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OFB Chapter 12 111/1/2004
Chapter 12Redox reactions and
Electrochemistry• 12-1 Balancing Redox
Equations• 12-2 Electrochemical Cells• 12-3 Stoichiometry in
Electrochemical Cells• 12-4 Metals and Metallurgy• 12-5 Electrometallurgy
Note: See course website for text book error corrections Chapter 12 p. A-63 problem 29
OFB Chapter 12 211/1/2004
• This entire Chapter deals with Oxidation, Reduction, electrochemistry and electron transfer
• Balancing Redox Reactions• Practice, Practice, Practice• All reactions take place in either
Acid (H+) or Basic (OH-) conditions
• Very important to remember the problem you are working. Is it acidic or basic conditions?
• We will be adding H2O and H+
for acid reactions or H2O and OH- for basic reaction in order to properly balance the reactions
• Practice, Practice, Practice
OFB Chapter 12 311/1/2004
Gains ElectronsDecreaseSubstance Reduced
Loses ElectronsIncreaseSubstance Oxidized
Supplies ElectronsIncrease
Reducing Agent, does the reducing
Picks Up electronsDecrease
Oxidizing Agent, does the oxidizing
Gain of ElectronsDecreaseReduction
Loss of ElectronsIncreaseOxidation
Electron Change
Oxidation Number ChangeTerm
OFB Chapter 12 411/1/2004
• Step 1 Write two unbalanced half-equations, one for the species that is oxidized and its product and one for the species that is reduced and its product
• Step 2 Insert coefficients to make the numbers of atoms of all elements except oxygen and hydrogen equal on the two sides of each half-equation.
• Step 3 Balance oxygen by adding H2O to the side deficient in O in each half-equation
• Step 4 Balance hydrogen. – For half-reaction in acidic solution, add H+ on
to the side deficient in hydrogen. – For a half-reaction in basic solution, add H2O
to the side that is deficient in hydrogen and an equal amount of OH- to the other side
OFB Chapter 12 511/1/2004
• Step 5 Balance charge by inserting e-
(electrons) as a reactant or product in each half-reaction.
• Step 6 Multiply the two half-equations by numbers chosen to make the number of electrons given off by the oxidation equal to the number taken up by the reduction. Then add the two half-equations and cancel out the electrons. If H+ ion, OH-
ion, or H2O appears on both sides of the final equation, cancel out the duplication
• Step 7 Check for balance
OFB Chapter 12 611/1/2004
Balancing Redox Reactions
• Dithionate ion reacting with chlorous acid in aqueous acidicconditions
S2O62-(aq) +HClO2(aq)
→ SO42-(aq) + Cl2(g)
Will have:
Oxidation half reaction
Reduction half reaction
OFB Chapter 12 711/1/2004
• Step 1 Write two unbalanced half-equations, one for the species that is oxidized and its product and one for the species that is reduced and its product
S2O62- → SO4
2-
Dithionate (we will see eventually that this is the
oxidation reaction)
HClO2 → Cl2Chlorous acid
(we will see eventually that this is the reduction reaction)
S2O62-(aq) +HClO2(aq)
→ SO42-(aq) + Cl2(g)
OFB Chapter 12 811/1/2004
• Step 2 Insert coefficients to make the numbers of atoms of all elements except oxygen and hydrogen equal on the two sides of each half-equation.
S2O62- → SO4
2-
HClO2 → Cl2
2
2
OFB Chapter 12 911/1/2004
• Step 3 Balance oxygen by adding H2O to the side deficient in O in each half-equation
S2O62- → 2 SO4
2-
2HClO2 → Cl2
(8Os)(6Os)
2H2O +(Need 2 Os)
(4Os)
+ 4H2O(Need 4 Os)
OFB Chapter 12 1011/1/2004
• Step 4 Balance hydrogen. – For half-reaction in acidic solution, add H+ on to the
side deficient in hydrogen. – For a half-reaction in basic solution, add H2O to the
side that is deficient in hydrogen and an equal amount of OH- to the other side
2H2O + S2O62- → 2SO4
2-
2HClO2 → Cl2 + 4H2O
Dithionate ion reacting with chlorous acid in aqueous acidic conditions
Need 4 Hs
Need 6 Hs (2 Hs) (8 Hs)
+ 4H+
6H+ +
(4 Hs)
OFB Chapter 12 1111/1/2004
• Step 5 Balance charge by inserting e- (electrons) as a reactant or product in each half-reaction.
