12-1 balancing redox equations • 12-2 electrochemical...

OFB Chapter 12 111/1/2004

Chapter 12Redox reactions and

Electrochemistry• 12-1 Balancing Redox

Equations• 12-2 Electrochemical Cells• 12-3 Stoichiometry in

Electrochemical Cells• 12-4 Metals and Metallurgy• 12-5 Electrometallurgy

Note: See course website for text book error corrections Chapter 12 p. A-63 problem 29


• This entire Chapter deals with Oxidation, Reduction, electrochemistry and electron transfer

• Balancing Redox Reactions• Practice, Practice, Practice• All reactions take place in either

Acid (H+) or Basic (OH-) conditions

• Very important to remember the problem you are working. Is it acidic or basic conditions?

• We will be adding H2O and H+

for acid reactions or H2O and OH- for basic reaction in order to properly balance the reactions

• Practice, Practice, Practice


Gains ElectronsDecreaseSubstance Reduced

Loses ElectronsIncreaseSubstance Oxidized

Supplies ElectronsIncrease

Reducing Agent, does the reducing

Picks Up electronsDecrease

Oxidizing Agent, does the oxidizing

Gain of ElectronsDecreaseReduction

Loss of ElectronsIncreaseOxidation

Electron Change

Oxidation Number ChangeTerm


• Step 1 Write two unbalanced half-equations, one for the species that is oxidized and its product and one for the species that is reduced and its product

• Step 2 Insert coefficients to make the numbers of atoms of all elements except oxygen and hydrogen equal on the two sides of each half-equation.

• Step 3 Balance oxygen by adding H2O to the side deficient in O in each half-equation

• Step 4 Balance hydrogen. – For half-reaction in acidic solution, add H+ on

to the side deficient in hydrogen. – For a half-reaction in basic solution, add H2O

to the side that is deficient in hydrogen and an equal amount of OH- to the other side


• Step 5 Balance charge by inserting e-

(electrons) as a reactant or product in each half-reaction.

• Step 6 Multiply the two half-equations by numbers chosen to make the number of electrons given off by the oxidation equal to the number taken up by the reduction. Then add the two half-equations and cancel out the electrons. If H+ ion, OH-

ion, or H2O appears on both sides of the final equation, cancel out the duplication

• Step 7 Check for balance


Balancing Redox Reactions

• Dithionate ion reacting with chlorous acid in aqueous acidicconditions

S2O62-(aq) +HClO2(aq)

→ SO42-(aq) + Cl2(g)

Will have:

Oxidation half reaction

Reduction half reaction



S2O62- → SO4

2-

Dithionate (we will see eventually that this is the

oxidation reaction)

HClO2 → Cl2Chlorous acid

(we will see eventually that this is the reduction reaction)

S2O62-(aq) +HClO2(aq)

→ SO42-(aq) + Cl2(g)



S2O62- → SO4

2-

HClO2 → Cl2

2

2



S2O62- → 2 SO4

2-

2HClO2 → Cl2

(8Os)(6Os)

2H2O +(Need 2 Os)

(4Os)

+ 4H2O(Need 4 Os)

OFB Chapter 12 1011/1/2004

• Step 4 Balance hydrogen. – For half-reaction in acidic solution, add H+ on to the

side deficient in hydrogen. – For a half-reaction in basic solution, add H2O to the

side that is deficient in hydrogen and an equal amount of OH- to the other side

2H2O + S2O62- → 2SO4

2-

2HClO2 → Cl2 + 4H2O

Dithionate ion reacting with chlorous acid in aqueous acidic conditions

Need 4 Hs

Need 6 Hs (2 Hs) (8 Hs)

+ 4H+

6H+ +

(4 Hs)

OFB Chapter 12 1111/1/2004

• Step 5 Balance charge by inserting e- (electrons) as a reactant or product in each half-reaction.

2H2O + S2O62-

→ 2SO42- + 4H+

6H+ + 2HClO2

→ Cl2 + 4H2O

-2

-4 +4 Need -2

+ 2e-

+6

Zero charge

Need -6

+ 6e-

Oxidation reaction, electrons are lost on the product side

Reduction reaction, electrons are gained on the reactant side

OFB Chapter 12 1211/1/2004

• .Step 6 Multiply the two half-equations by numberschosen to make the number of electrons given off by the oxidation equal to the number taken up by the reduction. Then add the two half-equations and cancel out the electrons. If H+ ion, OH- ion, or H2O appears on both sides of the final equation, cancel out the duplication

