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Unit 3: Presentation E

Redox Reactions

AP Chemistry

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REDOX Reactions

Photosynthesis is a classic REDOX reaction where electrons are transferred from water to carbon dioxide to create a new

sugar molecule.

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Oxidation and ReductionREDOX reactions involve the transfer of electrons from one

species to another. The substance that gains electrons is said to be reduced while the substance that loses electrons is said

to be oxidized.

Cu(s) + Ag+(aq) --> Cu+

(aq) + Ag(s)

In this reaction, a copper atom loses an electron to a silver ion.

Cu(s) got oxidized

Ag+(aq) got reduced

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Oxidation and Reduction

Oxidation ReductionLoses electrons (LEO) gains electrons (GER)

oxidation state increases oxidation state decreaseselectrons are product

Na(s) --> Na+(aq) + e-

electrons are reactant

Mg2+(aq) + 2e- --> Mg(s)

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Oxidation StatesIn order to determine whether a substance got oxidized or reduced in a reaction, the oxidation state of each species must be known.

If the substance is found in its neutral standard state, its oxidation state will be zero.

Na(s) = 0 or F2(g) = 0 or C(graphite) = 0

If the substance is a molecular compound made of more than one element, the sum of the oxidation states must equal 0

C O2 The more electronegative of the two receives a

charge equivalent to it's normal ionic charge.

-2The less electronegative of the two receives the +

charge necessary to cancel out the negative

charge

+4

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Oxidation StatesFor ionic species, the sum of the oxidation states must equal

the charge on the ion.

S O32-

+4 -2For ionic compounds, first separate the compound into its

ions, then determine charge on each element.

Mg Cr2O7+2 -2

Cr2 O7 + 6 -2

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Oxidation States

Peroxides and hydrides

In a peroxide (H2O2 or Na2O2) O carries an oxidation state of -1.

In a hydride (NaH or CaH2), H carries an oxidation state of -1.

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1 Oxidation/Reduction reactions study the transfer of ________ between atoms and molecules.

A protons

B neutrons

C H+ ions

D OH- ions

E electrons

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1 Oxidation/Reduction reactions study the transfer of ________ between atoms and molecules.

A protons

B neutrons

C H+ ions

D OH- ions

E electrons [This object is a pull tab]

Ans

wer

A

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2 Which of the substances found in the reaction below would contain an element with an oxidation number = 0?10Au(s) + 36H+(aq) + 6NO3-(aq) --> 3N2(g) + 18H2O(l) + 10Au3+(aq)

A AuB HNO3

C Au and N2

D Au and H2OE N2 and Au3+

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2 Which of the substances found in the reaction below would contain an element with an oxidation number = 0?10Au(s) + 36H+(aq) + 6NO3-(aq) --> 3N2(g) + 18H2O(l) + 10Au3+(aq)

A AuB HNO3

C Au and N2

D Au and H2OE N2 and Au3+ [This object is a pull tab]

Ans

wer

C


3 In the substance glucose (C6H12O6), the sum total of the oxidation states must equal _____.

A 0

B +1

C The sum total of the number of valence electrons present in each element

D +3E 24

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3 In the substance glucose (C6H12O6), the sum total of the oxidation states must equal _____.

A 0

B +1

C The sum total of the number of valence electrons present in each element

D +3E 24

[This object is a pull tab]

Ans

wer

A


4 In the compound IBr, which element would be the most electronegative and what would it's oxidation state be?

I Br

A I, -1

B Br, -1

C I, 0

D Br, 0

E Br, +1

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4 In the compound IBr, which element would be the most electronegative and what would it's oxidation state be?

I Br

A I, -1

B Br, -1

C I, 0

D Br, 0

E Br, +1


Ans

wer

B


5 The oxalate ion, C2O42-, is one of the common components of kidney stones! Which is the more electronegative atom in the ion and what is it's oxidation state?

A C, +4

B C, -2

C C, -4

D O, -2E O, -4

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5 The oxalate ion, C2O42-, is one of the common components of kidney stones! Which is the more electronegative atom in the ion and what is it's oxidation state?

A C, +4

B C, -2

C C, -4

D O, -2E O, -4


Ans

wer

D


6 The polyatomic ion cyanide (CN-) is toxic and used in the mining of gold in America's desert southwest. Which of the following could be a correct set of oxidation states for C and N in the ion?

A C = +4, N = -3B C = +4, N = 0C C = +2, N = -3

D C = 0, N = +1

E C = -3, N = +4

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6 The polyatomic ion cyanide (CN-) is toxic and used in the mining of gold in America's desert southwest. Which of the following could be a correct set of oxidation states for C and N in the ion?

