AP Chemistry: 9.8 Cell Potential and Free Energy – Exam Style questions with Answer- MCQ

Questions 

Standard reduction potentials for the half-reactions associated with the electrochemical cell shown above are given in the table below.

What is the value of E° for the cell?
(A) 0.04 V
(B) 0.84 V
(C) 1.56 V
(D) 2.36 V

▶️Answer/Explanation

Ans: C

To find the cell potential (Eocell) for the overall reaction, we use the following equation:

Eccell = Eccathode – Eoanode

Where: Eocathode is the standard reduction potential for the cathode (reduction) reaction Ecanode is the standard reduction potential for the anode (oxidation) reaction

From the data given: Cathode (reduction) reaction: Ag+(aq) + e- -> Ag(s), Eocathode = +0.80 V Anode (oxidation) reaction: Zn(s) -> Zn2+(aq) + 2e-, Ecanode = -0.76 V

Substituting in the equation: Eocell = Eocathode – Ecanode = +0.80 V – (-0.76 V) = +0.80 V + 0.76 V = 1.56 V

Therefore, the value of E° (standard cell potential) for this electrochemical cell is 1.56 V.

The correct answer is (C) 1.56 V.

Question

             

Pacemakers are electronic devices that help regulate the heart rate. Currently, lithium-iodine cells are commonly used to power pacemakers and have replaced zinc-mercury cells. Table 1 provides the operating cell potential, E , for each cell. Table 2 provides the standard reduction potentials for several half-reactions related to zinc-mercury and zinc-air cells. Based on the information given, which of the following is a major difference between the zinc-mercury cell and the lithium-iodine cell?

A During the initial cell operation, each reaction is thermodynamically favorable, but the larger operating potential of the lithium-iodine cell indicates that its cell reaction is less thermodynamically favorable.

B During the initial cell operation, each reaction is thermodynamically favorable, but the larger operating potential of the lithium-iodine cell indicates that its cell reaction is more thermodynamically favorable.

C During the initial cell operation, the oxidation of iodine is thermodynamically favorable but the oxidation of mercury is not.

D During the initial cell operation, the oxidation of mercury is thermodynamically favorable but not the oxidation of iodine is not.

▶️Answer/Explanation

Ans:B

During the initial cell operation, each redox reaction is thermodynamically favorable because ΔG\(=−nFE\) and each cell has a positive operating potential, meaning that ΔG<0, which is a criterion used to determine thermodynamic favorability. However, the larger, more positive operating potential of the lithium-iodine cell compared to the zinc-mercury cell results in a larger, more negative ΔG for the cell reaction and greater thermodynamic favorability.

Question

The operation of a hydrogen fuel cell under standard conditions relies on the chemical reaction represented above. The table provides the relevant reduction half-reactions and the standard reduction potentials. Based on the information given, which of the following equations can be used to calculate the standard reduction potential, in volts, of the half-reaction occurring at the cathode?

A \(E^{\circ}_{red}(cathode)=(\frac{−474}{4×96,500})\)

B \(E^{\circ}_{red}(cathode)=-(\frac{−474000}{4×96,500})\)

C \(E^{\circ}_{red}(cathode)=(\frac{4×96,500}{474})\)

D \(E^{\circ}_{red}(cathode)=-(\frac{4×96,500}{−474000})\)

▶️Answer/Explanation

Ans:B

Question

The reaction represented by the equation above serves as the basis for the construction of an electrochemical cell. The table gives the reduction half-reactions and their respective standard reduction potentials. Based on the overall reaction, which of the following identifies the type of electrochemical cell that was constructed based on the value of ΔG?

A A galvanic or voltaic cell, because ΔG=\(\frac{−96,500×(+0.46)}{1,000}=−44 kJ/mol_{rxn}\) and its operation generates a potential of 0.46V .

B A galvanic or voltaic cell, because ΔG=\(\frac{−2×96,500×(+0.46)}{1,000}=−89 kJ/mol_{rxn}\) and its operation generates a potential of 0.46V .

C An electrolytic cell, because ΔG=\(\frac{−96,500×(−0.46)}{1,000}=44kJ/mol_{rxn}\) and its operation requires a potential smaller than 0.46V to be supplied.

D An electrolytic cell, because ΔG=\(\frac{−2×96,500×(−0.46)}{1,000}=89kJ/mol_{rxn}\) and its operation requires a potential greater than 0.46V to be supplied.

▶️Answer/Explanation

Ans:D

Based on the standard reduction potentials given, because \(E^{\circ}_{cell}=E^{\circ}_{red}(cathode)-E^{\circ}_{red}(anode)=0.34-0.80=-0.46V\) and ΔG\(=−nFE=\frac{−2×96,500×(−0.46)}{1,000}=89kJ/mol_{rxn}\), the reaction is not thermodynamically favorable and requires an external source of energy to proceed as written. Electrolytic cells are constructed when the redox reaction has a negative cell potential and ΔG>0. An external potential greater than 0.46V will result in the formation of \(Ag^+(aq)\) and \(Cu(s)\).

Question

\(2H_{2}O_{2}(aq)\rightarrow 2H_{2}O(l)+O_{2}(g) E^{\circ} \)=0.55V

The equation and standard cell potential for the decomposition of H2O2(aq) in acidic solution at 25°C is given above. The reduction half-reactions for the process are listed below.

\(O_{2}(g)+4H^{+}(aq)+4e^{-}\rightarrow 2H_{2}O(l) E^{\circ} \)=1.23

\(O_{2}(g)+2H^{+}(aq)+2e^{-}\rightarrow H_{2}O_{2}O\)(aq) \(E^{\circ}\)=?

\(O_{2}(g)+2H^{+}(aq)+2e^{-}\rightarrow H_{2}O_{2}O\)(aq)

What is the  standard reduction potential for the half-reaction represented above?
(A) -1.78 V
(B) -0.68 V
(C) +0.68 V
(D) +1.78 V

▶️Answer/Explanation

Ans:A

Question

Which of the halogens in Table 20.1 is the strongest oxidizing agent?
A) \(Br_2\)
B) \(I_2\)
C) \(Cl_2\)
D) \(F_2\)
E) All of the halogens have equal strength as oxidizing agents.

▶️Answer/Explanation

Ans: D

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