2H2O + S2O62-
→ 2SO42- + 4H+
6H+ + 2HClO2
→ Cl2 + 4H2O
-2
-4 +4 Need -2
+ 2e-
+6
Zero charge
Need -6
+ 6e-
Oxidation reaction, electrons are lost on the product side
Reduction reaction, electrons are gained on the reactant side
OFB Chapter 12 1211/1/2004
• .Step 6 Multiply the two half-equations by numberschosen to make the number of electrons given off by the oxidation equal to the number taken up by the reduction. Then add the two half-equations and cancel out the electrons. If H+ ion, OH- ion, or H2O appears on both sides of the final equation, cancel out the duplication
Oxidation Reaction has 2 electrons
Reduction reaction has 6 electrons
∴Multiply the oxidation reaction by 3 this will give 6 electrons on both sides of the reaction
2H2O + S2O62-
→ 2SO42- + 4H+ + 2e-
3 times
OFB Chapter 12 1311/1/2004
• .Step 6 Multiply the two half-equations by numberschosen to make the number of electrons given off by the oxidation equal to the number taken up by the reduction. Then add the two half-equations and cancel out the electrons. If H+ ion, OH- ion, or H2O appears on both sides of the final equation, cancel out the duplication
3 times
6H2O + 3S2O62-
→ 6SO42- + 12H+ + 6e-
2H2O + S2O62-
→ 2SO42- + 4H+ + 2e-
equals
Now add both Re-Ox reactions
OFB Chapter 12 1411/1/2004
• .Step 6 Multiply the two half-equations by numberschosen to make the number of electrons given off by the oxidation equal to the number taken up by the reduction. Then add the two half-equations and cancel out the electrons. If H+ ion, OH- ion, or H2O appears on both sides of the final equation, cancel out the duplication
6H2O + 3S2O62-
→ 6SO42- + 12H+ + 6e-
6H+ + 2HClO2 + 6e-
→ Cl2 + 4H2O
6H2O + 3S2O62- + 6H+ +
2HClO2 + 6e-
→ 6SO42- + 12H+ + Cl2
+ 4H2O + 6e-
2
6
OFB Chapter 12 1511/1/2004
• Step 7 Check for balance
2H2O + 3S2O62- + 2HClO2
→ 6SO42- + 6H+ + Cl2
Left side
H 6
O 24
S 6
Cl 2
Right side
H 6
O 24
S 6
Cl 2If it doesn’t balance, look for simple mistakes first
Omitted subscripts
Omitted superscripts
OFB Chapter 12 1611/1/2004
Typical Exam Question
• Find the (integer) stoichiometry coefficients for the correctly balanced version of the following reaction
Cr2O72- + H+ + I-
→ Cr+3 + H2O + I3-
OFB Chapter 12 1711/1/2004
• Step 1 Write two unbalanced half-equations, one for the species that is oxidized and its product and one for the species that is reduced and its product
Cr2O72- + H+ + I-
→ Cr+3 + H2O + I3-
Cr2O72-→
Cr+3
I-→ I3-
OFB Chapter 12 1811/1/2004
• Step 2 Insert coefficients to make the numbers of atoms of all elements except oxygen and hydrogen equal on the two sides of each half-equation.
2
3
Cr2O72-→ Cr+3
I-→ I3-
OFB Chapter 12 1911/1/2004
• Step 3 Balance oxygen by adding H2O to the side deficient in O in each half-equation
(7Os)+ 7H2O
(Need 7 Os)
(No Oxygen, okay)
Cr2O72-→ 2Cr+3
3I-→ I3-
OFB Chapter 12 2011/1/2004
• Step 4 Balance hydrogen. – For half-reaction in acidic solution, add H+ on to the
side deficient in hydrogen. – For a half-reaction in basic solution, add H2O to the
side that is deficient in hydrogen and an equal amount of OH- to the other side
Dichromate ion reacting with iodide ion in aqueous acidic conditions
Need 14 Hs
14H+ +
(14 Hs)
Cr2O72-
→ 2Cr+3 + 7H2O
3I-→ I3-
(No Hydrogen, okay)
OFB Chapter 12 2111/1/2004
• Step 5 Balance charge by inserting e- (electrons) as a reactant or product in each half-reaction.
