Quiz 7

Quiz 7

Due: 11:59pm on Monday, November 26, 2012

Note: You will receive no credit for late submissions. To learn more, read your instructor's Grading Policy

Question 1

Determine the oxidation number for the indicated element in each of the following compounds.

Part A

in

Express your answer as an integer.

ANSWER:

Correct

Part B

in


ANSWER:

Correct

Part C

in


ANSWER:

4

2

3

Correct

Part D

in


ANSWER:

Correct

Part E

in


ANSWER:

Correct

Part F

in


ANSWER:

Correct

-2

3

6

Question 2

Part A

Which of the following are redox reactions?

Check all that apply.

ANSWER:

Correct

Part B

For those reactions that are redox, indicate which elements are oxidized.

Express your answers as chemical symbols separated by a commas.

ANSWER:

Correct

Part C

For those reactions that are redox, indicate which elements are reduced.

Express your answers as chemical symbols separated by a commas.

ANSWER:

P4(s) + 10HClO(aq) + 6H2O(l) ! 4H3PO4(aq) + 10HCl(aq)

Br2(l) + 2K(s) ! 2KBr(s)

CH3CH2OH(l) + 3O2(g) ! 3H2O(l) + 2CO2(g)

ZnCl2(aq) + 2NaOH(aq) ! Zn(OH)2(s) + 2NaCl(aq)

Correct

Part D

For those reactions that are not redox, indicate whether they are precipitation or neutralization reactions.

ANSWER:

Correct

Question 3

Redox reactions can be balanced using the method of half reactions. To balance a redox reaction in acidic solution,first break the reaction into two unbalanced half reactions. Each half reaction must then be balanced separatelyusing the following steps: (1) balance all atoms except H and O; (2) balance the oxygen atoms by adding water tothe side that needs more oxygen; (3) balance hydrogen atoms by adding hydrogen ions ( ) to the side that

needs more hydrogen atoms; (4) balance the charges on each side by adding electrons to the side that has thegreater positive charge. Once each half reaction is balanced, multiply both half reactions by a whole number inorder to make electrons lost in one half reaction equal the number of electrons gained in the other half reaction.Next, add both half reactions and simplify the equation. Finally, check your work by counting the atoms and totalcharges on each side of the equation.

Part A

Balance the reaction which occurs under acidic conditions:

Report the coefficients needed to balance the equation in the following order with each coefficientseparated by a comma: (example: 1,2,3,4,5,6)

ANSWER:

Correct

precipitation

neutralization

1,3,8,2,3,4

Part B

Balance the following redox reaction which occurs under acidic conditions:

Report the coefficients needed to balance the equation in the following order with each coefficientseparated by a comma:

ANSWER:

Correct

Question 4

Redox reactions can be balanced using the method of half reactions. To balance a redox reaction in basic solution,first break the reaction into two unbalanced half reactions. Balance each half reaction as if it occurred in acidicsolution: (1) balance all atoms except H and O; (2) balance the oxygen atoms by adding water to the side thatneeds more oxygen; (3) balance hydrogen atoms by adding hydrogen ions ( ) to the side that needs more

hydrogen atoms; (4) balance the charges on each side by adding electrons to the side that has the greater positivecharge. Once each half reaction is balanced, you must "correct" for the fact that the reaction actually occurs in basicsolution. For each present in the half reaction, add the same number of hydroxide ions ( ) to BOTH sides

of the equation. Combine the hydrogen and hydoxide ions that are found on the same side of the equation to formwater. Cancel out water molecules as needed to simplify. Now, multiply both half reactions by a whole number inorder to make electrons lost in one half reaction equal the number of electrons gained in the other half reaction.Next, add both half reactions and simplify the equation. Finally, check your work by counting the atoms and totalcharges on each side of the equation.

Part A

Balance the following equation in basic solution:

Report the coefficients needed to balance the equation in the following order with the numbersseparated by a comma: (Example: 1,2,3,4,5,6)

ANSWER:

Correct

Part B

5,2,16,10,2,8

4,3,4,4,3,8

Balance the following equation in basic solution:

Report the coefficients needed to balance the equation in the following order with the numbersseparated by a comma:

ANSWER:

Correct

Question 5

Indicate whether each of the following statements is true or false.

