Chapter 4
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Chapter 4. Red uction- Ox idation Reactions Redox Reactions. Sodium chloride. Sodium chloride. Na  Na + + e . Cl 2 + 2e   2Cl . Involves 2 processes: Oxidation = Loss of Electrons Na  Na + + e  Oxidation Half-Reaction Reduction = Gain of electrons

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Chapter 4

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Chapter 4

Chapter 4

Reduction-Oxidation Reactions

Redox Reactions


Chapter 4

Sodium chloride


Sodium chloride

Sodium chloride

Na  Na+ + e

Cl2 + 2e 2Cl


Oxidation reduction reactions

Involves 2 processes:

Oxidation = Loss of Electrons

Na  Na+ + eOxidation Half-Reaction

Reduction = Gain of electrons

Cl2 + 2e 2ClReduction Half-Reaction

Net reaction:

2Na + Cl2 2Na+ + 2Cl

Oxidation & reduction always occur together

Can't have one without the other

Oxidation–Reduction Reactions


Oxidation reduction reaction

Oxidation Reduction Reaction

Oxidizing Agent

- Substance that accepts e's

  • Accepts e's from another substance

  • Substance that is reduced

    Cl2 + 2e 2Cl–

    Reducing Agent

    - Substance that donates e's

  • Releases e's to another substance

  • Substance that is oxidized

    Na  Na+ + e–


Your turn

Your Turn!

Which species functions as the oxidizing agent in the following oxidation-reduction reaction?

Zn(s) + Pt2+(aq) Pt(s) + Zn2+(aq)

Pt(s)

Zn2+(aq)

Pt2+(aq)

Zn(s)

None of these, as this is not a redox reaction.


Redox reactions

Redox Reactions

  • Very common

    • Batteries—car, flashlight, cell phone, computer

    • Metabolism of food

    • Combustion

  • Chlorine Bleach

    • Dilute NaOCl solution

    • Cleans through redox

      reaction

    • Oxidizing agent

    • Destroys stains by oxidizing them


Redox reaction

Redox Reaction

Ex. Fireworks displays

Net: 2Mg + O2 2MgO

Oxidation:

Mg  Mg2+ + 2e

  • Loses electrons = Oxidized

  • Reducing agent

    Reduction:

    O2 + 4e 2O2

  • Gains electrons = Reduced

  • Oxidizing agent


Redox reaction1

Redox Reaction?

S + O2 SO2

  • Combustion: Oxidation in old sense, reaction with oxygen

  • But n no ions in SO2

  • How can we decide which loses and which gains!!

  • Oxidation number (state)!

  • If compound were ionic, what would the charges have been.


Rules for oxidation states numbers

Rules for Oxidation States (Numbers)

  • The sum of Oxidation numbers equals to the charge on molecule, formula unit or ion.

  • The oxidation state of elements is zero.

  • Oxidation state for monoatomic ions are the same as their charge.

  • In its compounds fluorine is always –1.

  • Hydrogen is assigned the oxidation state +1.

  • Oxygen is assigned an oxidation state of -2 in its covalent compounds (except as a peroxide).

  • If two rules conflict, apply higher rule.


Oxidation states

Oxidation States

  • Assign the oxidation states to each element in the following.

  • CO2

  • NO3-

  • H2SO4

  • Fe2O3

  • Fe3O4

  • Cr2O72-

  • O2F2

  • H2O2

  • LiH

  • BaO2


Oxidation reduction

Oxidation-Reduction

Transfer electrons, so the oxidation states change.

Ox

Ox

Oxidation

Increase in

Oxid. number

Reduction

decrease in

Oxid. number

Red

Red


Chapter 4

Ox

Ox

Red

Red

Red

C (CH4) oxidized →

CH4 reducing agent

O2 reduced →

O2 oxidizing agent


Chapter 4

Red

Ox

PbS has been oxidized, PbS is the reducing agent.

