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

Question

\( Cu^{+}(aq) + Au^{3+}(aq) \rightarrow 3 Cu ^{2+}(aq) + Au(s)\)

A student studying the reaction represented above constructs a voltaic cell as shown in the diagram below. The cell contains an Au(s) electrode in a beaker containing 1.00 M Au(NO3)3(aq) and a Pt(s) electrode in a beakercontaining 1.00 M \(CuNO_3(aq)\) and 1.00 M \(Cu(NO_3)_2\)(aq).

The standard reduction potentials associated with the cell are given in the following table.

(a) Calculate the value of the standard cell potential, E°.
(b) Does the mass of the Pt(s) electrode increase, decrease, or remain the same as the cell operates? Justify your answer.

▶️Answer/Explanation

An expanded view of the Au half-cell before the reaction occurs is shown in box A below. An expanded view of the Au electrode after the cell operates for a period of time is shown in box B below.

(c) Assuming that the representation in box A is accurate, explain what is wrong with the stoichiometry represented in box B.
(d) After the cell has operated for 30.0 minutes, 0.185 g of Au(s) is deposited on the Aunelectrode.

(i) Calculate the number of moles of Au(s) deposited.
(ii) Calculate the average current that passed through the cell during the deposition of Au(s). The student adds some NaCl(s) to the half-cell that contains \(Cu ^{+}\)(aq) and \(Cu^{2+}\)(aq), and a precipitate forms.
(e) Based on the information in the following table, determine the identity of the precipitate. Justify your answer with a calculation.

(f) Write the net ionic equation for the formation of the precipitate.
(g) The student predicts that the cell voltage will increase after the precipitate forms. Do you agree with the student’s prediction? Justify your answer.

(a) \(E^{\circ} = 1.50 V – 0.16 V = 1.34 V\)

(b) The mass of the Pt(s) electrode remains the same because the Pt does not react and no Cu atoms will be deposited on the Pt electrode.

(c) In a redox reaction, the number of electrons lost must equal the number gained, so the net charge must be zero. The total charge of the cations is less than the total charge of the anions.

d(i) \(0.185 Au \times \frac{1molAu}{196.97g Au}=9.39\times 10^{-4}mol Au\)

d(ii) 

(e) For CuCl:
\(CuCl(s) \rightarrow Cu^+(aq) + Cl^-(aq)\)
Ksp =\( [Cu^+][Cl^{-}]\)
\(1.7\times 10^{-7} = x^2\)
\(x=4.1\times 10^{-4} \)= solubility of CuCl
CuCl has a lower solubility compared to \(CuCl_2\) and will precipitate first.

(f) \(Cu^+(aq)+Cl^-(aq)\rightarrow CuCl(s)\)

(g) Disagree.
The precipitate will decrease\( [Cu^{+}]\), which will cause Q to increase and become closer to the value of K. Therefore, the voltage of the cell will decrease.

Question

A student is given 50.0 mL of a solution of Na2CO3 of unknown concentration. To determine the concentration of the solution, the student mixes the solution with excess 1.0 M Ca(NO3)2(aq), causing a precipitate to form.
The balanced equation for the reaction is shown below.
Na2CO3(aq) + Ca(NO3)2(aq) → 2 NaNO3(aq) + CaCO3(s)
(a) Write the net ionic equation for the reaction that occurs when the solutions of Na2CO3 and Ca(NO3)2 are mixed.
(b) The diagram below is incomplete. Draw in the species needed to accurately represent the major ionic species remaining in the solution after the reaction has been completed.

The student filters and dries the precipitate of CaCO3 (molar mass 100.1 g/mol) and records the data in the table below.

(c) Determine the number of moles of Na2CO3 in the original 50.0 mL of solution.
(d) The student realizes that the precipitate was not completely dried and claims that as a result, the calculated Na2CO3 molarity is too low. Do you agree with the student’s claim? Justify your answer.
(e) After the precipitate forms and is filtered, the liquid that passed through the filter is tested to see if it can conduct electricity. What would be observed? Justify your answer.

The student decides to determine the molarity of the same Na2CO3 solution using a second method. When
Na2CO3 is dissolved in water, CO32−(aq) hydrolyzes to form HCO3(aq), as shown by the following equation.
CO32−(aq) + H2O(l) ⇔  HCO3(aq) + OH(aq)      \(K_{b}=\frac{[HC{O_{3}}^{-}][OH^{-}]}{[{CO_{3}}^{2-}]}=2.1\times 10^{-4}\)

(f) The student decides to first determine [OH] in the solution, then use that result to calculate the initial concentration of CO32−(aq).
(i) Identify a laboratory method (not titration) that the student could use to collect data to determine [OH] in the solution.
(ii) Explain how the student could use the measured value in part (f)(i) to calculate the initial concentration of CO32−(aq). (Do not do any numerical calculations.)
(g) In the original Na2CO3 solution at equilibrium, is the concentration of HCO3(aq) greater than, less than, or equal to the concentration of CO32−(aq) ? Justify your answer.
(h) The student needs to make a CO32−/HCO3 buffer. Is the Na2CO3 solution suitable for making a buffer with a pH of 6? Explain why or why not.

