Download 25 Questions on Electrochemistry II with Solution - Quiz | CH 302 and more Quizzes Chemistry in PDF only on Docsity!
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001 10.0 points How many moles of Cl 2 (g) are produced by the electrolysis of concentrated sodium chlo- ride if 2.00 A are passed through the solution for 4.00 hours? The equation for this process (the “chloralkali” process) is 2 NaCl(aq)+ 2 H 2 O(ℓ) → 2 NaOH(aq) + H 2 (g) + Cl 2 (g)
- 0.00248 mol
- 0.298 mol
- 0.0745 mol
- 0.447 mol
- 0.149 mol correct
Explanation:
002 10.0 points A steel surface has been electroplated with 5.10 g of vanadium (V, molar mass = 51 g/mol). If 2.90× 104 C of charge were used, what was the original oxidation number of V?
- +3 correct
Explanation:
moles V =
5 .1 g 51 g/mol
= 0.1 mol
moles e−^ =
2. 90 × 104 C
- 65 × 104 C/mol
= 0.3 mol
ox. number =
0 .3 mol e− 0 .1 mol V
003 10.0 points How long will it take to deposit 0.00235 moles of gold by the electrolysis of KAuCl 4 (aq) us- ing a current of 0.214 amperes?
- 17.7 min
- 53.0 min correct
- 106 min
- 26.5 min
- 70.7 min
Explanation:
004 10.0 points Consider 3 electrolysis experiments: The first: 1 Faraday of electricity is passed through a solution of AgNO 3. The second: 2 Faradays of electricity are passed through a solution of Zn(NO 3 ) 2. The third: 3 Faradays of electricity are passed through a solution of Bi(NO 3 ) 3.
- Twice as many moles of metallic zinc are produced than metallic silver.
- Equal masses of all three metals are pro- duced.
- The reaction producing the smallest mass of metal is that of the silver solution.
- Equal numbers of moles of all three metals are produced. correct
Explanation: 1 F = 1 mol e−. The relevant half-reactions are
Ag+^ + 1 e−^ → Ag Zn2+^ + 2 e−^ → Zn Bi3+^ + 3 e−^ → Bi
Using the given amounts of electricity,
Exp. metal grams produced produced 1 1 mol Ag (1 mol Ag)
107 .87 g Ag mol Ag 2 1 mol Zn (1 mol Zn)
65 .39 g Zn mol Zn 3 1 mol Bi (1 mol Bi)
208 .98 g Bi mol Bi
Of the answer choices, the true statement says that equal moles of metals are produced.
005 10.0 points Sodium is produced by electrolysis of molten sodium chloride. What are the products at the anode and cathode, respectively?
- Cl 2 (g) and Na 2 O(ℓ)
- Na(ℓ) and O 2 (g)
- Cl−(aq) and Na 2 O(ℓ)
- Cl 2 (g) and Na(ℓ) correct
- O 2 (g) and Na(ℓ)
Explanation:
006 10.0 points What is the standard cell potential of the strongest battery that could be made using these half reactions? Br 2 + 2 e−^ −→ 2 Br− Fe3+^ + 3 e−^ −→ Fe Co3+^ + e−^ −→ Co2+ Zn2+^ + 2 e−^ −→ Zn
E◦^ = +1. 07
E◦^ = − 0. 04
E◦^ = +1. 80
E◦^ = − 0. 76
- 2.56 correct
Explanation: The strongest reducing agent is Co3+^ and the strongest oxidizing agent is Zn. The stan- dard cell potential of a battery built from these species would be:
E cell◦ = E cathode◦ − E anode◦ = 1. 80 − (− 0 .76) = 2. 56
007 10.0 points What would be the E◦^ cell of an electrolytic cell made from the half reactions AgCl(s) + e−^ −→ Ag(s) + Cl−(aq) E◦^ = +0.22 V Al3+(aq) + 3 e−^ −→ Al(s) E◦^ = − 1 .66 V
- − 1. 44
- 44
- − 1. 88 correct
- 88
Explanation:
E◦^ = E◦ cathode − E anode◦ = − 1. 66 − 0 .22 = − 1. 88
008 (part 1 of 3) 10.0 points
The galvanic cell below uses the standard half-cells Mg2+^ | Mg and Zn2+^ | Zn, and a salt bridge containing KCl(aq). The voltmeter gives a positive voltage reading.
Correct answer: 0.444884 V.
Explanation:
014 10.0 points Standard reduction potentials are established by comparison to the potential of which half reaction?
