(Pmccord) – HW12 Electrochemistry II – Mccord – (51520) 1 This Print-Out Should Have 25 Questions

Home , Chloralkali process

mccord (pmccord) – HW12 Electrochemistry II – mccord – (51520) 1 This print-out should have 25 questions. Multiple-choice questions may continue on 0.3 mol e− ox. number = = +3 the next column or page – ﬁnd all choices 0.1 mol V before answering. 003 10.0 points 001 10.0 points How long will it take to deposit 0.00235 moles How many moles of Cl (g) are produced by 2 of gold by the electrolysis of KAuCl (aq) us- the electrolysis of concentrated sodium chlo- 4 ing a current of 0.214 amperes? ride if 2.00 A are passed through the solution for 4.00 hours? The equation for this process 1. 26.5 min (the “chloralkali” process) is 2NaCl(aq)+ 2H O(ℓ) → 2 2. 17.7 min 2NaOH(aq) + H2(g) + Cl2(g) 3. 106 min 1. 0.149 mol correct 4. 70.7 min 2. 0.447 mol 5. 53.0 min correct 3. 0.298 mol Explanation: 4. 0.0745 mol 004 10.0 points 5. 0.00248 mol Consider 3 electrolysis experiments: Explanation: The ﬁrst: 1 Faraday of electricity is passed through a solution of AgNO3. 002 10.0 points The second: 2 Faradays of electricity are A steel surface has been electroplated with passed through a solution of Zn(NO3)2. 5.10 g of vanadium (V, molar mass = 51 The third: 3 Faradays of electricity are g/mol). If 2.90×104 C of charge were used, passed through a solution of Bi(NO3)3. what was the original oxidation number of V? 1. Equal numbers of moles of all three metals 1. +4 are produced. correct

2. +3 correct 2. Twice as many moles of metallic zinc are produced than metallic silver. 3. +5 3. The reaction producing the smallest mass 4. +6 of metal is that of the silver solution.

5. +1 4. Equal masses of all three metals are produced. 6. +2 Explanation: − Explanation: 1F=1mol e . The relevant half-reactions are 5.1g moles V = =0.1 mol 51 g/mol − Ag+ +1 e → Ag 4 2+ − − 2.90 × 10 C Zn +2 e → Zn moles e = =0.3 mol − 9.65 × 104 C/mol Bi3+ +3 e → Bi mccord (pmccord) – HW12 Electrochemistry II – mccord – (51520) 2 Using the given amounts of electricity, 6. 2.56 correct

Exp. metal grams produced produced Explanation: 107.87 g Ag The strongest reducing agent is Co3+ and 1 1molAg (1molAg) mol Ag the strongest oxidizing agent is Zn. The stan- 65.39 g Zn dard cell potential of a battery built from 2 1molZn (1molZn) mol Zn these species would be: 208.98 g Bi 3 1molBi (1molBi) mol Bi ◦ ◦ ◦ E = E − E Of the answer choices, the true statement cell cathode anode says that equal moles of metals are produced. =1.80 − (−0.76) =2.56 005 10.0 points Sodium is produced by electrolysis of molten sodium chloride. What are the products at 007 10.0 points the anode and cathode, respectively? ◦ What would be the E cell of an electrolytic cell made from the half reactions 1. Cl (g) and Na O(ℓ) − − 2 2 AgCl(s) + e −→ Ag(s) + Cl (aq) ◦ E = +0.22 V 2. Cl2(g) and Na(ℓ) correct − ◦ Al3+(aq)+ 3 e −→ Al(s) E = −1.66 V

3. Na(ℓ) and O2(g) 1. −1.44 − 4. Cl (aq) and Na2O(ℓ) 2. 1.44

5. O2(g) and Na(ℓ) 3. −1.88 correct Explanation: 4. 1.88 006 10.0 points What is the standard cell potential of the strongest battery that could be made using Explanation: these half reactions? − − ◦ Br2 +2 e −→ 2Br E = +1.07 − ◦ Fe3+ +3 e −→ Fe E = −0.04 ◦ ◦ ◦ 3+ − 2+ ◦ Co + e −→ Co E = +1.80 E = Ecathode − Eanode − ◦ Zn2+ +2 e −→ Zn E = −0.76 = −1.66 − 0.22 = −1.88 1. 1.11

2. 1.84 008 (part 1 of 3) 10.0 points 3. 1.83 The galvanic cell below uses the standard 4. 1.03 half-cells Mg2+ | Mg and Zn2+ | Zn, and a salt bridge containing KCl(aq). The voltmeter 5. 1.04 gives a positive voltage reading. mccord (pmccord) – HW12 Electrochemistry II – mccord – (51520) 3

voltmeter 011 10.0 points The electrolysis of an aqueous sodium chloride solution using inert electrodes produces gaseous chlorine at one electrode. At the − ← C D → + other electrode gaseous hydrogen is produced Salt and the solution becomes basic around the Bridge electrode. What is the equation for the cath- A B ode half-reaction in the electrolytic cell? E F − − 1. 2Cl → Cl2 +2 e

