Redox Reactions

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Redox Reactions
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• Oxidation-Reduction

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Changing the oxidation number of iron

• Place a cubic inch of steel wool into 2 cups of
  vinegar. Cover and let sit overnight.
• Take 30 mL of iron acetate solution and place
  into a container. Add 5 mL of ammonia
  (light-orange-brown to dark green).
• Add 5 mL of hydrogen peroxide (deep blood-clot
  red).
• Iron is changed from ferrous ion to ferric ion.

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But there is a lot more to "oxidation-reduction"

• Atoms all assigned oxidation state of zero
• Oxidation-reduction reactions involve change in
  oxidation state of atoms or ions
• "Leo goes Ger"
• Loss of electrons oxidation
• Gain of electrons reduction

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Process of oxidation cannot occur w/out
corresponding reduction reaction electrons
"lost" by 1 substance always "gained" by another
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• Label oxidation for all atoms
• Find oxidation change in atom(s)/both sides
  equation
• Metals always lose electrons (oxidized ? cations)
• Oxidation state always increases (more )
• Numerically equal to of electrons lost (valence
  es)
• Oxidation group (IA/1, IIA/2, IIIA/3,
  IVA/4)
• Ca ? Ca2 2e-
• Nonmetals always gain electrons (reduced ?
  anions)
• Oxidation state always decreases (more -)
• Depends on electrons gained
• Oxidation subtract valence electrons from 8.
  (IVA/-4, VA/-3, VIA/-2, VIIA/-1)
• O 2e- ? O2-

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Redox and electronegativity

• Metals lose electrons/nonmetals gain
• Molecular substances/polyatomic ions with
  covalent bonds
• More EN atom reduced by gaining electrons
• Less EN atom oxidized by losing electrons

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Rules for assigning oxidation numbers
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Oxidation States

• 0.01 M potassium permanganate solution (1.58 g/L
  water)
• 0.01 M sodium bisulfate (1.04 g/L water)
• While stirring, add 0.01 M sodium bisulfate to
  tubes and note color change.
• Fill in the chart on the next slide.

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Disruption of equilibrium (pg. 416/425)

• Add 8 g KOH to 300 mL of water in a flask, cool
  the solution, then dissolve 10 g of glucose. Add
  2-3 drops of methylene blue.
• Stopper the flask, shake and swirl. Adjust
  amounts if color change doesnt occur.
• When colorless, remove the stopper, pick it up
  and shake it.
• Is the solution blue or colorless at equilibrium?
  Explain.
• Why should you remove the stopper between
  shakings? What happens if you never remove the
  stopper?
• The blue bottle reaction is an oxidation-reduction
  reaction. What is the reducing agent in this
  reaction? What substance is reduced? What is
  the oxidizing agen in the reaction? What
  substance is oxidized?

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Homework

• Read 20.1, pp. 634-643
• Q pg. 643, 7-9
• Q pp. 658-659, 36-37, 44, 45, 46, 49, 51

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Oxidation number change method

• Redox equation balanced by comparing increases
  and decreases in oxidation
• Tip-off If you are asked to balance an equation
  and if you are not told whether the reaction is a
  redox reaction or not, you can use the following
  procedure

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• Step 1  Try to balance the atoms in the equation
  by inspection. If you successfully balance the
  atoms, go to Step 2. If you are unable to balance
  the atoms, go to Step 3.
• Step 2  Check to be sure that the net charge is
  the same on both sides of the equation. If it is,
  you can assume that the equation is correctly
  balanced. If the charge is not balanced, go to
  Step 3.
• Step 3  If you have trouble balancing the atoms
  and the charge by inspection, determine the
  oxidation numbers for the atoms in the formula,
  and go to Step 4.
• Step 4  Determine the net increase in oxidation
  number for the element that is oxidized and the
  net decrease in oxidation number for the element
  that is reduced.
• Step 5  Determine a ratio of oxidized to reduced
  atoms that would yield a net increase in
  oxidation number equal to the net decrease in
  oxidation number (number of electrons lost
  number of electrons gained).
• Step 6  Add coefficients to the formulas so as
  to obtain the correct ratio of the atoms whose
  oxidation numbers are changing.
• Step 7  Balance the rest of the equation by
  inspection.