-2+14
+6
Need -6
+ 2e-
-3 Need -2
6e- +
Oxidation reaction
Reduction reaction
14H+ + Cr2O72-
→ 2Cr+3 + 7H2O
3I-→ I3-
-1
OFB Chapter 12 2211/1/2004
• .Step 6 Multiply the two half-equations by numberschosen to make the number of electrons given off by the oxidation equal to the number taken up by the reduction. Then add the two half-equations and cancel out the electrons. If H+ ion, OH- ion, or H2O appears on both sides of the final equation, cancel out the duplication
Reduction reaction has 6 electrons
Oxidation Reaction has 2 electrons
Multiply the oxidation reaction by 3 this will give 6 electrons on both sides of the reaction
3 times
3I-→ I3- + 2e-
OFB Chapter 12 2311/1/2004
• .Step 6 Multiply the two half-equations by numberschosen to make the number of electrons given off by the oxidation equal to the number taken up by the reduction. Then add the two half-equations and cancel out the electrons. If H+ ion, OH- ion, or H2O appears on both sides of the final equation, cancel out the duplication
equals
Now add both Re-Ox reactions
3 times
3I-→ I3- + 2e-
9I-→ 3 I3- + 6 e-
OFB Chapter 12 2411/1/2004
• .Step 6 Multiply the two half-equations by numberschosen to make the number of electrons given off by the oxidation equal to the number taken up by the reduction. Then add the two half-equations and cancel out the electrons. If H+ ion, OH- ion, or H2O appears on both sides of the final equation, cancel out the duplication
6e- + 14H+ + Cr2O72-
→ 2Cr+3 + 7H2O
9I-→ 3 I3- + 6 e-
6e- + 14H+ + Cr2O72- + 9I-
→ 2Cr+3 + 7H2O + 3 I3- + 6 e-
equals
OFB Chapter 12 2511/1/2004
• Step 7 Check for balance
Left side
H 14
Cr 2
O 7
I 9
Right side
H 14
Cr 2
O 7
I 9
14H+ + Cr2O72- + 9I-
→ 2Cr+3 + 7H2O + 3 I3-
OFB Chapter 12 2611/1/2004
Basic SolutionMethod 1
• At Step 4– Instead of adding H+ on the
deficient side as did with the acid solution method
– Add H2O on the H deficient side and add an equal amount of OH- on the other side.
OFB Chapter 12 2711/1/2004
Typical Exam Question
• Note: this is not Example 12-2, which is similar
• Find the (integer) stoichiometry coefficients for the correctly balanced version of the following reaction conducted in a basic solution
Ag2S + Cr(OH)3
→ Ag0 + HS- + CrO42-
OFB Chapter 12 2811/1/2004
• Step 1 Write two unbalanced half-equations, one for the species that is oxidized and its product and one for the species that is reduced and its product
Ag2S + Cr(OH)3
→ Ag0 + HS- + CrO42-
Ag2S → Ag + HS-
Cr(OH)3 → CrO42-
Reduction reaction
+1 0
Oxidation reaction
+3 +6
OFB Chapter 12 2911/1/2004
• Step 2 Insert coefficients to make the numbers of atoms of all elements except oxygen and hydrogen equal on the two sides of each half-equation.
2Ag2S → Ag + HS-
Cr(OH)3 → CrO42-
Already Balanced
OFB Chapter 12 3011/1/2004
• Step 3 Balance oxygen by adding H2O to the side deficient in O in each half-equation
(4Os)
H2O +
(Oxygen not needed)
Ag2S → 2Ag + HS-
Cr(OH)3 → CrO42-
(3Os)(need 1O)
OFB Chapter 12 3111/1/2004
• Step 4 Balance hydrogen. – For half-reaction in acidic solution, add H+ on
to the side deficient in hydrogen. – For a half-reaction in basic solution, add H2O
to the side that is deficient in hydrogen and an equal amount of OH- to the other side
This problem is in aqueous basic conditions
Needs H
(1 H)
Ag2S
→ 2Ag + HS-
H2O +
+ OH-
(need O)
H2O + Cr(OH)3
→ CrO42-
(3 H)(2 H)
+ 5H2O
5OH- +
Needs H
Add OH-
OFB Chapter 12 3211/1/2004
• Step 5 Balance charge by inserting e- (electrons) as a reactant or product in each half-reaction.
