AP Chemistry: 4.1 Introduction for Reactions – Exam Style questions with Answer- FRQ

Question

                                                           

A student is given a standard galvanic cell, represented above, that has a Cu electrode and a Sn electrode. As current flows through the cell, the student determines that the Cu electrode increases in mass and the Sn electrode decreases in mass.

(a) Identify the electrode at which oxidation is occurring. Explain your reasoning based on the student’s observations.

(b) As the mass of the Sn electrode decreases, where does the mass go?

(c) In the expanded view of the center portion of the salt bridge shown in the diagram below, draw and label a particle view of what occurs in the salt bridge as the cell begins to operate. Omit solvent molecules and use arrows to show the movement of particles.

                                                                             

d) A nonstandard cell is made by replacing the 1.0 M solutions of Cu\((NO_3)_2\) and Sn\((NO_3)_2\) in the standard cell with 0.50 M solutions of Cu(NO3)2 and Sn\((NO_{3})_{2}\). The volumes of solutions in the nonstandard cell are identical to those in the standard cell.

(i) Is the cell potential of the nonstandard cell greater than, less than, or equal to the cell potential of the standard cell? Justify your answer.

(ii) Both the standard and nonstandard cells can be used to power an electronic device. Would the nonstandard cell power the device for the same time, a longer time, or a shorter time as compared with the standard cell? Justify your answer.

(e) In another experiment, the student places a new Sn electrode into a fresh solution of 1.0 M\( Cu(NO_{3})_{2}\) ·

                                   

(i) Using information from the table above, write a net-ionic equation for the reaction between the Sn electrode and the Cu\((NO_3)_2\) solution that would be thermodynamically favorable. Justify that the reaction is thermodynamically favorable.

(ii) Calculate the value of \(\Delta G^0\) for the reaction. Include units with your answer.

▶️Answer/Explanation

(a) Since the Sn electrode is losing mass, Sn atoms must be forming \(Sn^{2+}\)(aq). This process is oxidation. OR because the cell operates, E° must be positive and, based on the E° values from the table, it must be Sn that is oxidized.

(b) The atoms on the Sn electrode are going into the solution as \(Sn^{2+}\) ions.

(c) The response should show at least one \(K^+\) ion moving toward the Cu compartment on the left and at least one \(NO_{3}^-\) ion moving in the opposite direction.

d(i) It is the same. In the cell reaction \(Q= \frac{[Sn^{2+}]}{ [Cu^{2+}]}\) , and the concentrations of \(Sn^{2+}\) and \(Cu^{2+}\) are equal to each other in both cases.

d(ii) The nonstandard cell would power the device for a shorter time because the supply of \(Cu^{2+}\) ions will be exhausted more quickly. OR The nonstandard cell would power the device for a shorter time because the reaction will reach E=0 more quickly.

e (i) \(Cu^{2+}(aq) + Sn(s) → Cu(s) + Sn^{2+}\)(aq) E° is positive \((0.34 V + 0.14 V = 0.48 V)\), therefore the reaction is thermodynamically favorable. OR The cell observations from earlier parts of the question are evidence that the Sn is oxidized and Cu is reduced, therefore E° must be positive.

e.(ii) \(\Delta G^0=-nFE^0\)

             \(\Delta G^0=-\frac{2 mol e^-}{mol_{rxn}}\times \frac{96,485}{mol e^-}\times \frac{0.48}{c}=-93,000 J/mol_{rxn}\)

Question

                                                                                         \(CaCO_{3}(s)\rightleftharpoons CaO(s)+CO_{2}(g)\)

When heated, calcium carbonate decomposes according to the equation above. In a study of the decomposition of calcium carbonate, a student added a 50.0 g sample of powdered \(CaCO_{3}\)(s) to a 1.00 L rigid container. The student sealed the container, pumped out all the gases, then heated the container in an oven at 1100 K. As the container was heated, the total pressure of the \(CO_2\)(g) in the container was measured over time. The data are plotted in the graph below.

The student repeated the experiment, but this time the student used a 100.0 g sample of powdered \(CaCO_3\)(s). In this experiment, the final pressure in the container was 1.04 atm, which was the same final pressure as in the first experiment.

(a) Calculate the number of moles of \(CO_{2}(g)\) present in the container after 20 minutes of heating.

(b) The student claimed that the final pressure in the container in each experiment became constant because all of the\( CaCO_{3}\)(s) had decomposed. Based on the data in the experiments, do you agree with this claim? Explain.

(c) After 20 minutes some \(CO_2\)(g) was injected into the container, initially raising the pressure to 1.5 atm. Would the final pressure inside the container be less than, greater than, or equal to 1.04 atm? Explain your reasoning.

(d) Are there sufficient data obtained in the experiments to determine the value of the equilibrium constant, K,for the decomposition of \(CaCO_3\)(s) at 1100 K? Justify your answer.

▶️Answer/Explanation

(a) \(PV=nRT\)

      \(\frac{PV}{RT}=n=\frac{(1.04 atm)(1.00 L)}{(0.0821\frac{L atm}{mol K})(1100 K)}== 0.0115 mol CO_2\)

(b)  Do not agree with claim

Explanation I: In experiment 1, the moles of \(CaCO_3 = 50.0 g/100.09 g/mol = 0.500 mol CaCO_3\). If the reaction had gone to completion, 0.500 mol of \(CO_2\) would have been produced. From part (a) only 0.0115 mol was produced. Hence, the student’s claim was false.

Explanation II: The two different experiments (one with 50.0 g of \(CaCO_3\) and one with 100.0 g of \(CaCO_3\)) reached the same constant, final pressure of 1.04 atm. Since increasing the amount of reactant did not produce more product, there is no way that all of the \(CaCO\), reacted. Instead, an equilibrium condition has been achieved and there must be some solid \(CaCO_3\) in the container.

(c) The final pressure would be equal to 1.04 atm. Equilibrium was reached in both experiments; the equilibrium pressure at this temperature is 1.04 atm. As the reaction shifts toward the reactant, the amount of \(CO_2\)(g) in the container will decrease until the pressure returns to 1.04 atm.

(d) Yes. For the equilibrium reaction represented by the chemical equation in this problem, at a given temperature the equilibrium pressure of \(CO_2\) determines the equilibrium constant. Since the measured pressure of \(CO_2\) is also the equilibrium pressure of \(CO_2\), \(K_p= P_{co_2} =1.04\).

Note: If the response in part (b) indicates “yes”, that all of the \(CaCO_3\)(s) had decomposed, then the point can be earned by stating that the system did not reach equilibrium in either experiment and hence the value of \(K_p\), cannot be calculated from the data.

Question

Answer the following questions about the solubility of  Ca(OH)2 ( Ksp= 1.3 10-6 ) .
(a) Write a balanced chemical equation for the dissolution of Ca(OH)2(s) in pure water.
(b) Calculate the molar solubility of Ca(OH)2 in 0.10 M Ca(NO3)2 .
(c) In the box below, complete a particle representation diagram that includes four water molecules with proper orientation around the Ca2+ ion.
Represent water molecules as

▶️Answer/Explanation

Ans:

(a) 

\(Ca(OH)_{2}\rightleftharpoons Ca^{2+}+2 OH^{-}\)

(b)

Ksp = [Ca2+] [OH]2

1.3 × 10-6 = (0.10 + x) (2x)2 ≈ (0.10)  4x2 [assuming x << 0.10]

1.3  × 10-5 = 4x2 

x = 0.0018 M

Molar solubility of Ca(OH)2 = 0.0018 M

(c)

[The diagram should show the oxygen side of the water molecules oriented closer to the Ca2+ ion.]
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