Part A

If something is oxidized, it is formally losing electrons.

ANSWER:

Correct

Part B

For the reaction,

is the reducing agent and is the oxidizing agent.

ANSWER:

Correct

1,4,2,2,4,5

true

false

true

false

Part C

If there are no changes in the oxidation state of the reactants or products of a particular reaction, that reactionis not a redox reaction.

ANSWER:

Correct

Question 6

A voltaic cell similar to that shown in Figure 20.5 in the textbook is constructed. One electrode compartmentconsists of an aluminum strip placed in a solution of , and the other has a nickel strip placed in a solution

of . The overall cell reaction is

Part A

What is being oxidized?

Express your answer as a chemical expression.

ANSWER:

Correct

Part B

What is being reduced?

Express your answer as a chemical expression.

ANSWER:

true

false

Correct

Part C

Write the half-reactions that occur in the two electrode compartments.

Express your answer as a chemical equation. Identify all of the phases in your answer.

ANSWER:

Correct

Part D


ANSWER:

Correct

Part E

Which electrode is the anode, and which is the cathode?

ANSWER:

Correct

Part F

anode reaction:

cathode reaction:

electrode is the anode, electrode is the cathode.

electrode is the anode, electrode is the cathode.

Indicate the signs of the electrodes.

ANSWER:

Correct

Part G

Do electrons flow from the aluminum electrode to the nickel electrode, or from the nickel to the aluminum?

ANSWER:

Correct

Part H

In which directions do the cations and anions migrate through the solution? Assume the is not coated with

its oxide.

ANSWER:

Correct

Question 7

electrode is negative, electrode is positive.

electrode is negative, electrode is positive.

Electrons flow from the nickel electrode to the aluminum electrode.

Electrons flow from the aluminum electrode to the nickel electrode.

Cations and anions migrate to the nickel electrode.

Cations migrate to the nickel electrode, anions migrate to the aluminum electrode.

Cations migrate to the aluminum electrode, anions migrate to the nickel electrode.

Cations and anions migrate to the aluminum electrode.

Nickel and aluminum electrodes are used to build a galvanic cell. At 25 , the standard reduction potential for the

nickel(II) ion is 0.26 and that of the aluminum(III) ion is 1.66 .

Part A

What is the theoretical cell potential assuming standard conditions?

Express your answer with the appropriate units.

Hint 1. How to approach the problem

Recall that the potentials for nickel and aluminum found in a table are both written in reduction form. Weknow that, in any redox reaction, there must be only one reduction half-reaction and one oxidation half-reaction. Therefore, you will have to change the sign on one of the values to express it in oxidation form.Once you have done that, you simply add the two numbers:

Also recall that spontaneous redox reactions must have a positive sign for .

ANSWER:

Correct

Part B

Which metal is the cathode?


Reduction occurs at the cathode. In your calculation for Part A, you had to change the sign of one of thehalf-reactions to indicate that it is the oxidation half-reaction. The half-reaction whose sign you did notchange is therefore the reduction half-reaction.

ANSWER:

= 1.40

CorrectThe half-reaction involving nickel is a reduction reaction, thus making the cathode. Aluminum is

oxidized and is therefore the anode.

Part C

Which statement is true?

Hint 1. Electron flow

In any redox reaction, one component loses electrons and one component gains electrons. Therefore,electrons must be flowing toward the component that is gaining electrons.

Hint 2. Electrode mass

The oxidation half-reaction tells us that solid aluminum is being turned into aqueous aluminum ions. Inother words, the aluminum electrode is dissolving.

The reduction half-reaction tells us that nickel ions are being turned into solid nickel, which plates ontothe nickel electrode.

ANSWER:

Correct

Part D

Al

Ni

Electrons flow from Ni to Al.

Electrons flow from Al to Ni.

The nickel electrode loses mass.

The aluminum electrode gains mass.

Type the shorthand notation for this cell. Do not include concentrations.

For example, in shorthand notation your answer might look like Cu|Cu^+||Pd^(2+)|Pd. To enter avertical line, type shift+backslash.


