19.2.6 Electrochemicamical Determinants of Equilibria Flashcards

1
Q

Electrochemical Determinants of Equilibria

A
  • At equilibrium, the Nernst equation reduces to E ̊ cell = (R · T) / (n · F) · lnK eq .
  • This relationship can be used to calculate equilibrium constants (K eq ) based on electrochemical data.
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2
Q

note

A
  • At equilibrium, there is no driving force for a redox reaction to proceed in either direction, so the cell voltage (E cell ) is zero. This can be used to simplify the Nernst equation to E ̊ cell – (R · T) / (n · F) · lnQ. Since Q = K eq at equilibrium, this equation can be used to calculate K eq , given E ̊ cell .
  • One application of this is in determining the solubility product constant (K sp ) of a compound.
  • First, E ̊ cell must be calculated by breaking the reaction into half-reactions. Once this calculation has been made, finding K sp is simply a matter of plugging the numbers into the formula.
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3
Q

Which would best determine the equilibrium constant inside the cells of a car battery?

A

[H + ] and [HSO4− ]

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4
Q

Which best describes the reaction that occurs when a car battery discharges?

A

Oxidation-reduction

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5
Q

Which of the following is the oxidation half-reaction when a battery is discharging?

A

Pb(s) + HSO4− (aq) → PbSO4(s) + H + (aq) + 2e−

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6
Q

Which type of reaction occurs inside the cells of a car battery?

A

Oxidation

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7
Q

When does equilibrium occur in a battery?

A

Ecell = 0

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8
Q

Which best describes the relationship between the reaction quotient and the voltage of a car battery?

A

↓ Q: ↑ battery potential

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9
Q

Which of the following types of data have a triangular relationship?

A

Composition, electrochemical, and calorimetric

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10
Q

Which best represents an oxidation half-reaction?

A

Ag → Ag+ + e−

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11
Q

Which trends best describes what happens when a car battery discharges?

A

↓[H+] and ↓[HSO4-]

↑Q:↓ battery potential

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12
Q

How many half-reactions occur when a car battery discharges?

A

Two

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13
Q

Calculate the equilibrium constant for the following reaction at 25°C, given that
E°cell = +0.99 V

3Cu(s) + Cr2O72-(aq) + 14H+(aq) –> 3Cu2+(aq) + 2Cr3+(aq) + 7H2O(l)

A

2.18 × 10^100

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