Home / AP Chemistry: 9.10 Electrolysis and Faraday’s Law – Exam Style questions with Answer- MCQ

AP Chemistry: 9.10 Electrolysis and Faraday’s Law – Exam Style questions with Answer- MCQ

Question

Two identical spoons were electroplated with Ag or Cd through the use of the electrolytic cells illustrated above. A current of 5.00A was supplied to each cell for 600. seconds, and the masses of the spoons before and after the electroplating were recorded. Which of the following mathematical equations can best be used to account for the much larger increase in mass of the spoon electroplated with Ag compared with the spoon electroplated with Cd?

A ΔG°=ΔH°−TΔS°

B ΔG°=−RTlnK

C E=E°\(−(\frac{RT}{nF})lnQ\)

D \(I=\frac{q}{t}\)

▶️Answer/Explanation

Ans: D

Based on \(I=\frac{q}{t}\) , using the same current, I, and the same time for the electrolysis, t, results in the same number of coulombs of charge, q, supplied for both processes. As a result, the difference in the mass of the spoons is due to the difference in the number of moles of electrons consumed per mole of metal ion. Based on \(Ag^+(aq)+e^−→Ag(s)\), one mole of \(Ag^+\) is reduced per mole of electrons consumed and because \(Cd^{2+}(aq)+2e^−→Cd(s)\), one mole of \(Cd^{2+}\) is reduced for every two moles of electrons consumed.

Question

The diagrams above illustrate the equipment used to electroplate four identical objects with silver or zinc. The table provides the conditions used. Which of the following provides the basis for the identification of the object that requires the highest current, I , to complete the electroplating?

A Object 1, based on the molar mass of Ag and the time of operation.

B Object 2, based on the molar mass of Ag and the time of operation.

C Object 3, based on the coulombs of charge needed and the time of operation.

D Object 4, based on the coulombs of charge needed and the time of operation.

▶️Answer/Explanation

Ans:C

Using \(I=\frac{q}{t}\) where q is the coulombs of charge needed for the electroplating and t is the time in seconds, the greatest current, I, corresponds to the process that needs the largest q supplied in the smallest t. Based on the half-reactions involved in the electroplating, \(Ag^+(aq)+e^−→Ag(s)\) and \(Zn^{2+}(aq)+2e^−→Zn(s)\), the deposition of 1mol of Zn requires 2mole− and corresponds to the largest q. Therefore, the electroplating of object 3 required the conditions of q and t that maximize I compared to the other objects.

Question

Two different electrolytic cells are constructed to electroplate two identical objects based on the half-reactions given in the table above. An object will be plated with a certain number of moles of Au in one cell, and in the other cell another object will be plated with the same number of moles of Ag in the same amount of time. Which of the following mathematical relationships can be used to determine how much more current is needed to electroplate an object with Au than to electroplate it with Ag?

A The difference in standard reduction potential of \(Au^{3+}\)(aq) and \(Ag^+\) is 0.70V .

B Three times more electrons are needed to reduce \(Au^{3+}\)(aq) to \(Au(s)\) than are needed to reduce \(Ag^+\)(aq) to \(Ag(s)\).

C ΔG° is more negative for the reduction of \(Au^{3+}\)(aq) to \(Au(s)\) than for the reduction of \(Ag^+\)(aq) to \(Ag(s)\) .

D The molar-mass ratio of \(\frac{Au}{Ag}\) is \(\frac{196.97}{107.87}\).

▶️Answer/Explanation

Ans:B

The number of electrons needed to reduce one mole of \(Au^{3+}\) is three times the number of electrons needed to reduce one mole of \(Ag^+\). Thus, the electric charge, q, associated with reducing one mole of Au is three times the amount of charge associated with reducing one mole of Ag . Because electrical current is the rate of transfer of charge \(I=\frac{q}{t}\) and the time for depositing the same number of moles of each metal is the same, the current required for depositing the Au must be greater (three times as great) than the current required to deposit the Ag.

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.

Based on the information in the table above, which of the following shows the cell potential and the Gibbs free energy change for the overall reaction that occurs in a standard galvanic cell?

▶️Answer/Explanation

Ans:C

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