A C = +4, N = -3B C = +4, N = 0C C = +2, N = -3

D C = 0, N = +1

E C = -3, N = +4


Ans

wer

C


7 What is the oxidation state of O in the compound N2O (laughing gas) and what would be the total negative charge from all the oxygens?

N N OA -2, -2

B -1, 0

C -2, 0

D -2, -4E -2, -4

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7 What is the oxidation state of O in the compound N2O (laughing gas) and what would be the total negative charge from all the oxygens?

N N OA -2, -2

B -1, 0

C -2, 0

D -2, -4E -2, -4


Ans

wer

A


8 What would be the oxidation state of the fluorine atom in the compound OF2 and what would be the total negative charge from the Flourines?

A -1, -1B 0, -1C -1, -2D -1, -3E -1, 0

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8 What would be the oxidation state of the fluorine atom in the compound OF2 and what would be the total negative charge from the Flourines?

A -1, -1B 0, -1C -1, -2D -1, -3E -1, 0


Ans

wer

C


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10 What is the oxidation number of each oxygen atom in the compound MnO 2 ?

A -2

B -1

C 0

D +1

E +2

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10 What is the oxidation number of each oxygen atom in the compound MnO 2 ?

A -2

B -1

C 0

D +1

E +2[This object is a pull tab]

Ans

wer

A


11 What is the oxidation state of S in the S2O32- ion? The S2O32- ion can do great harm to metals like iron.

A -2

B +2

C -1D +4E 0

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11 What is the oxidation state of S in the S2O32- ion? The S2O32- ion can do great harm to metals like iron.

A -2

B +2

C -1D +4E 0


Ans

wer

B


12 What is the oxidation state of each atom in CaC2O4?

A Ca = 0, C = +2, O = -2

B Ca = +2, C = +2, O = -2C Ca = +4, C = +4, O =-2D Ca = +2, C = +3, O = -2E Ca = +2, C = +4, O = -2

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12 What is the oxidation state of each atom in CaC2O4?

A Ca = 0, C = +2, O = -2

B Ca = +2, C = +2, O = -2C Ca = +4, C = +4, O =-2D Ca = +2, C = +3, O = -2E Ca = +2, C = +4, O = -2


Ans

wer

D


13 In 2004, Coca cola had to recall their Dasani brand water because of the presence of the bromate ion in the drinking water. What would be the oxidation state of all atoms in CuBrO3?

A Cu = +2, Br = -1, O = -2

B Cu = +1, Br = +3, O = -2C Cu = +2, Br = +4, O = -2D Cu = +1, Br = +5, O =-2

E Cu = 0, Br = +6, O =-2

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13 In 2004, Coca cola had to recall their Dasani brand water because of the presence of the bromate ion in the drinking water. What would be the oxidation state of all atoms in CuBrO3?

A Cu = +2, Br = -1, O = -2

B Cu = +1, Br = +3, O = -2C Cu = +2, Br = +4, O = -2D Cu = +1, Br = +5, O =-2

E Cu = 0, Br = +6, O =-2 [This object is a pull tab]

Ans

wer

D


14 Chloroform, CHCl3, has wide uses. It is used as an anesthetic, a refrigerant, and in the making of teflon. What is the oxidation state of carbon in chloroform?

A -2

B -1

C 0

D +2

E -3

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14 Chloroform, CHCl3, has wide uses. It is used as an anesthetic, a refrigerant, and in the making of teflon. What is the oxidation state of carbon in chloroform?

A -2

B -1

C 0

D +2

E -3[This object is a pull tab]

Ans

wer

D


15 In which of the following compounds would oxygen have an oxidation state of -1?

A Li2O2

B Mg(OH)2

C CH3OH

D O2 E Li2O

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15 In which of the following compounds would oxygen have an oxidation state of -1?

A Li2O2

B Mg(OH)2

C CH3OH

D O2 E Li2O


Ans

wer

A


REDOX ReactionsTo determine which substances got oxidized or reduced, look at

the change in the substances oxidation state.

H+(aq) + I O3

-(aq) + F-

(g) --> I2(s) + F2(aq) + H2 O(l)

+1 +5 -2 -1 0 0 +1 -2

IO3-(aq) got reduced = +5 --> 0

F2(g) got oxidized = 0 --> -1

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16 Which of the following represents a reduction?

A An elements charge going from -3 --> -2

B An elements charge going from -2 -->-3

C An elements charge going from 0 --> +2

D An elements charge going from +1 --> +3

E None of these

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16 Which of the following represents a reduction?