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• HNO3(aq)  H3AsO3(aq)    NO(g)  H3AsO4(aq)  H2O
  (l)
• Step 1  Try to balance the atoms by inspection.
• H/O atoms difficult to balance in this equation.
• Step 3  Is the reaction redox?
• N atoms change from 5 to 2 (reduced)
• As atoms change from 3 to 5 (oxidized)
• Step 4  Determine net increase/decrease in
  oxidation numbers
• As  3 to 5     Net Change 2
• N  5 to 2      Net Change -3
• Step 5  Determine ratio of oxidized to reduced
  atoms that would yield a net increase in
  oxidation number equal to the net decrease in
  oxidation number
• 3 As lose 6 e-/2 N gain 6 e-
• Ratio of As atoms to N atoms is 32.
• Step 6 To get ratio identified in Step 5, add
  coefficients to formulas which contain elements
  whose oxidation number is changing
• 2HNO3(aq)  3H3AsO3(aq)     NO(g)  H3AsO4(aq) 
  H2O(l)
• Step 7  Balance rest of equation by inspection
• 2HNO3(aq)  3H3AsO3(aq)    2NO(g)  3H3AsO4(aq) 
   H2O(l)

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• Cu(s)  HNO3(aq)  ?  Cu(NO3)2(aq)
   NO(g)  H2O(l)
• Step 1  Try to balance atoms by inspection.
• N/O atoms difficult to balance by inspection.
• Step 3  Is the reaction redox?
• Cu 0 to 2/some N 5 to 2.
• Step 4  Determine net increase/decrease in
  oxidation numbers
• Cu  0 to 2      Net Change 2
• Some N  5 to 2     Net Change -3
• Step 5  Determine ratio of oxidized to reduced
  atoms that would yield net increase in oxidation
  number equal to the net decrease in oxidation
  number
• 3 Cu atoms (net change of 6) for every 2 N (net
  change of -6).
• Only put 2 in front of nitrogen atoms that are
  changing or have changed
• The 3 for the copper atoms can be placed in front
  of the Cu(s).
• Step 6 To get ratio identified in Step 5, add
  coefficients to formulas which contain elements
  whose oxidation number is changing
• 3Cu(s)  HNO3(aq)  ?  Cu(NO3)2(aq)  2NO(g)  H2O(
  l)
• Step 7  Balance rest of equation by inspection.
• 3Cu(s)  8HNO3(aq) ?3Cu(NO3)2(aq)  2NO(g)  4H2O(
  l)

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Some Examples

• Al(s)    MnO2(s)    Al2O3(s)    Mn(s)
• 4Al(s) 3MnO2(s) 2Al2O3(s) 3Mn(s)
• SO2(g)  HNO2(aq)  H2SO4(aq)   NO(g)
• SO2 2HNO2 H2SO4 2NO
• HNO3(aq)  H2S(aq)  NO(g)  S(s)  H2O(l)
• 2HNO3 3H2S 2NO 3S 4H2O
• Al(s)  H2SO4(aq)  Al2(SO4)3(aq)  H2(g)
• 2Al 3H2SO4 Al2(SO4)3 3H2

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Homework

• Read 20.2, pp. 644-649
• Q pg. 649, 20, 21
• Q pp. 658-659, 38-40, 54
• If you need more practice, do 56/57 for extra
  credit

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Balancing Redox Equations Run in Acidic
Conditions Using the Half-reaction Technique  

• Tip-off If you are asked to balance a redox
  equation and told that it takes place in an
  acidic solution, you can use the following
  procedure 

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• Cr2O72-(aq)  HNO2(aq) ? Cr3(aq) NO3-(aq) 
• (acidic)
• Step 1   Write the skeletons of the oxidation
  and reduction half-reactions.
• Cr2O72-  ?  Cr3
• HNO2   ?    NO3-
• Step 2    Balance all elements other than H and
  O.
• Cr2O72-  ?   2Cr3 (need 2 in front of chromium)
• HNO2   ?   NO3-
• Step 3    Balance the oxygen atoms by adding
  H2O molecules on the side of the arrow where O
  atoms are needed.
• The first half-reaction needs seven oxygen atoms
  on the right, so we add seven H2O molecules.
• Cr2O72-  ?    2Cr3    7H2O
• The second half-reaction needs one more oxygen
  atom on the left, so we add one H2O molecule.
• HNO2      H2O  ?   NO3- 

