AP Chemistry: 9.9 Cell Potential Under Nonstandard Conditions – Exam Style questions with Answer- MCQ

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.

On average, after one year of operation, the potential of a lithium-iodine cell decreases by 1%-2%. Which of the following best helps to explain the cause for the decrease in cell potential?

A \(E_{cell}=E^{\circ}_{red}(cathode)-E^{\circ}_{red}(anode)\) , and as the cell operates, \(E^{\circ}_{red}(cathode)\) decreases.

B \(E_{cell}=E^{\circ}_{red}(cathode)-E^{\circ}_{red}(anode)\) , and as the cell operates, \(E^{\circ}_{red}(cathode)\) increases.

C \(E_{cell}=E^{\circ}_{cell}-\frac{RT}{nF}lnQ\) , and as the cell operates, Q decreases.

D \(E_{cell}=E^{\circ}_{cell}-\frac{RT}{nF}lnQ\) , and as the cell operates, Q increases.

▶️Answer/Explanation

Ans:D

As the cell generates a potential over a period of time, the consumption of reactants and formation of products result in nonstandard conditions of operation. \(E_{cell}=E^{\circ}_{cell}-\frac{RT}{nF}lnQ\) , and for \(E_{cell}\) to decrease, the term \(-\frac{RT}{nF}lnQ\) must become more negative. Therefore, \(lnQ\) must become more positive over time, which occurs when Q increases.

Question

                                           

                                              \(2Ag^+(aq)+Zn(s)→2Ag(s)+Zn_2^+(aq)\)

Under standard conditions, the galvanic cell shown above has a cell potential of +1.56V using the reaction given. The salt bridge contains \(KNO_3\), which allows \(K^+\) ions and \(NO_3^−\) ions to move in the directions indicated. If \(KNO_3\) in the salt bridge is replaced with \(KOH\), some \(Zn(OH)_2(s)\) precipitates in the \(Zn-Zn(NO_3)_2\) half-cell. Which of the following best explains how the cell potential is affected as \(Zn(OH)_2(s)\) starts to precipitate, and why?

A The cell potential increases because the concentration of \(Zn^{2+}\)(aq) decreases and Q, \(\frac{[Zn^{2+}]}{[Ag^{+}]^{2}}\) , becomes smaller.

B The cell potential decreases because the concentration of \(Zn^{2+}\)(aq) decreases and Q, \(\frac{[Zn^{2+}]}{[Ag^+]^2}\) , becomes smaller.

C The cell potential increases because \(K^+\) ions replace \(Zn^{2+}\) ions and the reduction of K+ is more thermodynamically favored than the reduction of \(Zn^{2+}\) .

D The cell potential stays the same because \(Zn(OH)_2(s)\) is not part of the redox reaction responsible for the operation of the galvanic cell.

▶️Answer/Explanation

Ans:A

 The formation of Zn(OH)2(s) decreases the concentration of \(Zn^{2+}\)(aq) in solution, resulting in nonstandard operating conditions \([Zn^{2+}]<1M\). The Nernst equation can be used to calculate the cell potential: \(E_{cell}\)=E°\(_{cell}-\frac{RT}{nF}lnQ\), where Q=\(\frac{[Zn^{2+}]}{[Ag^{+}]^{2}}\). Because Q<1, \(lnQ\) is negative, making the term \(-\frac{RT}{nF}lnQ\) positive. As a result, the cell potential increases \(E_{cell}\)>E°=+1.56V). Alternatively, given that galvanic cells have very large values of K and that cell potential approaches zero as the cell reaches equilibrium (as Q approaches the very large value of K), any change that makes Q smaller than standard conditions (where Q=1) will increase the cell potential.

Question

                                               \(Ni^{2+}(aq)+Cd(s)→Ni(s)+Cd^{2+}(aq)\) 

                                                             E°\(_{cell}=+0.15V\)

Under standard conditions, a \(Ni-Cd\) galvanic cell generates a potential of \(+0.15V\) using the chemical reaction represented above. Which of the following best explains how \([Ni^{2+}]>1M\) affects the cell potential if this is the only change made to the operation of the galvanic cell?

A The ratio \(\frac{[Cd^{2+}]}{[Ni^{2+}]}\) is smaller if \([Ni^{2+}]>1M\), resulting in Q<1 and \(E_{cell}\)>E°\(_{cell}\).

B The ratio \(\frac{[Cd^{2+}]}{[Ni^{2+}]}\) is smaller if \([Ni^{2+}]>1M\), resulting in Q>1 and \(E_{cell}\)>E°\(_{cell}\).

C The ratio \(\frac{[Ni^{2+}]}{[Cd^{2+}]}\) is larger if \([Ni^{2+}]>1M\), resulting in Q>1 and \(E_{cell}\)<E°\(_{cell}\) .

D The ratio \(\frac{[Ni^{2+}]}{[Cd^{2+}]}\) is larger if \([Ni^{2+}]>1M\), resulting in Q<1 and \(E_{cell}\)<E°\(_{cell}\).

▶️Answer/Explanation

Ans: A

Based on the chemical reaction given, the mathematical expression for the reaction quotient, Q , is \(Q=\frac{[Cd^{2+}]}{[Ni^{2+}]}\) and if \([Ni^{2+}]>1M\), then Q<1. \(E_{cell}\)=E°\(_{cell}-\frac{RT}{nF}lnQ\) and when Q<1, lnQ is negative, making the term \(-\frac{RT}{nF}lnQ\) positive and \(E_{cell}\)>E°\(_{cell}\).

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) \(E^{\circ}\)=?

Which of the following is true for the decomposition of \(H_{2}O_{2}\)(aq) ?
(A) \(\Delta\)G° > 0 and\( K_{eq}\) > 1

(B) \(\Delta\)G°> 0 and \(K_{eq }\)< 1

(C) \( \Delta\)G° < 0 and \(K_{eq}\) > 1

(D)\( \Delta\)G° < 0 and \(K_{eq} \)< 1

▶️Answer/Explanation

Ans:A

Question

The standard cell potential (Eecell) for the voltaic cell based on the reaction below is __________ V.
\(Sn^{2+} (aq) + 2Fe^{3+} (aq) \rightarrow 2Fe^{2+} (aq) + Sn^{4+} (aq)\)
A) +1.39           B) +0.46          C) +0.617           D) +1.21            E) -0.46

▶️Answer/Explanation

Ans: C

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