A An elements charge going from -3 --> -2

B An elements charge going from -2 -->-3

C An elements charge going from 0 --> +2

D An elements charge going from +1 --> +3

E None of these


Ans

wer

B


17 Which of the following represents an oxidation?

A Al3+(aq) --> Al(s)

B F2(g) --> 2F-(aq)

C Mn7+ --> Mn2+

D Fe2+(aq) --> Fe3+

(aq)

E O-(aq) --> O2-

(aq)

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17 Which of the following represents an oxidation?

A Al3+(aq) --> Al(s)

B F2(g) --> 2F-(aq)

C Mn7+ --> Mn2+

D Fe2+(aq) --> Fe3+

(aq)

E O-(aq) --> O2-

(aq)


Ans

wer

D


18 In which of the following equations does O get oxidized?

A 2H2O2 -- 2H2O + O2

B 2H2O --> 2H2 + O2

C CH4 + 2O2 --> CO2 + 2H2OD 3Fe + 3O2 --> 2Fe2O3

E A and B

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18 In which of the following equations does O get oxidized?

A 2H2O2 -- 2H2O + O2

B 2H2O --> 2H2 + O2

C CH4 + 2O2 --> CO2 + 2H2OD 3Fe + 3O2 --> 2Fe2O3

E A and B


Ans

wer

E


19 Which of the following is NOT an oxidation/reduction reaction?

A Ca + 2H+ --> Ca2+ + H2

B H2 + Cl2 -->2HCl

C 4Al + 3O2 --> 2Al2O3

D CaO + CO2 --> CaCO3

E Mg3N2 --> 3Mg + N2

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19 Which of the following is NOT an oxidation/reduction reaction?

A Ca + 2H+ --> Ca2+ + H2

B H2 + Cl2 -->2HCl

C 4Al + 3O2 --> 2Al2O3

D CaO + CO2 --> CaCO3

E Mg3N2 --> 3Mg + N2


Ans

wer

D


20 Which of the following would be TRUE regarding the reaction that occurs below. Cr2O72-(aq) + 14H+(aq) + 6Fe2+(aq) --> 2Cr3+(aq) + 7H2O(l) + 6Fe3+(aq)

A Cr2O72- gets oxidized to Cr3+

B Fe2+ gets reduced to Fe3+

C H+ gets reduced to H2O

D Cr2O72- gets oxidized to H2O

E Fe2+ gets oxidized to Fe3+

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20 Which of the following would be TRUE regarding the reaction that occurs below. Cr2O72-(aq) + 14H+(aq) + 6Fe2+(aq) --> 2Cr3+(aq) + 7H2O(l) + 6Fe3+(aq)

A Cr2O72- gets oxidized to Cr3+

B Fe2+ gets reduced to Fe3+

C H+ gets reduced to H2O

D Cr2O72- gets oxidized to H2O

E Fe2+ gets oxidized to Fe3+[This object is a pull tab]

Ans

wer

E


Oxidizing and Reducing Agents

Any substance that gets oxidized is called a reducing agent or reducer becomes the electrons lost will be gained by another

species.

Any substance that gets reduced is called an oxidizing agent or oxidizer becomes the electrons gained will be lost by

another species.

Remember: Reducing Agents are Oxidized

Oxidation Agents are Reduced.

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Oxidizing and Reducing Agents

Good reducing agent must lose electrons easily. Neutral metals or substances with low ionization energies fit the bill.

For example: Na, I-, Fe2+

Good oxidizing agents must gain electrons easily. Neutral or positive non-metals or highly positive metals fit the bill here.

For example: MnO4-, CrO4

2-, F2, O2, HNO3

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21 Identify the oxidizing agent in the following reaction.

Al(s) + 6HCl(aq) --> AlCl3(aq) + H2(g)

A AlB AlCl3C HCl

D H2

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21 Identify the oxidizing agent in the following reaction.

Al(s) + 6HCl(aq) --> AlCl3(aq) + H2(g)

A AlB AlCl3C HCl

D H2


Ans

wer

C


22 Which substance is oxidized in the following reaction? (First, assign oxidation numbers.)

Cu + S ➝ CuS

A Cu

B S

C Cu and S

D CuS

E This is not a redox reaction.

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Cu + S ➝ CuS

A Cu

B S

C Cu and S

D CuS

E This is not a redox reaction.[This object is a pull tab]

Ans

wer

A


23 Which substance is the reducing agent below?

Cu + S ➝ CuS

A Cu

B S

C Cu and S

D CuS


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23 Which substance is the reducing agent below?