- 2 H 2 O + 2 e−^ −→ H 2 + 2 OH−
- Na+^ + e−^ −→ Na
- F 2 + 2 e−^ −→ 2 F−
- 2 H+^ + 2 e−^ −→ H 2 correct
- Li+^ + e−^ −→ Li
Explanation: The hydrogen electrode is the standard ref- erence electrode
015 10.0 points Consider the cell Pb(s) | PbSO 4 (s) | SO^24 − (aq , 0 .60 M) || H+(aq , 0 .70 M) | H 2 (g , 192 .5 kPa) | Pt. If E◦^ for the cell is 0.36 V at 25◦C, write the Nernst equation for the cell at this tem- perature.
- E = 0.36 − 0.01285 ln
[
]
- E = 0.36 + 0.01285 ln
[
(0.70)^2 (0.60)
]
- E = 0.36 + 0.01285 ln
[
(0.70)^2 (0.60)
]
- E = 0.36 − 0.02569 ln
[
(0.70)^2 (0.60)
]
- E = 0.36 − 0.01285 ln
[
(0.70)^2 (0.60)
]
correct
Explanation:
016 10.0 points A concentration cell consists of the same re- dox couples at the anode and the cathode, with different concentrations of the ions in
the respective compartments. Find the un- known concentration for the following cell. Pb(s) | Pb2+(aq, ?) || Pb2+(aq, 0 .1 M) | Pb(s) E = 0.066 V
Correct answer: 0.000587185 M. Explanation: Ecell = 0.066 V M = 0.1 M E cell◦ = 0 V F = 96485 C/mol R T F
= 0.025693 V
The cell reaction is Pb2+(aq, 0 .1 M) → Pb2+(aq, x) n = 2, Ecell = 0.066 V Using the Nernst equation,
Ecell = E cell◦ −
R T
n F
ln Q R T n F
ln
( (^) x M
= E cell◦ − Ecell
ln
( (^) x M
n F R T
(E◦ cell − Ecell)
x = M exp
[
n F R T
(E cell◦ − Ecell)
]
x = (0.1 M) exp
[
2(0 V − 0 .066 V)
0 .025693 V
]
= 0.000587185 M.
Thus [Pb2+] = 0.000587185 M.
017 10.0 points What is ratio of [Co2+] | [Ni2+] when a bat- tery built from the two half reactions Ni2+^ −→ Ni E◦^ = − 0 .25 V Co2+^ −→ Co E◦^ = − 0 .28 V reaches equilibrium?
- 10.23 correct
Explanation: E◦ cell = +0.03 V
Ecell = E cell◦ −
Ne
log Q
log
[Co2+] [Ni2+]
log
[Co2+] [Ni2+]
[Co2+] [Ni2+]
= 10^1.^01 = 10. 23
018 10.0 points If E◦^ for the disproportionation of Cu+(aq) to Cu2+(aq) and Cu(s) is +0.37 V at 25◦C, calculate the equilibrium constant for the re- action.
- 3 × 103
- 4 × 102
- 2 × 1012
- 7 × 1018
- 8 × 106 correct
Explanation: The disproportionation is 2 Cu+^ → Cu2++ Cu This corresponds to a ONE electron trans- fer from one Cu+^ ion to another Cu+^ ion. Therefore n = 1 for this reaction.
nF E = RT ln K K = enF E/RT^ = e1(0.37)/^0.^257 = 1. 8 × 106
019 10.0 points You turn on a flashlight containing brand new NiCad batteries and keep it lit for a minute or two. Which of the following can be considered TRUE regarding the chemical state of these batteries? I. ∆G for the battery reaction is negative. II. Ecell > 0. III. The batteries are at equilibrium. IV. Ecell is substantially decreasing during this time.
- III only
- All but III
- Maybe IV and I
- All but IV
- II and IV only
- I and II only correct
- All Explanation: (I and II: TRUE) ∆G must be negative because the light IS on and therefore a spon- taneous change is occuring. Same logic goes for the Ecell except opposite in sign (positive). (III: FALSE) The batteries are headed to- ward equilibrium but are not there yet. True equilibrium would mean a potential of zero volts - the ultimate DEAD battery. (IV: FALSE) The reason batteries work as well as they do is because they hold a very constant potential throughout most of the life of the battery. Only near the battery’s end does the potential start to fall off from the brand new potential. Why? Most batteries rely on solids in their equilibria which means constant voltage as long as the solid reactants (and products) are present (Q in the Nernst equation will equal 1). Once the solids are depleted, you see the great change in potential and usually the failure of the battery to work in its intended device.
020 10.0 points What is ∆G◦^ for the half reaction below? Reaction E◦ ClO− 3 + 6 H+(aq) −→
Cl 2 (g) + 3 H 2 O(ℓ) +1. 47
- −709 kJ · mol−^1 correct
- − 1 , 418 kJ · mol−^1
- 194 , 000 kJ · mol−^1
- 194 kJ · mol−^1
