Ion electron method
Download
1 / 41

Ion Electron Method - PowerPoint PPT Presentation


  • 140 Views
  • Uploaded on

Ion Electron Method. Ch 20. Drill. Use AP rev drill #. Objectives. SWBAT Work through the steps of the Ion Electron Method for solving Redox equations in acidic and basic conditions. Begin with slide 22. Write Half Reactions. Write an oxidation and a reduction half reaction.

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about ' Ion Electron Method' - cargan


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

Drill
Drill

  • Use AP rev drill #


Objectives
Objectives

  • SWBAT

  • Work through the steps of the Ion Electron Method for solving Redox equations in acidic and basic conditions.



Write half reactions
Write Half Reactions

  • Write an oxidation and a reduction half reaction.

    Sn2+→ Sn 4+

    Hg 2+ + Cl-1→ Hg2Cl2


Balance half reactions
Balance Half Reactions

  • Balance each half reaction in terms of atoms.

    Sn2+→ Sn 4+

    2Hg 2+ + 2Cl-1→ Hg2Cl2


Balance charges
Balance Charges

  • Balance charges on opposite sides of each half-reaction equation by adding electrons to the appropriate side.

    Sn2+→ Sn 4+ + 2e-

    2e- + 2Hg 2+ + 2Cl-1→ Hg2Cl2

    (The top reaction ends with a +2 charge on both sides.

    The bottom reaction has no overall charge after adding electrons)


Make electrons equal
Make Electrons Equal

  • The number of electrons lost in the oxidation half reaction must equal the number of electrons gained in the reduction half reaction.

  • If necessary, multiply each half reaction equation by a stoichiometric coefficient to equalize the number of electrons transferred.

    Sn 2+→ Sn 4+ + 2e-

    2e- + 2Hg 2+ + 2Cl-1→ Hg2Cl2


Add the reactions
Add the Reactions

  • Add the resulting half-reactions to obtain the balanced net ionic equation.

    Sn 2+→ Sn 4+ + 2e-

    2e- + 2Hg 2+ + 2Cl- → Hg2Cl2

    2e- + Sn 2+ + 2Hg 2+ + 2Cl-1→ Sn 4+ + Hg2Cl2 + 2e-


Cancel out
Cancel Out

  • Cancel out any species that are the same on both sides of the reaction.

    Sn 2+ + 2Hg 2+ + 2Cl-1→ Sn 4+ + Hg2Cl2

    Note: Both atoms and charges are balanced.


Additional info
Additional Info

  • In many oxidation-reduction reactions that take place in aqueous solution, water plays an activerole.

  • Any aqueous solution contains the species H20, H+, and OH-.

  • In acidic solutions the predominant species are H20 and H+

  • In basic solutions they are H20 and OH-


Practice problem
Practice Problem

NO + SO4– 2 NO3 –1 + SO2


Practice problem answer
Practice Problem Answer

NO  NO3-1

SO4– 2 SO2


Balance atoms
Balance Atoms

NO + 2H2O  NO3-1

NO + 2H2O  NO3-1 + 4H+1

SO4– 2 SO2 + 2H2O

4H+1 + SO4– 2 SO2 + 2H2O


Add electrons
Add Electrons

NO + 2H2O  NO3-1 + 4H+1 + 3e-1

4H+1 + SO4– 2 + 2e-1 SO2 + 2H2O

multiply the top rxn by 2

multiply the bottom rxn by 3

both rxns now have 6 e-1


Final answer
Final Answer

2 NO + 4 H+1 + 3 SO4– 2 2NO3-1 + 3 SO2 + 2 H2O


Wrap up
Wrap Up

  • Try the practice problems at the end of Ch 11 in the UEHB text.


Acidic solutions
Acidic Solutions

  • The next section focuses on reactions that occur in acidic solution.


If the reaction occurs in acidic solution
If the reaction occurs in acidic solution …

Cr2O7 2- + H2S → Cr 3+ + S

Write the half reactions:

H2S → S

Cr2O72-→ Cr 3+


Acidic solution
Acidic Solution

Balance the S atoms first.

Add H+ to balance the H in the reaction, then

balance the H+

H2S → S + 2H+

Balance the charge by adding electrons:

H2S → S + 2H+ + 2e-


Use h 2 o and h 1 to balance the equation
Use H2O and H+1 to Balance the Equation

Balance the chromium atoms:

Cr2O72-→ 2Cr 3+

Balance the oxygens on the left by adding water to the right side of the equation:

Cr2O72-→ 2Cr 3+ + H2O


Now add H+1 to the left:

H+1 + Cr2O72- → 2Cr 3+ + H2O

Balance the H’s and O’s:

14H+1 + Cr2O72- → 2Cr 3+ + 7H2O


Now add electrons to balance the charge:

14H+1 + Cr2O72-→ 2Cr 3+ + 7H2O

There is 14+ and 2- on the left (overall 12+)

There is 6+ on the right

Therefore, add 6e- to the left to balance the charge.