Cu + S ➝ CuS

A Cu

B S

C Cu and S

D CuS



Ans

wer

A



Ca + Fe3+ ➝ Ca2+ + Fe

A Ca

B Fe3+

C Ca2+

D Fe


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Ca + Fe3+ ➝ Ca2+ + Fe

A Ca

B Fe3+

C Ca2+

D Fe

E This is not a redox reaction.[This object is a pull tab]

Ans

wer

A


25 Which substance is the reducing agent?

3 K + Al(NO3)3 ➝ Al + 3 KNO3

A K

B Al(NO3)3

C KNO3

D This is not a redox reaction.

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25 Which substance is the reducing agent?

3 K + Al(NO3)3 ➝ Al + 3 KNO3

A K

B Al(NO3)3

C KNO3

D This is not a redox reaction.[This object is a pull tab]

Ans

wer

A


26 Which of the following would make the best oxidizing agent?

A H2O

B IO3-

C IO2-

D I2

E I-

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A H2O

B IO3-

C IO2-

D I2

E I-


Ans

wer

BReason: The oxidation state of Iodine in this ion is +5. Iodine is normally VERY electronegative. This high positive charge being

forced on the Iodine combined with its normally very electron hungry

nature makes Iodine VERY likely to take electrons from other chemical

species. That makes it a VERY good oxidizing agent!


27 Which of the following would make the best reducing agent?

A F2(g)

B Cl2(g)

C K(s)

D Al(s)

E C(s)

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A F2(g)

B Cl2(g)

C K(s)

D Al(s)

E C(s)


Ans

wer

C


28 Good oxidizing agents will have low ionization energies.

Yes

No

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28 Good oxidizing agents will have low ionization energies.

Yes

No


Ans

wer

NO



A HNO

B HNO2

C HNO3

D Na

E H2O

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A HNO

B HNO2

C HNO3

D Na

E H2O[This object is a pull tab]

Ans

wer

CReason: Remember to check the charge of the Nitrogen atoms. In

answer C, Nitrogen has an Oxidation state of +5, higher than any other element in all the other answers listed making it VERY

electron hungry.



A F-

B Cl-

C Br-

D I-

E O2-

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A F-

B Cl-

C Br-

D I-

E O2-[This object is a pull tab]

Ans

wer

D


Balancing REDOX Reactions

In a redox reaction, both the atoms AND electrons must be balanced.

Example: Cu2+(aq) + Ag(s) --> Cu(s) + Ag+

(aq)

Gained 2 e- Lost 1 e-

Even though there is one of each atom on each side, the equation is not balanced because the number of electrons lost does not equal the number gained. This can be changed by reacting more Ag atoms so more electrons are lost.

Balanced: Cu2+(aq) + 2Ag(s) --> Cu(s) + 2Ag+

(aq)

Gained 2 e- Lost 2 e-

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Example: Balance the following:

Al(s) + Fe2+(aq) --> Al3+

(aq) + Fe(s)

2Al(s) + 3Fe2+(aq) --> 2 Al3+(aq) + 3Fe(s)

Lost 6e- Gain 6e-

move for answer

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31 What is the correct set of coefficients when the reaction below is balanced?

Cu+(aq) + Mg(s) --> Cu(s) + Mg2+(aq)

A 1,2,1,2

B 1,1,1,1C 2,1,2,1D 2,2,1,1E 1,1,2,2

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31 What is the correct set of coefficients when the reaction below is balanced?

Cu+(aq) + Mg(s) --> Cu(s) + Mg2+(aq)

A 1,2,1,2

B 1,1,1,1C 2,1,2,1D 2,2,1,1E 1,1,2,2 [This object is a pull tab]

Ans

wer

C



REDOX reactions occur in solutions just like most of the reactions we do in Chemistry. REDOX reactions are different because the medium in which the reaction occurs affects the results of the reactions.

Since REDOX reactions depend on the movement of electrons, certain chemical species - like rogue H + ions (aka protons) - can have an impact on how the reaction occurs.

Therefore, when balancing a REDOX reaction it is important to note if there is an excess or deficit of H + ions. In other words, is the solution in which the reaction is occurring Acidic (excess of H +) or Basic (deficit of H+). Acidic and Basic solutions have slightly different rules for balancing in each scenario.

