729 = 0.5701 → mole fraction O₂

We replace in the formula: Mole fraction O₂ = Partial pressure O₂ / 4.80 atm

0.5701 . 4.80 atm = Partial pressure O₂ → 2.74 atm

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