▶️Answer/Explanation

Ans:

(a)

Ca2+(aq) + CO32-(aq)  → CaCO3(s)

(b)

The drawing shows one Ca2+ ion.

(c)

\(0.93 g CaCO_{3}\times \frac{1 mol CaCO_{3}}{100.1 g}=0.0093 mol CaCO_{3}\)

\(0.0093 mol CaCO_{3}\times \frac{1 mol Na_{2}CO_{3}}{1 molCaCO_{3} }=0.0093 mol Na_{2}CO_{3}\)

(d)

Disagree. The presence of water in the solid will cause the measured mass of the precipitate to be greater than the actual mass of CaCO3. As a result, the calculated number of moles of CaCO3 and moles of Na2CO3 will be greater than the actual moles present. Therefore the calculated concentration of Na2CO3(aq) will be too high.

(e)

The liquid conducts electricity because ions (Na+(aq), Ca2+(aq), and NO3(aq) are present in the solution.

(f) (i)

Determine the pH of the solution using a pH meter.

 (ii)

First determine [OH] using pOH = 14 – pH, then [OH] = 10-pOH.

Then, use the Kb expression and an ICE table (see example below to determine [CO32-] and [HCO3] at equilibrium. The initial concentration of CO32-, ci, is equal to the sum of the equilibrium concentrations of CO32- and HCO3.

(g)

Less than. The small value of Kb, 2.1 × 10-4, indicates that the reactants are favored.

(h)

No, the Na2CO3 solution is not suitable. The pKa of HCO3 is 10.32. Buffers are effective when the required pH is approximately equal to the pKa of the weak acid. An acid with a pKa of 10.32 is not appropriate to prepare a buffer with a pH of 6.

Question

M + I2 → MI2

To determine the molar mass of an unknown metal, M, a student reacts iodine with an excess of the metal to form the water-soluble compound MI2 , as represented by the equation above. The reaction proceeds until all of the I2 is consumed. The MI2(aq) solution is quantitatively collected and heated to remove the water, and the product is dried and weighed to constant mass. The experimental steps are represented below, followed by a data table.

(a) Given that the metal M is in excess, calculate the number of moles of I2 that reacted.
(b) Calculate the molar mass of the unknown metal M.
The student hypothesizes that the compound formed in the synthesis reaction is ionic.
(c) Propose an experimental test the student could perform that could be used to support the hypothesis. Explain how the results of the test would support the hypothesis if the substance was ionic.
The student hypothesizes that Br2 will react with metal M more vigorously than I2 did because Br2 is a liquid at room temperature.
(d) Explain why I2 is a solid at room temperature whereas Br2 is a liquid. Your explanation should clearly reference the types and relative strengths of the intermolecular forces present in each substance.
While cleaning up after the experiment, the student wishes to dispose of the unused solid I2 in a responsible manner. The student decides to convert the solid I2 to I(aq) anion. The student has access to three solutions, H2O2(aq), Na2S2O3(aq), and Na2S4O6(aq), and the standard reduction table shown below.

(e) Which solution should the student add to I2(s) to reduce it to I(aq)? Circle your answer below. Justify your answer, including a calculation of E° for the overall reaction.
                                             H2O2(aq)                        Na2S2O3(aq)                                       Na2S4O6(aq)
(f) Write the balanced net-ionic equation for the reaction between I2 and the solution you selected in part (e).

▶️Answer/Explanation

Ans:

(a)

127.570 – 126.549 = 1.021 g I2
\(1.021 g I_{2}\times \frac{1 molI_{2} }{253.80 g I_{2}}=0.004023 mol I_{2}\)

(b)

Number of moles of I2 = number of moles of M
1.284 g MI2 – 1.021 g I2 = 0.263 g M
Molar mass of \(\frac{0.263 g M}{0.004023 mol M}=65.4 g/mol\)

(c)

The student could dissolve the compound in water or melt the compound and see if the solution/melt conducts electricity. If the solution/melt conducts electricity, mobile ions capable of carrying charge must be present, thus the compound is likely to be ionic.
OR
The student could heat the compound until it melts or boils. If the melting/boiling point is very high, then the compound is likely to be ionic.

(d)

Both Br2 and I2 molecules are nonpolar molecules, therefore the only possible intermolecular forces are London dispersion forces.
The London dispersion forces are stronger in I2 because it is larger in size with more electrons and/or a more polarizable electron cloud. The stronger London dispersion forces in I2 result
in a higher melting point,which makes I2 a solid at room temperature.

(e) 

[Na2S2O3 (aq) should be circled.]
The reaction between S2O32- (aq) and I2 (s) will be thermodynamically favorable because E0 for the reaction is positive ( E0= 0.54 – 0.08 = +0.46 V), from which it follows that ΔG0 is negative because ΔG0 = -nFE0 .

(f)

I2 + 2 S2 O3 2- → 2 I + S4 O6 2-
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