-2
Need -2
+ 3e--5
Need -3
H2O + Ag2S
→ 2Ag + HS- + OH-
5OH- + H2O + Cr(OH)3
→ CrO42- + 5H2O
-1 -1
2e- +
Reduction reaction
Oxidation reaction
OFB Chapter 12 3311/1/2004
• .Step 6 Multiply the two half-equations by numberschosen to make the number of electrons given off by the oxidation equal to the number taken up by the reduction. Then add the two half-equations and cancel out the electrons. If H+ ion, OH- ion, or H2O appears on both sides of the final equation, cancel out the duplication
Reduction reaction has 2 electrons
Oxidation Reaction has 3 electrons2e x 3 = 6e
3e x 2 = 6e
OFB Chapter 12 3411/1/2004
• .Step 6 Multiply the two half-equations by numberschosen to make the number of electrons given off by the oxidation equal to the number taken up by the reduction. Then add the two half-equations and cancel out the electrons. If H+ ion, OH- ion, or H2O appears on both sides of the final equation, cancel out the duplication
6e- + 3H2O + 3Ag2S
→ 6Ag + 3HS- + 3OH-Reduction reaction
times Reduction Reaction =
10OH- + 2H2O + 2Cr(OH)3
→ 2CrO42- + 10H2O + 6e-
times Oxidation Reaction =
Oxidation reaction
OFB Chapter 12 3511/1/2004
• .Step 6 Multiply the two half-equations by numberschosen to make the number of electrons given off by the oxidation equal to the number taken up by the reduction. Then add the two half-equations and cancel out the electrons. If H+ ion, OH- ion, or H2O appears on both sides of the final equation, cancel out the duplication
Now add both Re-Ox reactions
6e- + 3H2O + 3Ag2S → 6Ag + 3HS- + 3OH-
10OH- + 2H2O + 2Cr(OH)3
→ 2CrO42- + 10H2O + 6e-
OFB Chapter 12 3611/1/2004
equals
Now add both Re-Ox reactions6e- + 3H2O + 3Ag2S → 6Ag + 3HS- + 3OH-
10OH- + 2H2O + 2Cr(OH)3
→ 2CrO42- + 10H2O + 6e-
6e- + 3H2O + 3Ag2S +10OH- + 2H2O + 2Cr(OH)3
→ 6Ag + 3HS- + 3OH- +2CrO4
2- + 10H2O + 6e-
OFB Chapter 12 3711/1/2004
simplify
6e- + 3H2O + 3Ag2S +10OH- + 2H2O + 2Cr(OH)3
→ 6Ag + 3HS- + 3OH- +2CrO4
2- + 10H2O + 6e-5
7
3Ag2S + 7OH- + 2Cr(OH)3
→6Ag + 3HS- + 2CrO4
2- + 5H2O
OFB Chapter 12 3811/1/2004
• Step 7 Check for balance
Left side
Ag 6
S 3
O 13
H 13
Cr 2
3Ag2S + 7OH- + 2Cr(OH)3
→6Ag + 3HS- + 2CrO4
2- + 5H2O
Right side
Ag 6
S 3
O 13
H 13
Cr 2NOTE: This problem is the reverse reaction of Example 12-2
Left side
H 14
Cr 2
O 7
I 9
Right side
H 14
Cr 2
O 7
I 9
OFB Chapter 12 3911/1/2004
Typical Exam QuestionAnswer
• Note: this is not Example 12-2, which is similar
• Find the (integer) stoichiometry coefficients for the correctly balanced version of the following reaction conducted in a basic solution
3 Ag2S + 2 Cr(OH)3+ 7 OH-→ 6 Ag + 3 HS- +
2CrO42- + 5H2O
OFB Chapter 12 4011/1/2004
Chapter 12Redox reactions and
Electrochemistry
• Practice• Practice• Practice
OFB Chapter 12 4111/1/2004
Balancing Disproportionation Reactions
• Same Steps• For example
Cl2 → ClO3- + Cl-
• Step 1– Oxidation:
Cl2 → ClO3-
– Reduction:Cl2 → Cl-
OFB Chapter 12 4211/1/2004
Chapter 12Redox reactions and
Electrochemistry• Review of Common Terms
Term Oxidation State Change Electron Change
1 Oxidation increase loss of electronse.g., Cu to Cu 2+ (0 to +2)
2 Reduction decrease gain of electronse.g., Ni + to Ni (+1 to 0)
3 Oxidizing Agent decrease picks up electronse.g., O 2 to oxide (0 to -2)
4 Reducing Agent increase supplies electronse.g., H 2 to H 2 O (0 to +1)
5 Substance Oxidized increase loses electronse.g., Mg to MgO (0 to +2)
6 Substance Reduced decrease gains electronse.g., CuO to Cu (+2 to 0)
OFB Chapter 12 4311/1/2004
Electrochemical Cells
Cu(s) + 2Ag+(aq) → Cu++(aq) + 2Ag(s)
Silver plates spontaneously
∴ ∆Gf < 0
Called a galvanic or voltaic cell
Oxidation Cu → Cu++ + 2e-
Reduction Ag+ + e-→ Ag
OFB Chapter 12 4411/1/2004
Cu(s) + 2Ag+(aq) → Cu++(aq) + 2Ag(s)
Oxidation occurs here at the anode
Reduction occurs here at the cathode
Chemical energy converted to Electrical Energy. This potential difference can be measured.
