AP Chemistry · Topic 9.9
Cell Potential Under Nonstandard Conditions Practice
Part of Applications of Thermodynamics.(ENE-6.C)
Practice questions
17
Sample questions
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Sample 1difficulty 2/5
A student builds a Cu concentration cell with two Cu electrodes immersed in 0.0010 M and 1.00 M Cu(NO3)2 with a salt bridge. At 25 C the measured cell potential is +0.088 V (left half-cell positive). E^o for the same metal in both half-cells is 0 V; the Nernst equation gives E_cell = -(0.0592/2) * log(Q).
What cell potential is predicted by the Nernst equation, and how does it compare to the measurement?
- Acheck_circle
E_calc = +0.089 V; agrees with the +0.088 V measurement
- B
E_calc = +0.18 V; cell is non-ideal
- C
E_calc = 0 V; the cell does not operate
- D
E_calc = -0.089 V; cell is reversed
Why
Q = [Cu^2+]_dilute / [Cu^2+]_conc = 0.001/1.00 = 1e-3. E = -(0.0592/2) log(1e-3) = -(0.0296)(-3) = +0.0888 V. Matches measurement.
- A
Sample 2difficulty 3/5
For Q > 1, why does the Nernst plot show E decreasing as T increases?
- A
Because Faraday's constant decreases
- Bcheck_circle
Because (RT/nF) ln Q grows with T when Q > 1
- C
Because E° decreases with T
- D
Because n increases with T
Why
With Q>1, ln Q>0 and (RT/nF)·ln Q rises with T, so E = E° − (RT/nF)ln Q falls.
- A
Sample 3difficulty 3/5
A galvanic cell uses the half-reactions Cu²⁺/Cu and Zn²⁺/Zn (E°_cell = +1.10 V). A student varies [Cu²⁺] while keeping [Zn²⁺] = 1.0 M.
These data are best explained by which trend?
- A
E is +1.10 V regardless of Q.
- B
E is independent of concentrations.
- C
E increases as [Cu²⁺] decreases.
- Dcheck_circle
E decreases as Q increases ([Cu²⁺] decreases pushes Q higher).
Why
Q = [Zn²⁺]/[Cu²⁺]. Lower [Cu²⁺] increases Q, and per Nernst equation E = E° − (0.0592/n)log Q, E drops as Q rises. Each ten-fold drop reduces E by ~0.030 V (n = 2), matching the data.
- A
Sample 4difficulty 3/5
A dead battery has
- A
E doubled
- Bcheck_circle
E = 0 (Q = K)
- C
E ≪ 0
- D
E = E°
Why
At equilibrium, Q = K and E = 0 → no driving force.
- A
Sample 5difficulty 3/5
A student builds a Cu concentration cell with two Cu electrodes immersed in 0.0010 M and 1.00 M Cu(NO3)2 with a salt bridge. At 25 C the measured cell potential is +0.088 V (left half-cell positive). E^o for the same metal in both half-cells is 0 V; the Nernst equation gives E_cell = -(0.0592/2) * log(Q).
Which electrode is the cathode?
- A
the Cu electrode in 0.0010 M Cu^2+
- Bcheck_circle
the Cu electrode in 1.00 M Cu^2+
- C
both function as cathodes simultaneously
- D
neither - the cell is at equilibrium
Why
Reduction of Cu^2+ -> Cu is favored where [Cu^2+] is greater. The cathode is the electrode in 1.00 M Cu^2+.
- A