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Guidelines for Balancing REDOX Reactions1. Determine Oxidation states of all elements

2. Determine which element was oxidized, which was reduced

3. Write half reactions for the element that was oxidized and the element that was reduced

4. Using coefficients, balance the elements (all non-hydrogen or non-oxygen elements) that are being oxidized/reduced in each half reaction

5. Balance remaining oxygen atoms in water molecules in each half reaction

6. Balance the remaining hydrogen atoms with H+ ions in each half reaction

7. Balance charge with electrons (e-) in each half reaction

8. Multiple coefficients in each half reaction by a number to make sure electrons are balanced

9. Recombine the two half reactions by adding them together. Cancel out any redundant chemical species - electrons should always cancel out

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MnO4-(aq) + Fe2+(aq) --> Fe3+(aq) + MnO2(s)

MnO4-

O:-2 Mn + (4*-2)=-1

Therefore:Mn: +7Fe: +2

MnO2

O:-2 Mn + (2*-2)=0

Therefore:Mn:+4Fe: +3

Step 1) Determine Oxidation states of all elements

Acidic REDOX Reactions Example

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MnO4-

O:-2 Mn + (4*-2)=-1

Therefore:Mn: +7Fe: +2

MnO2

O:-2 Mn + (2*-2)=0

Therefore:Mn:+4Fe: +3

Step 2) Determine which element got oxidized and which element got reduced

Because Mn went from an oxidation state of +7 to +4, it must have been reduced.

Because Fe went from an oxidation state of +2 to +3, it must have been oxidized. (It is able to accept more oxygen now since it is more positive)

Question: Mn is the oxidizing agent and Fe is the reducing agent.


Which agent is the oxidizing agent?

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Step 3) Write half reactions for the element that was oxidized and the element that was reduced

MnO4-(aq) --> MnO2(s)

Fe2+(aq) --> Fe3+(aq)

Reduction half reaction

Oxidation half reaction

Note: A half reaction is literally just one half of a reaction. It is useful sometimes in REDOX reactions to look at the reduction half and the oxidation half separately, and to write a half reaction just look at the change in the species reducing (reduction half reaction) and the change in the species oxidizing (oxidation half reaction).

REMEMBER: Since a half reaction is only PART of a chemical reaction it may include invalid chemical species (like lone electons, cations, or anions). When you originally write half reactions they may not be balanced. THAT IS OKAY!


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Step 4) Using coefficients, balance the elements that are being oxidized/reduced in each half reaction

MnO4-(aq) --> MnO2(s) Fe2+(aq) --> Fe3+(aq)

Reduction half reaction Oxidation half reaction

Mn is already balanced! Fe is already balanced!


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Step 5) Balance the remaining Oxygen atoms with water molecules in each half reaction

MnO4-(aq) --> MnO2(s) + 2H2O(l) Fe2+(aq) --> Fe3+(aq)

2 excess Oxygen atoms are balanced with 2 more water molecules

No Oxygen needs balancing!



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Step 6) Balance the remaining Hydrogen atoms with H+ ions in each half reaction

MnO4-(aq) + 4H+(aq) --> MnO2(s) + 2H2O(l) Fe2+(aq) --> Fe3+(aq)

4 excess Hydrogen atoms are balanced with 4 more H+ ions

No Hydrogen needs balancing



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Step 7) Balance the charge with electrons in each half reaction

MnO4-(aq) + 4H+(aq) --> MnO2(s) + 2H2O(l) Fe2+(aq) --> Fe3+(aq)

Charges:MnO4

-:-1*1 =-1H+: +1*4 =+4Total +4 + -1 =+3

Charges:MnO2

-: 0H+: 0Total 0

Charges:Fe2+:+2*1=+2Total +2

Charges:Fe3+: +3*1=+3Total +3

MnO4-(aq) + 4H+(aq) + 3e- --> MnO2(s) + 2H2O(l)

The charges can be balanced by adding 3e- to the reactants side of this half reaction

The charges can be balanced by adding 1e- to the products side of this half reaction

Fe2+(aq) --> Fe3+(aq) + e-



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Step 8) Multiply coefficients in each half reaction to make sure electrons are balanced

MnO4-(aq) + 4H+(aq) + 3e- --> MnO2(s) + 2H2O(l) Fe2+(aq) --> Fe3+(aq) + e-

Multiply the oxidation half reaction by 3 in order to balance out the 3 electrons in the reduction half reaction

3Fe2+(aq) --> 3Fe3+(aq) + 3e-



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Step 9) Recombine the two half reactions by adding them together. Cancel out any redundant chemical species

Reduction half reaction

+ Oxidation half reactionMnO4-(aq) + 4H+(aq) + 3e- --> MnO2(s) + 2H2O(l)

3Fe2+(aq) --> 3Fe3+(aq) + 3e-

MnO4-(aq) + 4H+(aq) + 3e- + 3Fe2+(aq)--> MnO2(s) + 2H2O(l) + 3Fe3+(aq) + 3e-

MnO4-(aq) + 4H+(aq) + 3Fe2+(aq)--> MnO2(s) + 2H2O(l) + 3Fe3+(aq)


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Balancing REDOX Reactions in Acid Solutions

MnO4-(aq) + Fe2+(aq) --> MnO2(s) + Fe3+(aq)

