ion electron method
DESCRIPTION
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. - PowerPoint PPT PresentationTRANSCRIPT
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Ion Electron Method
Ch 20
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Drill
• Use AP rev drill #
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Objectives
• SWBAT• Work through the steps of the Ion Electron
Method for solving Redox equations in acidic and basic conditions.
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• Begin with slide 22
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Write Half Reactions
• Write an oxidation and a reduction half reaction.
Sn2+ → Sn 4+
Hg 2+ + Cl-1 → Hg2Cl2
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Balance Half Reactions
• Balance each half reaction in terms of atoms.
Sn2+ → Sn 4+
2Hg 2+ + 2Cl-1 → Hg2Cl2
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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)
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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
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Add the Reactions
• Add the resulting half-reactions to obtain the balanced net ionic equation. Sn 2+ → Sn 4+ + 2e-2e- + 2Hg 2+ + 2Cl- → Hg2Cl22e- + Sn 2+ + 2Hg 2+ + 2Cl-1 → Sn 4+ + Hg2Cl2 + 2e-
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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.
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Additional Info• In many oxidation-reduction reactions that
take place in aqueous solution, water plays an active role.
•
• 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-
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Practice Problem
NO + SO4 – 2 NO3 – 1 + SO2
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Practice Problem Answer
NO NO3 -1
SO4 – 2 SO2
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Balance Atoms
NO + 2H2O NO3 -1 NO + 2H2O NO3 -1 + 4H+1
SO4 – 2 SO2 + 2H2O 4H+1 + SO4 – 2 SO2 + 2H2O
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Add ElectronsNO + 2H2O NO3 -1 + 4H+1 + 3e-1
4H+1 + SO4 – 2 + 2e-1 SO2 + 2H2O
multiply the top rxn by 2multiply the bottom rxn by 3both rxns now have 6 e-1
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Final Answer
2 NO + 4 H+1 + 3 SO4 – 2 2NO3 -1 + 3 SO2 + 2 H2O
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Wrap Up
• Try the practice problems at the end of Ch 11 in the UEHB text.
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Acidic Solutions
• The next section focuses on reactions that occur in acidic solution.
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If the reaction occurs in acidic solution …
Cr2O7
2- + H2S → Cr 3+ + S
Write the half reactions:H2S → SCr2O7 2- → Cr 3+
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Acidic Solution
Balance the S atoms first.Add H+ to balance the H in the reaction, thenbalance the H+
H2S → S + 2H+
Balance the charge by adding electrons:H2S → S + 2H+ + 2e-
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Use H2O and H+1 to Balance the Equation
Balance the chromium atoms: Cr2O7 2- → 2Cr 3+
Balance the oxygens on the left by adding water to the right side of the equation:
Cr2O7 2- → 2Cr 3+ + H2O
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Now add H+1 to the left:H+1 + Cr2O7 2- → 2Cr 3+ + H2O
Balance the H’s and O’s:14H+1 + Cr2O7 2- → 2Cr 3+ + 7H2O
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Now add electrons to balance the charge:14H+1 + Cr2O7 2- → 2Cr 3+ + 7H2O
There is 14+ and 2- on the left (overall 12+)There is 6+ on the rightTherefore, add 6e- to the left to balance the
charge.6e- + 14H+1 + Cr2O7 2- → 2Cr 3+ + 7H2O
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Add the 2 half reactions together
3 (H2S → S + 2H+ + 2e-)6e- + 14H+ + Cr2O7
2- → 2Cr 3+ + 7H203H2S + 14H+ + Cr2O7
2- + 6e- → 3S + 6H+ + 2Cr 3+ + 7H20 + 6e-
Cancel out anything that is the same on both sides:3H2S + 8H+ + Cr2O7
2- → 3S + 2Cr 3+ + 7H20
Note: notice how some of the H+ ions cancel out.
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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 andthen two H+ ions to the opposite side to remove the hydrogen imbalance.
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Practice Problems
Practice Problem #1:Balance the following equation in acidic
solution:
Fe+2 + Cr2O7 -2 → Fe+3 + Cr+3
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Practice Problem # 1 Answer
6Fe+2 + 14 H+1 + Cr2O7 -2 → 6Fe+3 + 2Cr+3 + 7H2O
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Basic Solutions
• The next section focuses on reactions that occur in basic solution.
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If the reaction occurs in basic solution …
• Although you can use H2O and OH- directly, the simplest technique is to firstbalance 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.
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Balance the Reaction in a Basic SolutionPb → 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.
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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-
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• 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-
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• 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-
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Practice Problem #2 (in basic solution)
MnO4 -1 + I-1 → MnO2 + I2
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Practice Problem #2 Answer
2MnO4 -1 + 6 I-1 + 4H2O → 2MnO2 + 3 I2 + 8OH-
1
Worked example is on the next several slides
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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
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• Balance the charge:3e- + MnO4 -1 + 2H2O → MnO2 + 4OH-1
2 I-1 → I2 + 2e-
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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-
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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
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Website to Check Out• http://fac.swic.edu/clercdg/
Chem101_Redox_IonElectronMethod.PDF
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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.