Current = charge/time or I=Q/t
Units amperes = coulombs/sec.
OFB Chapter 12 4511/1/2004
Cu(s) + 2Ag+(aq) → Cu++(aq) + 2Ag(s)
Convention for this reaction
Cu(s) | Cu+2(aq) ¦¦ Ag+(aq) | Ag(s)
OFB Chapter 12 4611/1/2004
If an opposing electric force is applied the
reverse reaction occurs, called an electrolytic cell
Cu++(aq) + 2Ag(s)→ Cu(s) +2Ag+(aq)
The electrodes are reversed
Ag becomes the anode
Cu becomes the cathode
Note that the electron flow is still anode to cathode
Ag+(aq) | Ag(s) ¦¦ Cu(s) | Cu+2(aq)Ag+(aq) | Ag(s) ¦¦ Cu(s) | Cu+2(aq)
OFB Chapter 12 4711/1/2004
Stoichiometry in Electrochemical Cells
• Faraday’s Law– The quantities of substances
produced or consumed at the electrodes are directly proportional to the amount of electric charge passed through the cell.
– When a given amount of electric charge passes through a cell, the quantity of a substance produced or consumed at an electrode is proportional to its molar mass divided by the number of moles of electrons required to produce or consume one mole of the substances.
OFB Chapter 12 4811/1/2004
1 Faraday = 96,485 Coulombs of charge per mole of electrons
Faraday Constant (F ) = the electric charge carried by 1 mole of electrons.
Charge 1 e = 1.602X10-19 Coulombs
X Avogadro’s number
OFB Chapter 12 4911/1/2004
Problem 12-30What number of coulombs of electricity
is produced in each of the following oxidations?
• 1.00 mol of H2O2(aq) to O2(g)
CxCOH
OH 5-
22
-
22 1093.1e mol
485,96 mol
e mol 2 mol00.1 =
×
×
OFB Chapter 12 5011/1/2004
Problem 12-30What number of coulombs of electricity
is produced in each of the following oxidations?
• 1.70 mol of H2S (aq) to 0.850 mol of H2S2O3(aq)
H2S → H2S2O3
Cx
C
5
-2
-
2
1056.6
e mol485,96
SH mol 2e mol 8SH .7mol1
=
×
×
OFB Chapter 12 5111/1/2004
Stoichiometry in Electrochemical Cells
Ag+ + e-→ Ag
187.107M
=en
Cu → Cu++ + 2e-
254.63
nM
e
=
electrons #per massmolar M/nMassMolar M
e ==
OFB Chapter 12 5211/1/2004
Chapter 12Redox reactions and
Electrochemistry
Example
How many grams of Cl2 can be produced by the electrolysis of molten NaCl at a current of 10.0 amps for 5 minutes?
2Cl- → Cl2 + 2e-
grams 11.1produced Cl485,96
29.70minsec/60min5amps0.10
)(Cl
96,485nM t I
mass
2
2
e
=
•••=
••=
grams
OFB Chapter 12 5311/1/2004
Reminder
1 mole electrons reduces 1 mol of Ag+
2 moles electrons reduces 1 mole of Mg2+
3 moles electrons reduces 1 mole of Al3+
OFB Chapter 12 5411/1/2004
ExampleHow long would it take to plate 5.43 grams Nickel on to an electrode from a solution of NiCl2 at a current of 12.34 amps? Molar mass of Nickel is 58.693 g mol-1.
Ni+2 + 2e- → Ni
OFB Chapter 12 5511/1/2004
Chapter 12Redox reactions and
Electrochemistry
Summary• Balancing Redox reactions
–––
• Understand electrochemical cells and the Faraday law calculations