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• Step 4    Balance H atoms by adding H ions on
  the side of the arrow where H atoms are needed.
• 1st half-reaction needs 14 H atoms on left to
  balance 14 hydrogen atoms in 7 H2O molecules-add
  14 H ions to left. 
• Cr2O72-    14H   ?   2Cr3    7H2O
• 2nd half-reaction needs 3 H atoms on right to
  balance 3 hydrogen atoms on left-add 3 H ions to
  the right. 
• HNO2      H2O   ?   NO3-    3H
• Step 5    Balance the charge by adding
  electrons.
• Sum of charges on left side of chromium
  half-reaction is 12 (-2 for the Cr2O72- plus 14
  for the 14 H).
• Sum of charges on right side of chromium
  half-reaction is 6 (for the 2 Cr3).
• Add 6 electrons to left side-sum on each side
  becomes 6.
• 6e-    Cr2O72-    14H  ?  2Cr3    7H2O
• Sum on the left side of the nitrogen
  half-reaction is zero.
• Sum on the right side of the nitrogen
  half-reaction is 2 (-1 for the nitrate plus 3
  for the 3 H).
• Add 2 es to right side-sum on each side becomes
  zero.
• HNO2      H2O   ?   NO3-    3H    2e-

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• Step 6    If es lost in oxidation half ?
  es in reduction half, multiply one or both
  reactions by making es gained lost.
• For Cr half-reaction to gain 6 electrons, N
  half-reaction must lose 6 electrons-multiply
  coefficients in N half-reaction by 3.
• 6e-    Cr2O72-    14H   ?   2Cr3    7H2O
• 3(HNO2      H2O   ?   NO3-    3H    2e-)
• 3HNO2      3H2O   ?   3NO3-    9H    6e-
• Step 7    Add the 2 half-reactions.
• 3 H2O in 2nd half-reaction cancel 3 of 7 H2O in
  1st half-reaction to yield 4 H2O on the right of
  the final equation.
• 9 H on right of 2nd half-reaction cancel 9 of 14
  H on left of 1st half-reaction leaving 5 H on
  left of final equation.
• Cr2O72-    3HNO2     5H   ?   2Cr3   
  3NO3-     4H2O
• Step 8    Check to make sure that the atoms and
  the charge balance.
• The atoms in our example balance and the sum of
  the charges is 3 on each side, so our equation
  is correctly balanced.
• Cr2O72-(aq)  3HNO2(aq)   5H(aq) ?
• 2Cr3(aq)  3NO3- (aq)  4H2O(l)

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Example

• H2O2 I- ? I2 H2O
• I- is oxidized (oxidation state increases from -1
  to 0).O (oxygen) is reduced (oxidation state
  decreases from -1 to -2).
• The two half-reactions w/balanced number of key
  atoms are
• 2 I- ? I2 lt--- (oxidation)
• H2O2 ? 2 H2O lt--- (reduction of oxygen)
• Add electrons to compensate for changes of
  oxidation state
• 2 I- ? I2 2 e- (oxidation)
• H2O2 2 e- ? 2 H2O (reduction)
• Obviously, H should be added to the reduction
  half-reaction, and the balanced equations are
• 2 I- ? I2 lt--- (oxidation)
• H2O2 2H ? 2 H2O (reduction)

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Some Examples

• MnO4-(aq) Br(aq) MnO2(s) BrO3-(aq)
• MnO4-(aq) HSO3- (aq) Mn2(aq) SO42-(aq)
• Cr2O72-(aq)  C2O42-(aq)Cr3(aq) CO2(g)
• Mn(s)   HNO3(aq)  Mn2(aq)  NO2(g)
• ClO3 SO2 ? SO42 Cl
• H2S NO3 ? S8 NO
• MnO4 H2S ? Mn2 S8
• Cu SO42 ? Cu2 SO2

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• MnO4 CH3OH ? CH3COOH Mn2
• Cr2O72 Fe2 ? Cr3 Fe3
• HNO2 ? NO NO2
• H2C2O4 MnO4 ? CO2 Mn2
• O2 As ? HAsO2 H2O
• NO3 I2 ? IO3 NO2
• 11) HBr SO42 ? SO2 Br2
• 12) H5IO6 Cr ? IO3 Cr3

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• Fe HCl ? HFeCl4 H2
• NO3 H2O2 ? NO O2
• BrO3 Fe2 ? Br Fe3
• Cr2O72 C2H4O ? CH3COOH Cr3
• MnO4 C2H4O ? CH3COOH MnO2
• Zn NO3 ? NH4 Zn2
• NO2 H2O ? HNO3 NO
• S2 NO3 ? NO S8

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Balancing Redox Equations Run in Basic Conditions
Using the Half-reaction Technique  

• Tip-off If you are asked to balance a redox
  equation and told that it takes place in a basic
  solution, you can use the following procedure

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• Steps 1-7 Begin by balancing the equation as if
  it were in acid solution. If you have H ions in
  your equation at the end of these steps, proceed
  to Step 8. Otherwise, skip to Step11.
• Step 8   Add enough OH- ions to each side to
  cancel the H ions. (Be sure to add the OH- ions
  to both sides to keep the charge and atoms
  balanced.)
• Step 9  Combine the H ions and OH- ions that
  are on the same side of the equation to form
  water.
• Step 10  Cancel or combine the H2O molecules.
• Step 11  Check to make sure that the atoms and
  the charge balance. If they do balance, you are
  done. If they do not balance, re-check your work
  in Steps 1-10.