Oxidation Reaction has 2 electrons

Reduction reaction has 6 electrons

∴Multiply the oxidation reaction by 3 this will give 6 electrons on both sides of the reaction

2H2O + S2O62-

→ 2SO42- + 4H+ + 2e-

3 times

OFB Chapter 12 1311/1/2004


3 times

6H2O + 3S2O62-

→ 6SO42- + 12H+ + 6e-

2H2O + S2O62-

→ 2SO42- + 4H+ + 2e-

equals

Now add both Re-Ox reactions

OFB Chapter 12 1411/1/2004


6H2O + 3S2O62-

→ 6SO42- + 12H+ + 6e-

6H+ + 2HClO2 + 6e-

→ Cl2 + 4H2O

6H2O + 3S2O62- + 6H+ +

2HClO2 + 6e-

→ 6SO42- + 12H+ + Cl2

+ 4H2O + 6e-

2

6

OFB Chapter 12 1511/1/2004


2H2O + 3S2O62- + 2HClO2

→ 6SO42- + 6H+ + Cl2

Left side

H 6

O 24

S 6

Cl 2

Right side

H 6

O 24

S 6

Cl 2If it doesn’t balance, look for simple mistakes first

Omitted subscripts

Omitted superscripts

OFB Chapter 12 1611/1/2004

Typical Exam Question

• Find the (integer) stoichiometry coefficients for the correctly balanced version of the following reaction

Cr2O72- + H+ + I-

→ Cr+3 + H2O + I3-

OFB Chapter 12 1711/1/2004


Cr2O72- + H+ + I-

→ Cr+3 + H2O + I3-

Cr2O72-→

Cr+3

I-→ I3-

OFB Chapter 12 1811/1/2004


2

3

Cr2O72-→ Cr+3

I-→ I3-

OFB Chapter 12 1911/1/2004


(7Os)+ 7H2O

(Need 7 Os)

(No Oxygen, okay)

Cr2O72-→ 2Cr+3

3I-→ I3-

OFB Chapter 12 2011/1/2004

• Step 4 Balance hydrogen. – For half-reaction in acidic solution, add H+ on to the

side deficient in hydrogen. – For a half-reaction in basic solution, add H2O to the

side that is deficient in hydrogen and an equal amount of OH- to the other side

Dichromate ion reacting with iodide ion in aqueous acidic conditions

Need 14 Hs

14H+ +

(14 Hs)

Cr2O72-

→ 2Cr+3 + 7H2O

3I-→ I3-

(No Hydrogen, okay)

OFB Chapter 12 2111/1/2004


-2+14

+6

Need -6

+ 2e-

-3 Need -2

6e- +

Oxidation reaction

Reduction reaction

14H+ + Cr2O72-

→ 2Cr+3 + 7H2O

3I-→ I3-

-1

OFB Chapter 12 2211/1/2004



Oxidation Reaction has 2 electrons

Multiply the oxidation reaction by 3 this will give 6 electrons on both sides of the reaction

3 times

3I-→ I3- + 2e-

OFB Chapter 12 2311/1/2004


equals


3 times

3I-→ I3- + 2e-

9I-→ 3 I3- + 6 e-

OFB Chapter 12 2411/1/2004


6e- + 14H+ + Cr2O72-

→ 2Cr+3 + 7H2O

9I-→ 3 I3- + 6 e-

6e- + 14H+ + Cr2O72- + 9I-

→ 2Cr+3 + 7H2O + 3 I3- + 6 e-

equals

OFB Chapter 12 2511/1/2004


Left side

H 14

Cr 2

O 7

I 9

Right side

H 14

Cr 2

O 7

I 9

14H+ + Cr2O72- + 9I-

→ 2Cr+3 + 7H2O + 3 I3-

OFB Chapter 12 2611/1/2004

Basic SolutionMethod 1

• At Step 4– Instead of adding H+ on the

deficient side as did with the acid solution method

– Add H2O on the H deficient side and add an equal amount of OH- on the other side.

