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Potential, Free Energy, and Equilibrium Quiz

12 questions chemistry Grades 9-12

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  1. What was defined as the maximum work that could be performed by a system?

    • Cell potential, E°cell
    • Faraday's constant, F
    • Equilibrium constant, K
    • Standard free energy change, ΔG°
    Reveal answer

    Answer: Standard free energy change, ΔG°

    Source evidence

    PDF page 921: The standard free energy change of a process, ΔG°, was defined in a previous chapter as the maximum work that could be performed by a system, wmax. In the case of a redox reaction taking place within a galvanic cell under standard state conditions, essentially all the work is associated with transferring the electrons from reductant-to-oxidant, welec:

  2. Which equation correctly relates ΔG° to standard cell potential?

    • ΔG° = nF/E°cell
    • ΔG° = −RT ln E°cell
    • ΔG° = −nFE°cell
    • ΔG° = nFE°cell
    Reveal answer

    Answer: ΔG° = −nFE°cell

    Source evidence

    PDF page 921: elec The work associated with transferring electrons is determined by the total amount of charge (coulombs) transferred and the cell potential: ° ΔG° = w = −nFE elec cell ° ΔG° = −nFE cell where n is the number of moles of electrons transferred, F is Faraday’s constant, and E°cell is the standard cell potential. The relation between free energy change and standard cell potential confirms the sign conventions and spontaneity criteria previously discussed for both of these properties: spontaneous redox reactions exhibit positive potentials and negative free energy changes.

  3. In ΔG° = −nFE°cell, what does n represent?

    • Number of half-cells
    • Number of coulombs transferred
    • Number of moles of electrons transferred
    • Number of product molecules
    Reveal answer

    Answer: Number of moles of electrons transferred

    Source evidence

    PDF page 921: elec The work associated with transferring electrons is determined by the total amount of charge (coulombs) transferred and the cell potential: ° ΔG° = w = −nFE elec cell ° ΔG° = −nFE cell where n is the number of moles of electrons transferred, F is Faraday’s constant, and E°cell is the standard cell potential. The relation between free energy change and standard cell potential confirms the sign conventions and spontaneity criteria previously discussed for both of these properties: spontaneous redox reactions exhibit positive potentials and negative free energy changes.

  4. Spontaneous redox reactions exhibit which combination?

    • Positive potential, positive ΔG
    • Negative potential, positive ΔG
    • Negative potential, negative ΔG
    • Positive potential, negative ΔG
    Reveal answer

    Answer: Positive potential, negative ΔG

    Source evidence

    PDF page 921: elec The work associated with transferring electrons is determined by the total amount of charge (coulombs) transferred and the cell potential: ° ΔG° = w = −nFE elec cell ° ΔG° = −nFE cell where n is the number of moles of electrons transferred, F is Faraday’s constant, and E°cell is the standard cell potential. The relation between free energy change and standard cell potential confirms the sign conventions and spontaneity criteria previously discussed for both of these properties: spontaneous redox reactions exhibit positive potentials and negative free energy changes.

  5. Which relation combines ΔG° with K and E°cell?

    • ΔG° = RT ln K = nFE°cell
    • ΔG° = −RT/K = −nFE°cell
    • ΔG° = RT/K = nF/E°cell
    • ΔG° = −RT ln K = −nFE°cell
    Reveal answer

    Answer: ΔG° = −RT ln K = −nFE°cell

    Source evidence

    PDF page 921: ΔG° = −RT ln K = −nFE

  6. Redox reactions with large positive E°cell values will:

    • Proceed far towards completion
    • Not reach equilibrium
    • Remain at initial composition
    • Favor reactants at equilibrium
    Reveal answer

    Answer: Proceed far towards completion

    Source evidence

    PDF page 921: cell ° ⎛RT ⎞ E = ln K cell ⎝nF ⎠ This equation indicates redox reactions with large (positive) standard cell potentials will proceed far towards completion, reaching equilibrium when the majority of reactant has been converted to product. A summary of the relations between E°, ΔG° and K is depicted in Figure 17.7, and a table correlating reaction spontaneity to values of these properties is provided in Table 17.2.

  7. When K > 1, the values of ΔG° and E°cell are:

    • ΔG° > 0, E°cell < 0
    • ΔG° < 0, E°cell > 0
    • ΔG° < 0, E°cell < 0
    • ΔG° > 0, E°cell > 0
    Reveal answer

    Answer: ΔG° < 0, E°cell > 0

    Source evidence

    PDF page 922: K ΔG° E°cell > 1 < 0 > 0 Reaction is spontaneous under standard conditions Products more abundant at equilibrium

  8. When K < 1, the reaction under standard conditions is:

    • Non-spontaneous
    • Complete
    • Spontaneous
    • At equilibrium
    Reveal answer

    Answer: Non-spontaneous

    Source evidence

    PDF page 922: < 1 > 0 < 0 Reaction is non-spontaneous under standard conditions Reactants more abundant at equilibrium

  9. When ΔG° = 0 and E°cell = 0, the value of K is:

    • Equal to zero
    • Greater than 1
    • Less than 1
    • Equal to 1
    Reveal answer

    Answer: Equal to 1

    Source evidence

    PDF page 922: = 1 = 0 = 0 Reaction is at equilibrium under standard conditions Reactants and products equally abundant

  10. When a reaction is at equilibrium under standard conditions, reactants and products are:

    • Reactants more abundant
    • Equally abundant
    • Products more abundant
    • Fully converted
    Reveal answer

    Answer: Equally abundant

    Source evidence

    PDF page 922: = 1 = 0 = 0 Reaction is at equilibrium under standard conditions Reactants and products equally abundant

  11. For 2Ag+ + Fe ⇌ 2Ag + Fe2+, what standard cell potential was calculated?

    • +0.7996 V
    • +0.291 V
    • +1.247 V
    • −0.447 V
    Reveal answer

    Answer: +1.247 V

    Source evidence

    PDF page 922: 2+ − = −0.447 V anode (oxidation): Fe(s) ⟶ Fe (aq) + 2e E° 2+ Fe /Fe + − = 0.7996 V cathode (reduction): 2 × (Ag (aq) + e ⟶ Ag(s)) E° + Ag /Ag E° = E° − E° = = +1.247 V E° + − E° 2+ cell cathode anode Ag /Ag Fe /Fe With n = 2, the equilibrium constant is then

  12. For 2Ag+ + Fe ⇌ 2Ag + Fe2+, the equilibrium constant K was approximately:

    • 10^0.0592
    • 10^-42
    • 10^2
    • 10^42
    Reveal answer

    Answer: 10^42

    Source evidence

    PDF page 923: 0.0592 V E° = log K cell n n × E° /0.0592 V cell K = 10 2 × 1.247 V/0.0592 V K = 10 42.128 K = 10 42

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