AP Chemistry · Topic 9.8
Cell Potential and Free Energy Practice
Part of Applications of Thermodynamics.(ENE-6.B)
Practice questions
24
Sample questions
5 of 24 — sign in to practice the rest with adaptive difficulty and mastery tracking.
Sample 1difficulty 1/5
A student constructs a Zn|Zn^2+(1.0 M) || Cu^2+(1.0 M)|Cu galvanic cell with a KNO3 salt bridge and reads the voltage with a high-impedance voltmeter. Standard reduction potentials: Cu^2+/Cu = +0.34 V; Zn^2+/Zn = -0.76 V. The measured cell voltage at 25 C is 1.08 V.
What is the standard cell potential E_cell^o, and how does the measurement compare?
- A
E^o = +1.10 V; measured value should be exactly 1.10 V
- B
E^o = -1.10 V; cell is non-spontaneous
- C
E^o = +0.42 V; measurement is too high
- Dcheck_circle
E^o = +1.10 V; measured (1.08 V) is slightly low, consistent with internal resistance / non-ideal conditions
Why
E^o_cell = E^o_cath - E^o_an = 0.34 - (-0.76) = +1.10 V. Measured 1.08 V is within typical experimental error.
- A
Sample 2difficulty 2/5
Which species is the strongest reducing agent shown?
- A
Au(s)
- B
Pb(s)
- C
Cu(s)
- Dcheck_circle
Mg(s)
Why
The most negative E° corresponds to the species most easily oxidized; Mg → Mg²⁺ has E°ox = +2.37 V, so Mg is the best reducing agent.
- A
Sample 3difficulty 3/5
Adding more Cu²⁺ to a Zn–Cu cell at standard conditions causes E to
- A
Decrease
- B
Drop to 0
- C
Stay at E°
- Dcheck_circle
Increase
Why
Higher [Cu²⁺] makes Q smaller; Nernst correction adds to E°.
- A
Sample 4difficulty 3/5
Compute ΔG° in kJ.
- A
+212 kJ
- B
-1.10 kJ
- C
-106 kJ
- Dcheck_circle
-212 kJ
Why
ΔG° = -(2)(96485)(1.10) ≈ -212270 J ≈ -212 kJ.
- A
Sample 5difficulty 3/5
Cu²⁺ + 2e⁻ → Cu E° = +0.34 V Zn²⁺ + 2e⁻ → Zn E° = -0.76 V
For the galvanic cell Zn|Zn²⁺ || Cu²⁺|Cu, what is E°cell?
- Acheck_circle
+1.10 V
- B
-1.10 V
- C
-0.42 V
- D
+0.42 V
Why
E°cell = E°cathode - E°anode = (+0.34) - (-0.76) = +1.10 V.
- A