6e- + 14H+1 + Cr2O72-→ 2Cr 3+ + 7H2O


Add the 2 half reactions together
Add the 2 half reactions together

3 (H2S → S + 2H+ + 2e-)

6e- + 14H+ + Cr2O72-→ 2Cr 3+ + 7H20

3H2S + 14H+ + Cr2O72- + 6e- → 3S + 6H+ + 2Cr 3+ + 7H20 + 6e-

Cancel out anything that is the same on both sides:

3H2S + 8H+ + Cr2O72-→ 3S + 2Cr 3+ + 7H20

Note: notice how some of the H+ ions cancel out.


Summary
Summary

  • In summary, when balancing half-reactions in acid solution:

  • To balance a hydrogen atom we add a hydrogen ion, H+, to the side of the equation without any H’s.

  • To balance an oxygen atom we add a water molecule to the side deficient in oxygen and

    then two H+ ions to the opposite side to remove the hydrogen imbalance.


Practice problems
Practice Problems

Practice Problem #1:

Balance the following equation in acidic solution:

Fe+2 + Cr2O7 -2→ Fe+3 + Cr+3


Practice problem 1 answer
Practice Problem # 1 Answer

6Fe+2 + 14 H+1 + Cr2O7 -2→ 6Fe+3 + 2Cr+3 + 7H2O


Basic solutions
Basic Solutions

  • The next section focuses on reactions that occur in basic solution.


If the reaction occurs in basic solution
If the reaction occurs in basic solution …

  • Although you can use H2O and OH- directly, the simplest technique is to first

    balance the reaction as if it occurred in acidic solution, and then perform the "conversion" (described on the next slide) to adjust it to conform to conditions in basic solution.


Balance the reaction in a basic solution
Balance the Reaction in a Basic Solution

Pb → PbO

  • First we balance it as if it occurred in an acidic solution.

    H20 + Pb → PbO + 2H+ + 2e-

    Add water to balance the oxygens, add H+ to balance the H’s then add e- to balance the charge.


The conversion to basic solution follows these three steps
The conversion to basic solution follows these three steps:

  • Step 1

  • For each H+ that must be eliminated from the equation, add an OH- to both sides of the equation.

  • In this example, we have to eliminate 2H+, so we add 2OH- to each side.

    H20 + Pb + 2OH- → PbO + 2H+ + 2OH- + 2e-


  • Step 2

  • Combine H+ and OH- to form H20.

  • We have 2H+ and 2OH- on the right, which creates 2H20.

    H20 + Pb + 2OH- → PbO + 2H2O + 2e-


  • Step 3

  • Cancel any H20 that are the same on both sides.

  • We can cancel one H20 from each side.

  • The final balanced half-reaction in basic solution is:

  • Pb + 2OH- → PbO + H2O + 2e-


Practice problem 2 in basic solution
Practice Problem #2 (in basic solution)

MnO4-1 + I-1→ MnO2 + I2


Practice problem 2 answer
Practice Problem #2 Answer

2MnO4-1 + 6 I-1 + 4H2O→ 2MnO2 + 3 I2 + 8OH-1

Worked example is on the next several slides


Practice problem 2 answer1
Practice Problem #2 Answer

Separate the reaction into 2 half reactions:

MnO4-1→ 2MnO2

I -1→ I2

Balance the atoms:

MnO4-1 + 2H2O→ MnO2 + 4OH-1

2 I-1→ I2



Multiply to make the e- the same in both reactions:

2(3e- + MnO4-1 + 2H2O→ MnO2 + 4OH-1)

3(2 I-1→ I2 + 2e-)

The half reactions become:

6e- + 2MnO4-1 + 4H2O→ 2MnO2 + 8OH-1

6 I-1→ 3I2 + 6e-


Final answer1
Final Answer

  • Add the reactions together:

    6e- + 2MnO4-1 + 4H2O→ 2MnO2 + 8OH-1

    6 I-1→ 3I2 + 6e-______________________________________

    2MnO4-1 + 4H2O + 6 I-1→ 2MnO2 + 8OH-1 + 3I2


Website to check out
Website to Check Out

  • http://fac.swic.edu/clercdg/Chem101_Redox_IonElectronMethod.PDF


Wrap up1
Wrap Up

  • Over the weekend, try the reaction prediction questions at the end of Ch 11 in you UEHB textbook.

  • Do as much as you can.

  • If you get frustrated, please stop.


ad