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# Ion Electron Method - PowerPoint PPT Presentation

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.

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### Ion Electron Method

Ch 20

• Use AP rev drill #

• SWBAT

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

• Write an oxidation and a reduction half reaction.

Sn2+→ Sn 4+

Hg 2+ + Cl-1→ Hg2Cl2

• Balance each half reaction in terms of atoms.

Sn2+→ Sn 4+

2Hg 2+ + 2Cl-1→ Hg2Cl2

• 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)

• 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 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 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.

• 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-

NO + SO4– 2 NO3 –1 + SO2

NO  NO3-1

SO4– 2 SO2

NO + 2H2O  NO3-1

NO + 2H2O  NO3-1 + 4H+1

SO4– 2 SO2 + 2H2O

4H+1 + SO4– 2 SO2 + 2H2O

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

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

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

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

Cr2O7 2- + H2S → Cr 3+ + S

Write the half reactions:

H2S → S

Cr2O72-→ Cr 3+

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 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

H+1 + Cr2O72- → 2Cr 3+ + H2O

Balance the H’s and O’s:

14H+1 + Cr2O72- → 2Cr 3+ + 7H2O

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

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.

• 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 Problem #1:

Balance the following equation in acidic solution:

Fe+2 + Cr2O7 -2→ Fe+3 + Cr+3

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

• The next section focuses on reactions that occur 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.

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.

• 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-

MnO4-1 + I-1→ MnO2 + I2

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

Worked example is on the next several slides

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-

6e- + 2MnO4-1 + 4H2O→ 2MnO2 + 8OH-1

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

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

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

• 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.