ion electron method ch 20. drill use ap rev drill #
TRANSCRIPT
Objectives
• SWBAT• Work through the steps of the Ion Electron
Method for solving Redox equations in acidic and basic conditions.
Write Half Reactions
• Write an oxidation and a reduction half reaction.
Sn2+ → Sn 4+
Hg 2+ + Cl-1 → Hg2Cl2
Balance Half Reactions
• Balance each half reaction in terms of atoms.
Sn2+ → Sn 4+
2Hg 2+ + 2Cl-1 → Hg2Cl2
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• 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 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-
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• 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-
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
If the reaction occurs in acidic solution …
Cr2O7 2- + H2S → Cr 3+ + S
Write the half reactions:
H2S → S
Cr2O7 2- → Cr 3+
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 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
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
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
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
• 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 Problem #1:
Balance the following equation in acidic solution:
Fe+2 + Cr2O7 -2 → Fe+3 + Cr+3
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 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.
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 Answer
2MnO4 -1 + 6 I-1 + 4H2O → 2MnO2 + 3 I2 + 8OH-
1
Worked example is on the next several slides
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 Answer
• Add the reactions together:
6e- + 2MnO4 -1 + 4H2O → 2MnO2 + 8OH-1
6 I-1 → 3I2 + 6e-
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2MnO4 -1 + 4H2O + 6 I-1 → 2MnO2 + 8OH-1 + 3I2