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• Cr(OH)3(s)  ClO3-(aq)  ?  CrO42-(aq)  Cl-(aq)   
  (basic)
• Step 1
• Cr(OH)3   ?  CrO42-
• ClO3-    ?    Cl-
• Step 2 (Not necessary for this example)
• Step 3
• Cr(OH)3    H2O   ?   CrO42-
• ClO3-    ?    Cl-    3H2O
• Step 4
• Cr(OH)3    H2O  ?   CrO42-    5H
• ClO3-    6H    ?    Cl-    3H2O
• Step 5
• Cr(OH)3    H2O   ?   CrO42-    5H    3e-
• ClO3-    6H    6e-  ?    Cl-    3H2O
• Step 6  
• 2(Cr(OH)3    H2O   ?   CrO42-    5H    3e- )
• 2Cr(OH)3    2H2O   ?   2CrO42-    10H    6e-
• ClO3-    6H    6e-    ?    Cl-    3H2O
• Step 7

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• Step 8 Because there are 4 H on the right side
  of our equation above, we add 4 OH- to each side
  of the equation.
• 2Cr(OH)3  ClO3-  4OH- ?
• 2CrO42-  Cl-  H2O  4H  4OH-
• Step 9 Combine the 4 H ions and the 4 OH- ions
  on the right of the equation to form 4 H2O.
• 2Cr(OH)3  ClO3-  4OH- ?
• 2CrO42-  Cl-  H2O  4H2O
• Step 10  Cancel or combine the H2O molecules.
• 2Cr(OH)3(s)    ClO3-(aq)    4OH-(aq)    ?
  2CrO42-(aq)    Cl-(aq)    5H2O(l)
• Step 11 The atoms in our equation balance, and
  the sum of the charges in each side is -5. Our
  equation is balanced correctly.

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Example

• ClO2 OH- ? ClO2- ClO3-
• In this reaction, Cl from ClO2 is both oxidized
  and reduced.
• The two half-reactions are
• ClO2 ? ClO2- (reduction)
• ClO2 ? ClO3- (oxidation)
• Add electrons to compensate for the oxidation
  changes
• ClO2 e- ? ClO2- (reduced, 4 -gt 3 for Cl)
• ClO2 ? ClO3- e- (oxidized, 4 -gt 5)
• Add H, OH-, or H2O to balance both equations
  results in
• ClO2 e- ? ClO2-
• ClO2 2 OH- ? ClO3- e- H2O
• Now add the two half reactions together to give
  the overall reaction 2 ClO2 2 OH- ? ClO2-
  ClO3- H2O

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Examples

• CrO42-(aq)  S2-(aq) Cr(OH)3(s)  S(s)
• MnO4-(aq)  I-(aq)  MnO2(s)  IO3-(aq)
• H2O2(aq)  ClO4-(aq)  O2(g) ClO2-(aq)
• S2-(aq)  I2(s)  SO42-(aq)  I-(aq)
• Cr(OH)3(s)  ClO3-( aq)  ?  CrO42-(aq)  Cl-(aq)
• NH3 ClO ? N2H4 Cl
• Au O2 CN ? Au(CN)2 H2O2
• Br MnO4 ? MnO2 BrO3

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• AlH4 H2CO ? Al3 CH3OH
• Se Cr(OH)3 ? Cr SeO32
• H2O2 Cl2O7 ? ClO2 O2
• Fe NiO2 ? Fe(OH)2 Ni(OH)2
• MnO4 H2O2 ? MnO2 O2
• Zn BrO4 ? Zn(OH)42 Br
• MnO4 S2 ? MnO2 S8
• Fe(OH)2 CrO42 ? Fe2O3 Cr(OH)4

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Homework

• Read 20.3, pp. 650-656
• Q pg. 653, 30
• Q pg. 659, 59, 60
• Test practice, pg. 661, all questions
• Use link for quiz and submit as before.
• http//www.glencoe.com/qe/science.php?qi1004