OFB Chapter 12 2711/1/2004

Typical Exam Question

• Note: this is not Example 12-2, which is similar

• Find the (integer) stoichiometry coefficients for the correctly balanced version of the following reaction conducted in a basic solution

Ag2S + Cr(OH)3

→ Ag0 + HS- + CrO42-

OFB Chapter 12 2811/1/2004


Ag2S + Cr(OH)3

→ Ag0 + HS- + CrO42-

Ag2S → Ag + HS-

Cr(OH)3 → CrO42-

Reduction reaction

+1 0

Oxidation reaction

+3 +6

OFB Chapter 12 2911/1/2004


2Ag2S → Ag + HS-

Cr(OH)3 → CrO42-

Already Balanced

OFB Chapter 12 3011/1/2004


(4Os)

H2O +

(Oxygen not needed)

Ag2S → 2Ag + HS-

Cr(OH)3 → CrO42-

(3Os)(need 1O)

OFB Chapter 12 3111/1/2004

• Step 4 Balance hydrogen. – For half-reaction in acidic solution, add H+ on

to the side deficient in hydrogen. – For a half-reaction in basic solution, add H2O

to the side that is deficient in hydrogen and an equal amount of OH- to the other side

This problem is in aqueous basic conditions

Needs H

(1 H)

Ag2S

→ 2Ag + HS-

H2O +

+ OH-

(need O)

H2O + Cr(OH)3

→ CrO42-

(3 H)(2 H)

+ 5H2O

5OH- +

Needs H

Add OH-

OFB Chapter 12 3211/1/2004


-2

Need -2

+ 3e--5

Need -3

H2O + Ag2S

→ 2Ag + HS- + OH-

5OH- + H2O + Cr(OH)3

→ CrO42- + 5H2O

-1 -1

2e- +

Reduction reaction

Oxidation reaction

OFB Chapter 12 3311/1/2004



Oxidation Reaction has 3 electrons2e x 3 = 6e

3e x 2 = 6e

OFB Chapter 12 3411/1/2004


6e- + 3H2O + 3Ag2S

→ 6Ag + 3HS- + 3OH-Reduction reaction

times Reduction Reaction =

10OH- + 2H2O + 2Cr(OH)3

→ 2CrO42- + 10H2O + 6e-

times Oxidation Reaction =

Oxidation reaction

OFB Chapter 12 3511/1/2004



6e- + 3H2O + 3Ag2S → 6Ag + 3HS- + 3OH-

10OH- + 2H2O + 2Cr(OH)3

→ 2CrO42- + 10H2O + 6e-

OFB Chapter 12 3611/1/2004

equals

Now add both Re-Ox reactions6e- + 3H2O + 3Ag2S → 6Ag + 3HS- + 3OH-

10OH- + 2H2O + 2Cr(OH)3

→ 2CrO42- + 10H2O + 6e-

6e- + 3H2O + 3Ag2S +10OH- + 2H2O + 2Cr(OH)3

→ 6Ag + 3HS- + 3OH- +2CrO4

2- + 10H2O + 6e-

OFB Chapter 12 3711/1/2004

simplify

6e- + 3H2O + 3Ag2S +10OH- + 2H2O + 2Cr(OH)3

→ 6Ag + 3HS- + 3OH- +2CrO4

2- + 10H2O + 6e-5

7

3Ag2S + 7OH- + 2Cr(OH)3

→6Ag + 3HS- + 2CrO4

2- + 5H2O

OFB Chapter 12 3811/1/2004


Left side

Ag 6

S 3

O 13

H 13

Cr 2

3Ag2S + 7OH- + 2Cr(OH)3

→6Ag + 3HS- + 2CrO4

2- + 5H2O

Right side

Ag 6

S 3

O 13

H 13

Cr 2NOTE: This problem is the reverse reaction of Example 12-2

Left side

H 14

Cr 2

O 7

I 9

Right side

H 14

Cr 2

O 7

I 9

OFB Chapter 12 3911/1/2004

Typical Exam QuestionAnswer

• Note: this is not Example 12-2, which is similar

• Find the (integer) stoichiometry coefficients for the correctly balanced version of the following reaction conducted in a basic solution

3 Ag2S + 2 Cr(OH)3+ 7 OH-→ 6 Ag + 3 HS- +

2CrO42- + 5H2O

OFB Chapter 12 4011/1/2004


Electrochemistry

• Practice• Practice• Practice

OFB Chapter 12 4111/1/2004

Balancing Disproportionation Reactions

• Same Steps• For example

Cl2 → ClO3- + Cl-

• Step 1– Oxidation:

Cl2 → ClO3-

– Reduction:Cl2 → Cl-

OFB Chapter 12 4211/1/2004


Electrochemistry• Review of Common Terms

Term Oxidation State Change Electron Change

1 Oxidation increase loss of electronse.g., Cu to Cu 2+ (0 to +2)

2 Reduction decrease gain of electronse.g., Ni + to Ni (+1 to 0)

3 Oxidizing Agent decrease picks up electronse.g., O 2 to oxide (0 to -2)

4 Reducing Agent increase supplies electronse.g., H 2 to H 2 O (0 to +1)

5 Substance Oxidized increase loses electronse.g., Mg to MgO (0 to +2)

6 Substance Reduced decrease gains electronse.g., CuO to Cu (+2 to 0)

OFB Chapter 12 4311/1/2004

Electrochemical Cells

Cu(s) + 2Ag+(aq) → Cu++(aq) + 2Ag(s)