Identify A and write the half-reaction that − − 2. 2H O+2 e → H +2OH correct occurs in that compartment. 2 2 − − − 3. Cl +2 e → 2Cl 1. Zn(s); Zn2+(aq)+ 2 e → Zn(s) 2 − − − 4. H +2OH → 2H O+2 e 2. Mg(s); Mg2+(aq)+ 2 e → Mg(s) 2 2

− 5. None of the other answers listed is cor- 3. Zn(s); Zn(s) → Zn2+(aq)+ 2 e rect. − 4. Mg(s); Mg(s) → Mg2+(aq) + 2 e cor- Explanation: rect 012 10.0 points Explanation: ◦ What is the Ecell of 009 (part 2 of 3) 10.0 points What happens to the size of the electrode Zn(s) | Zn2+(aq) || Ce4+(aq) | Ce3+(aq) A during the operation of the cell? 2+ − ◦ Zn +2 e → Zn Ered = −0.76 − ◦ Ce4+ + e → Ce3+ E = +1.61 1. increases red 1. −0.85 2. No change 2. +1.61 3. decreases correct Explanation: 3. −2.37

010 (part 3 of 3) 10.0 points 4. +2.37 correct What is the voltmeter reading? 5. +0.85 1. +4.30 V Explanation: 2. +3.40 V 013 10.0 points Calculate the cell potential for a cell based on 3. +2.50 V the reaction 4. +0.50 V Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s) when the concentrations are as follows: [Ag+] 5. +1.60 V correct =0.51 M, [Cu2+]=0.9 M. (The temperature ◦ Explanation: is 25 C and E0 =0.4624 V.) mccord (pmccord) – HW12 Electrochemistry II – mccord – (51520) 4 the respective compartments. Find the un- Correct answer: 0.446443 V. known concentration for the following cell. 2+ Explanation: Pb(s) | Pb (aq, ?) || Pb2+(aq, 0.1 M) | Pb(s) E =0.025 V 014 10.0 points Standard reduction potentials are established Correct answer: 0.0142836 M. by comparison to the potential of which half Explanation: reaction? Ecell =0.025 V M =0.1 M ◦ − − Ecell =0V F = 96485 C/mol 1. F2 + 2 e −→ 2F R T =0.025693 V − F 2. Li+ + e −→ Li The cell reaction is Pb2+(aq, 0.1 M) → Pb2+(aq,x) + − 3. 2H + 2 e −→ H2 correct n =2,Ecell =0.025 V Using the Nernst equation, − − 4. 2H2O+2e −→ H2 +2OH ◦ R T − Ecell = Ecell − ln Q 5. Na+ + e −→ Na n F R T x ◦ Explanation: ln = Ecell − Ecell The hydrogen electrode is the standard ref- n F M erence electrode

x n F ◦ 015 10.0 points ln = (Ecell − Ecell) Consider the cell M R T 2− n F ◦ Pb(s) | PbSO4(s) | SO4 (aq , 0.60 M) || x = M exp (E − E ) + cell cell H (aq , 0.70 M) | H2(g , 192.5 kPa) | Pt . R T ◦ ◦ If E for the cell is 0.36 V at 25 C, write 2(0 V − 0.025 V) x = (0.1 M) exp the Nernst equation for the cell at this tem- 0.025693 V perature. =0.0142836 M .

1.90 2+ 1. E = 0.36 − 0.01285 ln Thus [Pb ]=0.0142836 M . (0.70)2(0.60) correct 017 10.0 points 2+ 2+ 192.5 What is ratio of [Co ] | [Ni ] when a bat- 2. E = 0.36 − 0.02569 ln 2 tery built from the two half reactions (0.70) (0.60) ◦ Ni2+ −→ Ni E = −0.25 V 1.90 ◦ 3. E = 0.36 − 0.01285 ln Co2+ −→ Co E = −0.28 V (0.70)(0.60) reaches equilibrium? 1.9 4. E = 0.36 + 0.01285 ln (0.70)2(0.60) 1. 3.20 192.5 5. E = 0.36 + 0.01285 ln (0.70)2(0.60) 2. 0.10 Explanation: 3. 0.31 016 10.0 points 4. 10.23 correct A concentration cell consists of the same re- dox couples at the anode and the cathode, Explanation: ◦ with diﬀerent concentrations of the ions in Ecell = +0.03 V mccord (pmccord) – HW12 Electrochemistry II – mccord – (51520) 5