Check that all atoms and charges are balanced

4H+(aq) + MnO4

-(aq) + 3Fe2+

(aq) --> MnO2(s) + 3Fe3+(aq) + 2H2O(l)

Reactants Products

Fe atoms = 3 Fe atoms = 3

Mn atoms = 1 Mn atoms = 1

O atoms = 4 O atoms = 4

H atoms = 4 H atoms = 4

Total charge = 9+ Total charge = 9+

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NO3-(aq) + I2(s) --> IO3-(aq) + NO2(g)

Determine Oxidation states of all elements

NO3-(aq) + I2(s) --> IO3

-(aq) + NO2(g)

5, -2 0 5,-2 4,-2

Determine which element was oxidized, which was reduced

Reduced: NO3-(N went from +5 -->+4) Gained 1 e-

Oxidized: I2 (I went from 0 -->+5) Lost 5 e -

move for answer

move for answer


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Using coefficients, balance the elements that are being oxidized/reduced

NO3-(aq) + I2(s) --> IO3-(aq) + NO2(g)

NO3-(aq) + I2(s) --> 2IO3

-(aq) + NO2(g)

Multiply coefficients by a number to make sure electrons are balanced

10NO3-(aq) + I2(s) --> 2IO3

-(aq) + 10NO2(g)

Gain 10 e- Lose 10 e-

move for answer

move for answer


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Balance remaining oxygen atoms with water molecules

NO3-(aq) + I2(s) --> IO3-(aq) + NO2(g)

10NO3-(aq) + I2(s) --> 2IO3

-(aq) + 10NO2(g) + 4H2O(l)

Balance remaining H atoms with H+ ions

8H+(aq) + 10NO3

-(aq) + I2(s) --> 2IO3

-(aq) + 10NO2(g) + 4H2O(l)move for answer

move for answer


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NO3-(aq) + I2(s) --> IO3-(aq) + NO2(g)

Reactants Products

N atoms = 10 N atoms = 10

I atoms = 2 I atoms = 2



Total charge = 2- Total charge = 2-

Check that all atoms and charges are balanced

8H+(aq) + 10NO3

-(aq) + I2(s) --> 2IO3

-(aq) + 10NO2(g) + 4H2O(l)

Move for answer


Slide 60 / 83

32 What is the proper coefficient in front of CO2 after the equation below is balanced in an acidic medium?H+(aq) + MnO4− (aq) + C2O42− (aq) ➝ Mn2+ (aq) + CO2 (aq) + H2O(l)

A 1B 2

C 4D 6E 10

Slide 61 / 83

32 What is the proper coefficient in front of CO2 after the equation below is balanced in an acidic medium?H+(aq) + MnO4− (aq) + C2O42− (aq) ➝ Mn2+ (aq) + CO2 (aq) + H2O(l)

A 1B 2

C 4D 6E 10


Ans

wer

E


33 What is the coefficient of the bromide ion after the following REDOX reaction is balanced?

A 1

B 2

C 4

D 5

E 6

N2H4(l) + BrO3-(aq) --> N2(g) + Br-(aq) + H2O(l)

Slide 62 / 83

33 What is the coefficient of the bromide ion after the following REDOX reaction is balanced?

A 1

B 2

C 4

D 5

E 6

N2H4(l) + BrO3-(aq) --> N2(g) + Br-(aq) + H2O(l)


Ans

wer

B


34 For the reaction represented below, how many electrons will be transferred when the equation is properly balanced?

Cu + NO3- --> NO2 + Cu2+

A 2B 4

C 5D 6E 8

Slide 63 / 83

34 For the reaction represented below, how many electrons will be transferred when the equation is properly balanced?

Cu + NO3- --> NO2 + Cu2+

A 2B 4

C 5D 6E 8 [This object is a pull tab]

Ans

wer

A


Guidelines for Balancing Basic REDOX Reactions

To balance REDOX reactions that occur under basic conditions, follow the same steps for acidic conditions with one key difference.

Recall, in an acidic solution there are excess H+ ions, which is why it is okay to leave H+ in the final reaction

ex: MnO4-(aq) + 4H+(aq) + 3Fe2+(aq)--> MnO2(s) + 2H2O(l) + 3Fe3+(aq)

However, in a basic solution there are no excess H+ ions, so we must eliminate the H+ ions. We do that by adding something that exists in excess in a basic solution - OH- ions.

Balancing Basic REDOX Reactions BONUS STEP:

10. Add OH- ions to both sides of your reaction to cancel out H+ ions and convert them to water. *Remember, cancel out any redundant chemical species you create in doing this!