O2 has been reduced, O2 is the oxidizing agent.


Chapter 4

Red

Ox

PbO has been reduced, PbO is the oxidizing agent.

CO has been oxidized, CO is the reducing agent.


Identify the

Identify the

1) Oxidizing agent

2) Reducing agent

3) Substance oxidized

4) Substance reduced

in the following reactions

  • Fe(s) + O2(g) ® Fe2O3(s)

  • Fe2O3(s)+ 3 CO(g) ® 2 Fe(l) + 3 CO2(g)

  • SO3- + H+ + MnO4- ® SO4- + H2O + Mn+2


Balancing redox reactions ion electron method acidic solution

Balancing Redox ReactionsIon-Electron Method – Acidic Solution

1. Divide equation into 2 half-reactions

2. Balance atoms other than H & O

3. Balance O by adding H2O to side that needs O

4. Balance H by adding H+ to side that needs H

5. Balance net charge by adding e–

6. Make e– gain equal e– loss; then add half-reactions

7. Cancel anything that is the same on both sides


Chapter 4

Balance in Acidic Solution

Cr2O72– + Fe2+ Cr3+ + Fe3+

1.Break into half-reactions

Cr2O72 Cr3+

Fe2+ Fe3+

2.Balance atoms other than H & O

Cr2O722Cr3+

  • Put in 2 coefficient to balance Cr

    Fe2+ Fe3+

  • Fe already balanced


Chapter 4

3. Balance O by adding H2O to the side that needs O.

Cr2O72 2Cr3+

  • Right side has 7 O atoms

  • Left side has none

  • Add 7 H2O to left side

    Fe2+ Fe3+

  • No O to balance

+ 7 H2O


Chapter 4

4.Balance H by adding H+ to side that needs H

Cr2O72  2Cr3+ + 7H2O

  • Left side has 14 H atoms

  • Right side has none

  • Add 14 H+ to right side

    Fe2+ Fe3+

  • No H to balance

14H++


Chapter 4

5.Balance net charge by adding electrons.

14H+ + Cr2O72 2Cr3+ + 7H2O

  • 6 electrons must be added to reactant side

    Fe2+  Fe3+

  • 1 electron must be added to product side

  • Now both half-reactions balanced for mass & charge

  • 6e +

    Net Charge =

    14(+1) (–2) = 12

    Net Charge =

    2(+3)+7(0) = 6

    + e


    Chapter 4

    6.Make e– gain equal e– loss; then add half-reactions

    6e + 14H++ Cr2O72– 2Cr3+ + 7H2O

    Fe2+ Fe3++ e

    7. Cancel anything that's the same on both sides

    6[

    ]

    6Fe2+ + 14H+ + Cr2O72

    6Fe3++ 2Cr3+

    + 7H2O

    6e + 6Fe2++ 14H+ + Cr2O72

    6Fe3++ 2Cr3++ 7H2O+ 6e

    

    


    Practice

    Practice

    • The following reactions occur in acidic aqueous solution. Balance them:

    • MnO4-+ Fe2+® Mn2++ Fe3+

    • Cu + NO3- ®Cu2++ NO(g)

    • Pb + PbO2 + SO42-® PbSO4

    • Mn2++ NaBiO3®Bi3++ MnO4-

    • Cr2O72- + C2H5OH ® Cr3+ + CO2


    Ion electron method in basic solution

    Ion-Electron method in Basic Solution

    • The simplest way to balance an equation in basicsolution

      Use steps 1-7 above, then

      8. Add the same number of OH– to both sides of the equation as there are H+.

      9. CombineH+ & OH– to form H2O

      10. Cancel any H2O that you can from both sides


    Basic solution

    Basic Solution

    • Ag + CN- +O2® Ag(CN)2-

    • Cr(OH)3 + OCl- + OH-® CrO42- + Cl- + H2O

    • CrI3+ Cl2® CrO4-+ IO4- + Cl-


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