◦ 0.05916 2. II and IV only Ecell = Ecell − log Q Ne 0.05916 [Co2+] 3. III only 0=0.03 − log 2 [Ni2+] 4. All but IV [Co2+] log =1.01 [Ni2+] 5. All but III [Co2+] = 101.01 = 10.23 [Ni2+] 6. Maybe IV and I

018 10.0 points 7. I and II only correct ◦ If E for the disproportionation of Cu+(aq) ◦ Explanation: to Cu2+(aq) and Cu(s) is +0.37 V at 25 C, (I and II: TRUE) ∆G must be negative calculate the equilibrium constant for the re- because the light IS on and therefore a spon- action. taneous change is occuring. Same logic goes for the Ecell except opposite in sign (positive). 1. 1.3 × 103 (III: FALSE) The batteries are headed to- ward equilibrium but are not there yet. True 2. 3.2 × 1012 equilibrium would mean a potential of zero volts - the ultimate DEAD battery. 3. 2.4 × 102 (IV: FALSE) The reason batteries work as well as they do is because they hold a very 4. 5.7 × 1018 constant potential throughout most of the life of the battery. Only near the battery’s end 5. 1.8 × 106 correct does the potential start to fall oﬀ from the Explanation: brand new potential. Why? Most batteries The disproportionation is rely on solids in their equilibria which means 2 Cu+ → Cu2++ Cu constant voltage as long as the solid reactants This corresponds to a ONE electron trans- (and products) are present (Q in the Nernst fer from one Cu+ ion to another Cu+ ion. equation will equal 1). Once the solids are Therefore n = 1 for this reaction. depleted, you see the great change in potential and usually the failure of the battery to work nFE = RT ln K in its intended device. K = enF E/RT = e1(0.37)/0.257 =1.8 × 106 020 10.0 points ◦ 019 10.0 points What is ∆G for the half reaction below? ◦ You turn on a ﬂashlight containing brand new Reaction E NiCad batteries and keep it lit for a minute or − + ClO3 +6H (aq) two. Which of the following can be considered 1 TRUE regarding the chemical state of these −→ Cl2(g)+3H2O(ℓ) +1.47 batteries? 2 − I. ∆G for the battery reaction is negative. 1. 194, 000 kJ · mol 1 II. Ecell > 0. − III. The batteries are at equilibrium. 2. −1, 418 kJ · mol 1 IV. Ecell is substantially decreasing during − this time. 3. −709 kJ · mol 1 correct

− 1. All 4. 194 kJ · mol 1 mccord (pmccord) – HW12 Electrochemistry II – mccord – (51520) 6 Explanation: −8 ∆G = −neFE 2. 1.7 × 10 = −5(96, 485)(1.47) 17 − 3. 8.0 × 10 = −709, 165 J · mol 1 −1 = −709 kJ · mol 4. 6.0 × 107 correct 021 10.0 points ◦ 3 For the reduction of Cu2+ by Zn, ∆G = 5. 7.7 × 10 ◦ − 212 kJ/mol and E = +1.10 V. If the Explanation: coeﬃcients in the chemical equation for this ◦ reaction are multiplied by 2, ∆G = − 424 ◦ 024 10.0 points kJ/mol. This means E = +2.20 V. The standard voltage of the cell

1. False correct − Ag(s) | AgBr(s) | Br (aq) || Ag+(aq) | Ag(s) 2. True ◦ is +0.73 V at 25 C. Calculate the equilibrium Explanation: constant for the cell reaction.

022 10.0 points 1. 2.2 × 1012 correct Consider the cell − 2. 4.6 × 10 13 2+ 2+ Zn(s) | Zn (aq) || Fe (aq) | Fe(s) 3. 5.1 × 1014 at standard conditions. Calculate the value of −15 ◦ 4. 2.0 × 10 ∆Gr for the reaction that occurs when current − is drawn from this cell. 5. 3.9 × 10 29 − 1. − 62 kJ · mol 1 correct Explanation:

− 2. + 230 kJ · mol 1 025 10.0 points The equilibrium constant for the reaction − − 2+ 3. − 31 kJ · mol 1 2Hg(ℓ)+2Cl (aq) + Ni (aq) → Ni(s) + Hg2Cl2(s) − − ◦ 4. + 62 kJ · mol 1 is 5.6 × 10 20 at 25 C. Calculate the value of ◦ E for a cell utilizing this reaction. − 5. − 230 kJ · mol 1 1. + 1.14 V Explanation: 2. − 0.57 V correct 023 10.0 points The standard potential of the cell 2− 3. + 0.57 V Pb(s) | PbSO4(s) | SO4 (aq) || Pb2+(aq) | Pb(s) ◦ 4. − 1.14 V is +0.23 V at 25 C. Calculate the equilibrium constant for the reaction of 1 M Pb2+(aq) − 5. − 0.25 V with 1 M SO2 (aq). 4 Explanation: 1. 3.7 × 1016