Version 044 – Exam 4 MWF ONLY – sutcliffe – (51465) - PSP ...

Version 044 – Exam 4 MWF ONLY – sutcliffe – (51465) 1

This print-out should have 26 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.This exam is ONLY to be taken by students

enrolled in the MWF section. DO NOT takethis exam if you are NOT in the MWF sectionas your exam will NOT be graded. Instead,notify a TA immediately.

001 10.0 points

Arrange the half reactions shown belowI) Na+ + e− → Na E◦

red = −2.71II) Ba2+ + 2 e− → Ba E◦

red = −2.91III) I2 + 2 e− → 2 I− E◦

red = +0.54IV) Au3+ + 3 e− → Au E◦

red = +1.50V) Ti3+ + e− → Ti2+ E◦

red = −0.37so that they show an increasing order of re-ducing agent strength. (For example: a choicethat listed I before II would be stating thatthe reducing agent shown in reaction I was aweaker reducing agent than the one shown inreaction II.)

1. IV, III, V, I, II correct

2. I, II, V, III, IV

3. I, II, III, IV, V

4. II, V, III, IV, I

5. V, IV, III, II, I

6. IV, I, V, II, III

Explanation:

Reducing agents cause other species to bereduced and so themselves are oxidized in theprocess. Standard reduction potential equa-tions like those given in the question show thetendency of a reactant species to be reduced.The more negative the standard reductionpotential value, the less this reactant speciesdesires to gain electrons, so the weaker an ox-idizing agent it is. Therefore the strongestreducing agents are shown in standard reduc-tion reactions as products and the reactionitself has the most negative standard reduc-tion potential values.

002 10.0 points

003 10.0 points

Rank the following salts from least to greatestmolar solubility:A) KIO4 Ksp = 0.000371

B) CdS Ksp = 3.6× 10−29

C) CuS Ksp = 8.5× 10−45

D) CaCr2O4 Ksp = 0.023

1. C < D < A < B

2. B < C < A < D

3. B < C < D < A

4. C < B < A < D correct

Explanation:

All of the salts are composed of a singlecation and anion, so the solubility of each isthe square root of its Ksp.


004 10.0 points

005 10.0 points

Note by excess, we mean much more is addedthan could possibly dissolve.BaF2(s) is slightly soluble in water at 25◦C,

Ksp = 4.7× 10−6. If excess BaF2(s) is put inthe water, what is the equilibrium concentra-tion [Ba2+]?

1. 1.08× 10−3 M

2. 1.06× 10−2 M correct

3. None of these

4. 2.4× 10−2 M

5. 2.4× 10−6 M

Explanation:

006 10.0 points

Consider the half-reaction2 H+(aq) + 2 e− ⇀↽ H2(g) E0 = 0.000 VAssuming the pressure of the hydrogen gas

is 1 atm, what is the pH of the cell if thereduction potential under non-standard con-centration is −0.1775 V?

1. 3.25

2. 3.00 correct

3. 2.00

4. 2.67

5. 1.50

6. 6.00

Explanation:

E = E0 −0.05916 V

2log

(

1

[H+]2

)

−0.1775 V = (−0.02958 V)(

−2 log[H+])

6 = −2 log[H+]

pH = 3

007 10.0 points

If 289,500 Coulombs is passed through a solu-tion of lead(II) nitrate (Pb(NO3)2) how manymoles of metallic lead will be produced?

1. 0.667 mol

2. 1.5 mol correct

3. 3 mol

4. 2 mol

Explanation:

q = 289500 C1 F = 1 mol e− = 96485 C.Write the lead half reaction:

Pb2+ + 2 e− → Pb

nPb = (289500 C) ·1 mol e−

96485 C·1 mol Pb

2 mol e−

= 1.50023 mol Pb


008 10.0 points

What are the units of a rate constant that is1.5 order in A and 1.0 order in B?

1. M−2.5sec−1

2. M−0.5sec−1

3. M1.5sec−1

4. M−1.5sec−1 correct

5. M−1sec−1

Explanation:

009 10.0 points

010 10.0 points

011 10.0 points

Ksp for iron(III) iodate (Fe(IO3)3) is 1.0 ×

10−14. We mix two solutions, one containingFe3+ and one containing IO−

3 . At the instant

of mixing, [Fe3+] = 10−4 M and [IO−

3 ] = 10−5

M. Which of the following statements is true?


1. No precipitate forms because Qsp is lessthan Ksp. correct

2.A precipitate forms because Qsp is greaterthan Ksp.