Slide 64 / 83

Basic REDOX Reactions ExampleStep 10) Add OH- ions to both sides of your reaction to cancel out

excess H+ ions and convert them to water. *Remember, cancel out any redundant chemical species you create in doing this!

MnO4-(aq) + 4H+(aq) + 3Fe2+

(aq)--> MnO2(s) + 2H2O(l) + 3Fe3+(aq)

4 OH- ions are needed on each side to cancel out the H+ ionsMnO4

-(aq) + 4H+

(aq) + 3Fe2+(aq) + 4OH-(aq)--> MnO2(s) + 2H2O(l) + 3Fe3+

(aq)+ 4OH-(aq)

OH- ions combine with H+ ions to form water

MnO4-(aq) + 3Fe2+

(aq) + 4H2O(l)--> MnO2(s) + 2H2O(l) + 3Fe3+(aq)+ 4OH-

(aq)

Cancle out the redundant speciesMnO4

-(aq) + 3Fe2+

(aq) + 4H2O(l)--> MnO2(s) + 2H2O(l) + 3Fe3+(aq)+ 4OH-

(aq)

MnO4-(aq) + 3Fe2+(aq) + 2H2O(l)--> MnO2(s) + 3Fe3+(aq)+ 4OH-(aq)

Final Answer!

Slide 65 / 83

Zn + NO3- --> Zn2+ + NH4+

Follow procedure for balancing in acidic solution

10H+ + 4Zn + NO3- --> 4Zn2+ + NH4+ + 3H2O

After adding H+ ions, add equivalent number of OH- to each side

10 OH- + 10H+ + 4Zn + NO3- --> 4Zn2+ + NH4+ + 3H2O + 10 OH-

Balancing REDOX Reactions in Basic Solutions

Slide 66 / 83

Zn + NO3- --> Zn2+ + NH4+

The H+ and OH- will react to make water, cancel out water from each side if necessary

10 H2O + 4Zn + NO3- --> 4Zn2+ + NH4+ + 3H2O + 10 OH-

7H2O + 4Zn + NO3- --> 4Zn2+ + NH4+ + 10 OH-

so after canceling 3 H2O from each side

Balancing Redox Reactions in Basic Solutions

Slide 67 / 83

Zn + NO3- --> Zn2+ + NH4+

Check atoms and charges

7H2O + 4Zn + NO3- --> 4Zn2+ + NH4+ + 10 OH-

Reactants Products

Zn atoms = 4 Zn atoms = 4

N atoms = 1 N atoms = 1



Total charge = -1 Total charge = -1


move for answer

Slide 68 / 83

Follow procedure for balancing in acidic solution

PbO2(s) + Cl-(aq) --> ClO-(aq) + Pb(OH)3-(aq)

2H2O(l) + PbO2(s) + Cl-(aq) --> ClO-(aq) + Pb(OH)3- + H+(aq)

After adding H+ ions, add equivalent number of OH- to each side

OH-(aq) + 2H2O(l) + PbO2(s) + Cl-(aq) --> ClO-(aq) + Pb(OH)3- + H+(aq) + OH-(aq)

move for answer

move for answer


Slide 69 / 83

The H+ and OH- will react to make water, cancel out water from each side if necessary

OH-(aq) + 2H2O(l) + PbO2(s) + Cl-(aq) --> ClO-(aq) + Pb(OH)3- + H2O(l)

And after crossing out 1 H2O molecule from each side

OH-(aq) + H2O(l) + PbO2(s) + Cl-(aq) --> ClO-(aq) + Pb(OH)3-


move for answer


Slide 70 / 83

Check atoms and charges

OH-(aq) + H2O(l) + PbO2(s) + Cl-(aq) --> ClO-(aq) + Pb(OH)3- (aq)


Reactants Products

Pb atoms = 1 Pb atoms = 1

Cl atoms = 1 Cl atoms = 1



move for answer


Slide 71 / 83

35 When the following reaction is balanced in basic solution, what is the proper coefficient in front of the hydroxide ion and which side of the reaction is it on?

MnO4-(aq) + IO3-(aq) --> MnO2(s) + IO4-(aq)

A 3, product sideB 2, product sideC 2, reactant sideD 3, reactant sideE 1 reactant side

Slide 72 / 83

35 When the following reaction is balanced in basic solution, what is the proper coefficient in front of the hydroxide ion and which side of the reaction is it on?

MnO4-(aq) + IO3-(aq) --> MnO2(s) + IO4-(aq)

A 3, product sideB 2, product sideC 2, reactant sideD 3, reactant sideE 1 reactant side


Ans

wer

B


36 When the equation below is balanced in basic solution, what are the proper coefficients for the hydroxide and bromide ions respectively?