In shorthand notation, a single vertical line (|) represents a phase boundary, such as that between asolid electrode and an aqueous solution, and the double vertical line (||) denotes a salt bridge. Theshorthand for the anode half-cell is always written on the left of the salt-bridge symbol, followed on theright by the shorthand for the cathode half-cell. The electrodes are written on the extreme left (anode)and the extreme right (cathode), so that each half-cell appears as reactant followed by product.

ANSWER:

Correct

Question 8

Using standard reduction potentials (Appendix E in the textbook), calculate the standard emf for each of thefollowing reactions.

Part A

.

Express your answer using three significant figures.

ANSWER:

Correct

Part B

.


Al|Al^(3+)||Ni^(2+)|Ni

= 0.823

ANSWER:

Correct

Part C

.

Express your answer using four significant figures.

ANSWER:

Correct

Part D

.

Express your answer using two significant figures.

ANSWER:

Correct

Question 9

Part A

Assuming standard conditions, arrange the following in order of decreasing strength as oxidizing agents inacidic solution: .

Rank from strongest to weakest. To rank items as equivalent, overlap them.

ANSWER:

= 1.89

= 1.211

= 0.62

Answer Requested

Part B

Arrange the following in order of decreasing strength as reducing agents in acidic solution: .

Rank from strongest to weakest. To rank items as equivalent, overlap them.

ANSWER:

Correct

Question 10

The equilibrium constant, , for a redox reaction is related to the standard potential, , by the equation

where is the number of moles of electrons transferred, (the Faraday constant) is equal to 96,500 ,

(the gas constant) is equal to 8.314 , and is the Kelvin temperature.

Standard reduction potentials

Reduction half-reaction ( )

0.80

0.34

0.15

0

0.26

0.45

0.76

1.66

2.37

Part A

Use the table of standard reduction potentials given above to calculate the equilibrium constant at 25 for the

following reaction:

Express your answer numerically using two significant figures.


First, determine for the reaction using values from the table. Next, determine the value of n by

looking at the balanced overall reaction. Finally, convert to kelvin and solve the following equation for

:

Recall that if , then .

Hint 2. Determine the value of Eº

What is the value of for this reaction?

Express your answer numerically in volts.


Determine the potential for each half-reaction using the table, then add them together:

Be aware that this table only shows reduction half-reactions. Therefore, the potential for the

corresponding oxidation half-reaction has the opposite sign of the value given.

Hint 2. Determine the reduction potential

What is the value of for the following half-reaction?


ANSWER:

Hint 3. Determine the oxidation potential




By reversing any of the reduction half-reactions, you get the corresponding oxidation half-reaction for which has the opposite sign of .

ANSWER:

ANSWER:

Hint 3. Determine the value of n

What is the value of for the following reaction?

Express your answer numerically as an integer.

= -0.26

= 0.45

= 0.19


The value for is equal to the number of moles of electrons transferred in the balanced overallreaction. Here are some examples:

Now determine the value of for the reaction in question:

ANSWER:

Hint 4. Convert the temperature to kelvin

What Kelvin temperature corresponds to 25 ?

Express your answer numerically in kelvins.

ANSWER:

ANSWER:

CorrectWhen and the reaction favors the products.

= 2

= 298

= 2.7"106

Part B

Calculate the standard cell potential ( ) for the reaction

if T = 25 and = 1.31"10#3.

Express your answer numerically in volts using two significant figures.


First, determine the value of n. Then, use the following equation to find :

Keep in mind that standard temperature is 298 .

Hint 2. Determine the value of n

What value of n should be used to calculate ?

Express your answer numerically as an integer.


The value for is equal to the number of moles of electrons transferred in the balanced overallreaction. Here are some examples:


ANSWER:

ANSWER:

All attempts used; correct answer displayedWhen and the reaction favors reactants.

Question 11

Given the following reduction half-reactions:

Part A

Write balanced chemical equation for the oxidation of by .


ANSWER:

Correct

Part B

= 1

= -0.17

Calculate for this reaction at 298 .