3. A precipitate forms because Qsp is lessthan Ksp.

4. None of the statements is true.

5. No precipitate forms because Qsp isgreater than Ksp.

Explanation:

012 10.0 points

013 10.0 points

Read carefully! The question says in theEXTERNAL circuit.Consider the voltaic cellIn(s) | In3+ (0.010 M) || Ru3+ (1.0 M),

Ru2+ (0.010 M) | C(s, graphite)

In3+ + 3 e− → In E0 = −0.34 VRu3+ + 1 e− → Ru2+ E0 = −0.08 VThe electron flow in the external circuit is

from

1. In3+ to Ru2+.

2. In to Ru3+.

3. In to C. correct

4. Ru3+ to In.

5. In3+ to C.

Explanation:


014 10.0 points

Consider the following statements:Z1) In electrolytic cells, the flow of electrons

is spontaneous.Z2) In electrolytic cells, the electrons flow in

the external circuit (through the wire)from the anode to cathode.

Z3) In electrolytic cells, the cathode is thenegative electrode.

Which of the following is true?

1. Z1 and Z3 only

2. Z2 and Z3 only correct

3. Z3 only

4. Z1 only

5. Z1, Z2 and Z3

Explanation:

Z1: FALSE; In electrolytic cells, the elec-tron flow is not spontaneous. An externalsource of energy must be supplied in order forthe process to take place.Z2: TRUE; Reduction occurs at the cath-

ode, so electrons must be supplied here. Thusthe flow of electrons must be from the anodeto the cathode.Z3: TRUE; As electrons (negatively

charged) are supplied at the cathode.

015 10.0 points

The rate of formation of oxygen in the follow-ing reaction is 2.28 M/s.

2N2O5(g) → 4NO2(g) + O2(g)

What is the rate of formation of NO2?

1. 1.14 M/s

2. 4.56 M/s

3. 2.28 M/s

4. 9.12 M/s correct

5. 0.57 M/s

Explanation:

The NO2 forms at 4× the rate of the O2.

016 10.0 points

017 10.0 points

Three separate experiments were performedon the rate of the reaction

3A2 + 2B → 2A3B

The measured initial concentrations of A2 (in


moles per liter) are shown below along withthe measured initial rates of formation of A3B(moles per liter per second).

Initial Initial InitialTrial [A2]0 [B]0 rate

M M M/s

1 1.5 3.0 7.0 × 10−8

2 1.5 1.5 1.75× 10−8

3 4.5 3.0 21.0 × 10−8

What is the order of the reaction?

1. None of these is correct.

2. first order in [A2] and second order in [B]correct

3. third order in [A2] and second order in[B]

4. first order in [A2] and first order in [B]

5. second order in [A2] and first order in [B]

Explanation:

Set up a ratio of the concentrations and therates to fit the expression: rate = k [A2]

x orrate = k [B]y and figure out the order of x ory to make the equation true.Hold [B] in trials 1 and 3 constant; then

Rate3Rate1

= k

(

[A2 rxn3]

[A2 rxn1]

)x

21.0× 10−8

7.0× 10−8= k

(

4.5

1.5

)x

3 = k (3)x

x = 1

The reaction is 1st order with respect to A2.Hold [A2] constant in trials 1 and 2; then

Rate2Rate1

= k

(

[Brxn2]

[Brxn1]

)y

1.75× 10−8

7.0× 10−8= k

(

1.5

3.0

)y

1

4= k

(

1

2

)y

y = 2

The reaction is 2nd order with respect to B.

018 10.0 points


019 10.0 points

Given the standard electrode potentialsFe3+(aq) + 1 e− → Fe2+(aq) E0 = +0.771 VI2(s) + 2 e− → 2 I−(aq) E0 = +0.535 Vcalculate Kc at 25

◦C for the reaction

2 Fe2+(aq) + I2(s) → 2Fe3+(aq) + 2 I−(aq) .

1. 1.05× 10−8 correct

2. 9.53× 107

3. 1.02× 10−4

4. 1.61× 1012

5. 9.76× 103

Explanation:

020 10.0 points

For a redox reaction to be SPONTANEOUSunder standard conditions, the following istrue

1. ∆G < 0 K < 1 Ecell < 0

2. ∆G < 0 K > 1 Ecell > 0 correct


3. ∆G > 0 K > 1 Ecell < 0

4. ∆G > 0 K < 1 Ecell > 0

Explanation:

∆G is negative for a spontaneous reaction.

K =[products]

[reactants], so when the equilibrium fa-

vors the products (making the reaction spon-

taneous), K > 1. Also, E0 =0.05916 logK

n.

So when K > 1, E0 is positive.

021 10.0 points

Zinc Silver

1.56 V

V

Voltmeter

e− e−

1 M Zn2+(aq) 1 M Ag+(aq)

Saltbridgeto carryions

In this electrochemical cell, what is the an-ode?