Br2(l) + OH-(aq) --> Br-(aq) + BrO3-(aq) + H2O(l)

A 6OH- and 5Br-

B 3OH- and 2Br-

C 6OH- and 4Br-

D 2OH- and 2Br-

E 5OH- and 3Br-

Slide 73 / 83

36 When the equation below is balanced in basic solution, what are the proper coefficients for the hydroxide and bromide ions respectively?

Br2(l) + OH-(aq) --> Br-(aq) + BrO3-(aq) + H2O(l)

A 6OH- and 5Br-

B 3OH- and 2Br-

C 6OH- and 4Br-

D 2OH- and 2Br-

E 5OH- and 3Br- [This object is a pull tab]

Ans

wer

A


37 What is the coefficient of the hydroxide ion after the following REDOX reaction has been balanced?

A 2

B 4

C 6

D 8

E 10

Cl2(g) + Cr(OH)3(s) + OH-(aq) --> Cl-(aq) + CrO42-(aq) + H2O(l)

Slide 74 / 83

37 What is the coefficient of the hydroxide ion after the following REDOX reaction has been balanced?

A 2

B 4

C 6

D 8

E 10

Cl2(g) + Cr(OH)3(s) + OH-(aq) --> Cl-(aq) + CrO42-(aq) + H2O(l)


Ans

wer

E


ApplicationThe breathalyzer test is a REDOX reaction that uses a color change to

see how much alcohol (C2H5OH) is present in a driver's breath.

Cr2O72-(aq) + C2H5OH(aq) --> CH3COOH(aq) + Cr3+(aq)

orange colorless colorless blue/green

The officer can then see how much alcohol is present by looking for a color change from orange to blue/green.

Now, let's balance it in acid solution!

2Cr2O72-(aq) + 3C2H5OH(aq) + 16H+(aq) --> 3CH3COOH(aq) + 4Cr3+(aq) + 11H2O(l)move for answer

Slide 75 / 83

REDOX StoichiometryTitrations can be utilized in REDOX reactions to determine the

concentration of an analyte.

By oxidizing or reducing an analyte with a standardized reductant or oxidant titrant, the amount of analyte can be determined.

standardized titrant

unknown analyte

Slide 76 / 83

REDOX StoichiometryThe end point of a titration is often accompanied by a color

change as a substance is either formed or created. It is then important to know some colors for some important oxidants and

reductants.

MnO4-(aq) = purple

CrO42-

(aq) = yellow

Cr2O72-

(aq) = orange

Slide 77 / 83

REDOX Stoichiometry

Transition metal ions also form complex ions that have particular colors.

Fe3+(aq) = yellow/brown

Co2+(aq) = pink

Cu2+(aq) = blue

Cr3+(aq) = green

Slide 78 / 83

REDOX Stoichiometry

Let's examine a titration of potassium permanganate with iron(II) nitrate in acidic medium

8H+(aq) + MnO4

-(aq) + 5Fe2+

(aq) --> Fe3+(aq) + Mn2+

(aq) + 4H2O(l)

As the permanganate disappears, so does the purple color and as the Fe3+ is made, the yellow/brown color appears.

When the purple is gone, the permanganate has all reacted.

Excess MnO4-(aq) Excess Fe3+

(aq)

Slide 79 / 83

38 How many mL of 2M HCl would need to be added to 10 mL of 0.34 M solution of potassium permanganate in order for all of the permanganate to be titrated by the reaction below?

8H+(aq) + MnO4-(aq) + 5Fe2+(aq) --> Fe3+(aq) + Mn2+(aq) + 4H2O(l)

Slide 80 / 83

38 How many mL of 2M HCl would need to be added to 10 mL of 0.34 M solution of potassium permanganate in order for all of the permanganate to be titrated by the reaction below?



Ans

wer

13.6 mL


39 How many mL of 0.44 M iron(II) nitrate would be needed to titrate a 340 gram sample of potassium permanganate by the reaction below?


Slide 81 / 83

39 How many mL of 0.44 M iron(II) nitrate would be needed to titrate a 340 gram sample of potassium permanganate by the reaction below?



Ans

wer

24,390 mL


40 What would be the expected color change in the analyte given the following titration?

A red to orange

B purple to clear

C yellow to green

D orange to clear

E the color will remain the same

CrO42-(aq) + I-(aq) --> I2(s) + Cr3+(aq) analyte titrant

Slide 82 / 83

40 What would be the expected color change in the analyte given the following titration?

A red to orange

B purple to clear

C yellow to green

D orange to clear

E the color will remain the same

CrO42-(aq) + I-(aq) --> I2(s) + Cr3+(aq) analyte titrant


Ans

wer

C


Slide 83 / 83

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