ANSWER:

Correct

Part C

Calculate the equilibrium constant for this reaction at 298 .

Express your answer using one significant figure.

ANSWER:

Correct

Part D



ANSWER:

Correct

Part E



ANSWER:

= 33

= 2"10#6

Correct

Part F

Calculate the equilibrium constant for this reaction at 298 .


ANSWER:

Answer Requested

Part G



ANSWER:

Correct

Part H



ANSWER:

= 490

= 1"10#86

= -22

Correct

Part I

Calculate the equilibrium constant for this reaction at 298 . Be sure to use extra significant figures in the

value for that you calculated in the previous part to avoid rounding errors.


ANSWER:

Correct

Question 12

If the equilibrium constant for a two-electron redox reaction at 298 is 1.5"10#4, calculate the corresponding

and under standard conditions.

Part A


ANSWER:

Correct

Part B


ANSWER:

= 8000

= 22

= -0.11

Correct

Question 13

Learning Goal:

To learn how to use the Nernst equation.

The standard reduction potentials listed in any reference table are only valid at standard-state conditions of 1

and, typically, 25 . To calculate the cell potential at non-standard-state conditions, one uses the Nernst equation,

where is the potential in volts, is the standard potential in volts, is the gas constant,

is the temperature in kelvin, is the number of moles of electrons transferred, is the

Faraday constant, and is the reaction quotient.

Consider the reaction

at 89 , where 2.90 and 0.310 .

Part A

What is the value for the reaction quotient, , for the cell?

Express your answer numerically.


In the reaction quotient, the concentrations of the aqueous products appear in the numerator, and theconcentrations of the aqueous reactants appear in the denominator. Just like an equilibrium constant,

, the coefficients in the balanced equation must also be considered.

Hint 2. Determine the setup for Q

In this problem, what setup is used to calculate the value for the reaction quotient, ?

ANSWER:

Correct

ANSWER:

Correct

Part B

What is the value for the temperature, , in kelvin?

Express your answer to three significant figures and include the appropriate units.

Hint 1. Do an example conversion

What Kelvin temperature corresponds to the Celsius temperature of 0 ?


ANSWER:

ANSWER:

= 0.107

= 273

Correct

Part C

What is the value for ?

Express your answer as an integer and include the appropriate units.


The value for is equal to the number of moles of electrons transferred in the balanced overall reaction.Here are some examples:


ANSWER:

Correct

Part D

Calculate the standard cell potential for

= 362

= 2



Using the table below determine the potential for each half-reaction, then add them together:

Reduction half-reaction ( )

0.80

0.34

0.15

0

0.26

0.45

0.76

1.66

2.37

Be aware that this table only shows reduction half-reactions. Therefore, the potential for the oxidationhalf-reaction has the opposite sign of the value given.

Hint 2. Determine the oxidation potential



ANSWER:

Hint 3. Determine the reduction potential


= 2.37

Express your answer to two decimal places and include the appropriate units.

ANSWER:

ANSWER:

Correct

Part E

What is the cell potential for the reaction

at 89 when 2.90 and 0.310 .



Use the Nernst equation

to calculate . Plug in the values of , , , and that you calculated in the previous parts.

ANSWER:

Correct

Question 14

A voltaic cell is constructed that uses the following reaction and operates at 298 :

= -0.45

= 1.92

= 1.95

.

Part A

What is the emf of this cell under standard conditions?


ANSWER:

Correct

Part B

What is the emf of this cell when 3.20 and 0.180 ?


ANSWER:

Correct

Part C

What is the emf of the cell when 0.200 and 0.810 ?


ANSWER:

Correct

Question 15

= 0.48

= 0.52

= 0.47

A voltaic cell utilizes the following reaction:.

Part A

What is the emf of this cell under standard conditions?


ANSWER:

Correct

Part B

What is the emf of this cell when 2.0 , 1.9"10#2 , 0.58 and the of the

solution in the cathode compartment is 2.5?


ANSWER:

Answer Requested

Score Summary:

Your score on this assignment is 89.1%.You received 13.37 out of a possible total of 15 points.

= 0.46

= 0.43

Quiz 7

Documents

Transcript of Quiz 7