1. the solid silver electrode

2. the solid zinc electrode correct

3. the Ag+(aq) ions in the 1 M solution

4. the Zn2+(aq) ions in the 1 M solution

Explanation:

Zn(s) + 2Ag+(aq) → Zn2+(aq) + 2Ag(s)

Oxidation occurs at the anode. In this cellthe oxidation half reaction is

Zn(s) → Zn2+(aq) + 2 e− .

Zn atoms in the solid Zn electrode areoxidized to Zn2+ ions; the electrode slowlyerodes.

022 10.0 points


023 10.0 points

When concentrated sulfuric acid is pouredonto sugar in a beaker and stirred, a vigorousexothermic reaction eventually occurs. A lotof steam is produced, and a black ’snake’consisting mainly of carbon rises out of thebeaker. As this reaction takes a while toreally get going, which of the actions listedbelow might help speed up this process?

I. switching from granulated sugar to pow-dered sugar

II. switching from powdered sugar to gran-ulated sugar

III. warming up the sulfuric acid

IV. putting the sulfuric acid container in anicebath before use

1. I only2. IV only3. II and IV4. III only5. II and III6. I and IV7. II only

8. I and III CORRECT

Explanation:

A chemical reaction can be speeded up byincreasing the surface area of the reactantsand raising their temperature.

024 10.0 points

Consider the half-reactions and the balancedequation for the cell reaction represented bythe skeletal equation

Mn(s) + Ti2+(aq) → Mn2+(aq) + Ti(s) .

What is the proper cell diagram for this reac-tion?

1. Mn2+(aq) |Mn(s) ||Ti(s) |Ti2+(aq)

2. Ti2+(aq) |Ti(s) ||Mn(s) |Mn2+(aq)

3. Ti(s) |Ti2+(aq) ||Mn2+(aq) |Mn(s)

4. Mn(s) |Mn2+(aq) ||Ti2+(aq) |Ti(s) cor-

rect

Explanation:

The two half-reactions, written as reduc-tions, are

Mn2+(aq) + 2 e− → Mn(s)

Ti2+(aq) + 2 e− → Ti(s)

Equate e−:Ti2+(aq) + 2 e− → Ti(s)

Mn(s) → Mn2+(aq) + 2 e−

Add the balanced half reactions:

Mn(s) + Ti2+(aq) → Mn2+(aq) + Ti(s)

To write the cell diagram we place thespecies involved with the reduction reaction(at the cathode) on the right of the salt bridgeand those involved with the oxidation reaction(at the anode) on the left of the bridge. Elec-trons are not shown, and the extreme left andright species listed are the electrode materialsthe anode and cathode respectively, are madeof. As neither reaction involves solely gases or


aqueous ions, we can use the elemental metalsinvolved in the half reactions.The cell diagram is

Mn(s) |Mn2+(aq) ||Ti2+(aq) |Ti(s)

025 10.0 points

Consider the lead storage battery and thechemical reaction that drives it forward ondischarge. There are various substancespresent as reactants and products. Whichof the following substances is reduced as thebattery discharges.

1. Pb (s)

2. PbCl2 (s)

3. H2SO4 (aq)

4. Pb-Ni(CO3)2 (s)

5. PbO2 (s) correct

6. PbSO4 (s)

Explanation:

PbO2 has lead in the +4 oxidation state.This will accept 2 electrons and thus be re-duced to PbSO4 as the battery discharges. Pbmetal is oxidized to PbSO4.

026 10.0 points

Galvanizing is a process done to steel itemssuch as trashcans, buckets, etc. to help pre-vent rusting (corrosion) of the iron. Whichstatement would be correct about galvaniz-ing?

1. Galvanizing involves bolting a lump ofzinc to the side of the steel item; the zinccorrodes in preference to the iron in the steel.2. Corrosion of iron occurs when iron is

exposed to air and moisture, and the ironbecomes reduced.3. Galvanizing involves coating the item

with a layer of aluminum, which forms a toughoxide coating on exposure to air.4. If a small hole develops in the zinc layer,

the zinc still oxidizes in preference to the iron

in the steel as zinc is more reactive than iron.CORRECT

Explanation:

Galvanizing involves completely coating asteel or iron item with a thin layer of zinc byeither dipping in molten zinc or electroplat-ing. As zinc is more reactive than iron, it willoxidize in preference to the iron. However, iftoo large of a hole develops, there will not beenough zinc to corrode in preference, which iswhy old trashcans are usually rusting on thecorners, edges and rivets - places likely to besubject to rougher handling. A solid block ofzinc attached to a steel object (or connectedto it via a wire) is referred to as a sacrificialanode. The zinc again corrodes in preferenceto the iron, but the block of zinc (anode) isintended to be replaced at regular intervals.

Version 044 – Exam 4 MWF ONLY – sutcliffe – (51465) - PSP ...

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