Silver plates spontaneously

∴ ∆Gf < 0

Called a galvanic or voltaic cell

Oxidation Cu → Cu++ + 2e-

Reduction Ag+ + e-→ Ag

OFB Chapter 12 4411/1/2004

Cu(s) + 2Ag+(aq) → Cu++(aq) + 2Ag(s)

Oxidation occurs here at the anode

Reduction occurs here at the cathode

Chemical energy converted to Electrical Energy. This potential difference can be measured.

Current = charge/time or I=Q/t

Units amperes = coulombs/sec.

OFB Chapter 12 4511/1/2004

Cu(s) + 2Ag+(aq) → Cu++(aq) + 2Ag(s)

Convention for this reaction

Cu(s) | Cu+2(aq) ¦¦ Ag+(aq) | Ag(s)

OFB Chapter 12 4611/1/2004

If an opposing electric force is applied the

reverse reaction occurs, called an electrolytic cell

Cu++(aq) + 2Ag(s)→ Cu(s) +2Ag+(aq)

The electrodes are reversed

Ag becomes the anode

Cu becomes the cathode

Note that the electron flow is still anode to cathode

Ag+(aq) | Ag(s) ¦¦ Cu(s) | Cu+2(aq)Ag+(aq) | Ag(s) ¦¦ Cu(s) | Cu+2(aq)

OFB Chapter 12 4711/1/2004

Stoichiometry in Electrochemical Cells

• Faraday’s Law– The quantities of substances

produced or consumed at the electrodes are directly proportional to the amount of electric charge passed through the cell.

– When a given amount of electric charge passes through a cell, the quantity of a substance produced or consumed at an electrode is proportional to its molar mass divided by the number of moles of electrons required to produce or consume one mole of the substances.

OFB Chapter 12 4811/1/2004

1 Faraday = 96,485 Coulombs of charge per mole of electrons

Faraday Constant (F ) = the electric charge carried by 1 mole of electrons.

Charge 1 e = 1.602X10-19 Coulombs

X Avogadro’s number

OFB Chapter 12 4911/1/2004

Problem 12-30What number of coulombs of electricity

is produced in each of the following oxidations?

• 1.00 mol of H2O2(aq) to O2(g)

CxCOH

OH 5-

22

-

22 1093.1e mol

485,96 mol

e mol 2 mol00.1 =

×

×

OFB Chapter 12 5011/1/2004

Problem 12-30What number of coulombs of electricity

is produced in each of the following oxidations?

• 1.70 mol of H2S (aq) to 0.850 mol of H2S2O3(aq)

H2S → H2S2O3

Cx

C

5

-2

-

2

1056.6

e mol485,96

SH mol 2e mol 8SH .7mol1

=

×

×

OFB Chapter 12 5111/1/2004

Stoichiometry in Electrochemical Cells

Ag+ + e-→ Ag

187.107M

=en

Cu → Cu++ + 2e-

254.63

nM

e

=

electrons #per massmolar M/nMassMolar M

e ==

OFB Chapter 12 5211/1/2004


Electrochemistry

Example

How many grams of Cl2 can be produced by the electrolysis of molten NaCl at a current of 10.0 amps for 5 minutes?

2Cl- → Cl2 + 2e-

grams 11.1produced Cl485,96

29.70minsec/60min5amps0.10

)(Cl

96,485nM t I

mass

2

2

e

=

•••=

••=

grams

OFB Chapter 12 5311/1/2004

Reminder

1 mole electrons reduces 1 mol of Ag+

2 moles electrons reduces 1 mole of Mg2+

3 moles electrons reduces 1 mole of Al3+

OFB Chapter 12 5411/1/2004

ExampleHow long would it take to plate 5.43 grams Nickel on to an electrode from a solution of NiCl2 at a current of 12.34 amps? Molar mass of Nickel is 58.693 g mol-1.

Ni+2 + 2e- → Ni

OFB Chapter 12 5511/1/2004


Electrochemistry

Summary• Balancing Redox reactions

–––

• Understand electrochemical cells and the Faraday law calculations

12-1 balancing redox equations • 12-2 electrochemical...

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