Balancing Redox Equations

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Balancing Redox Equations
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A little background material courtesy of the
College Board.

• Traditionally, this is the first topic in AP
  Chemistry that is totally new.
• However, the College Board has reworked their
  syllabus and has deemphasized balancing redox
  equations.
• In its place, there is more emphasis on the
  theory of an oxidation-reduction reaction.
• We are still going to take a look at them because
  it clarifies some of the more difficult aspects
  of redox.

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Oxidation Reduction Reactions

• Remember that although we talk about oxidation
  and reduction as if they were two completely
  separate processes, they always must be happening
  in unison.
• This is very evident when balancing redox
  equations.
• What this means, is that we are going to split
  each equation in half a reduction half and an
  oxidation half.
• They will be balanced individually and then
  recombined at the end.

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Before we work the problem

• Also, it makes quite a bit of difference if the
  reaction is taking place in either acidic or
  basic medium.
• How do you know which one, you ask ?
• Well, sometimes they will actually just come out
  and tell you.
• Other times, there is a known acidic or basic
  species in the problem. In other words, if the
  reaction is swimming in Sulfuric Acid (H2SO4)
  then we can be confident that it is in acidic
  medium.

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Example Problem 1 (acidic)

• Lets walk through a redox problem. Each step
  will be explained as we go.
• Example MnO4-1 Fe2 ? Fe3
  Mn2 (acidic)
• First we need to determine the oxidation states
  of each atom in the problem.
• Reactants Mn ? 7 O ? - 2 Fe ?
  2
• Products Fe ? 3 Mn ? 2
• We split this into a reduction and an oxidation,
  which is usually termed the reduction
  half-reaction and the oxidation half-reaction.
• We will do the reduction half-reaction first. we
  see that the Mn is being reduced.

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Example Problem 1 First Step

• So our half-reaction is going to look like.
• Reduction MnO4-1 ? Mn2 Mn goes from 7
  ? 2
• FIRST STEP Since the atoms of Mn are balanced,
  we then balance the Oxygen by adding water. That
  means 4 water molecules on the product side. Our
  equation becomes
• MnO4-1 ? Mn2 4 H2O
• Notice the water adds H1 ions that are not
  present on the other side. No problem we will
  take care of that in the next step.

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Example Problem 1 Second Step

• Here is what we have so far.
• MnO4-1 ? Mn2 4 H2O
• SECOND STEP Add Hydrogen Ions to the other side
  to balance the Hydrogen. We need to add 8 H1
  to the reactant side to balance the 8 Hydrogen
  atoms in water.
• 8 H1 MnO4-1 ? Mn2 4 H2O

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Example Problem 1 Third Step

• Here is what we have so far.
• 8 H1 MnO4-1 ? Mn2 4 H2O
• THIRD STEP We need to balance according to
  charge. Total up the charges on each side and
  then add electrons so the charges balance.
• Remember that redox is about changing charges on
  atoms.
• The reactant side is a 7 and the product side
  is 2. So we need to add 5 e- to the reactant
  side.
• 5 e-1 8 H1 MnO4-1 ? Mn2 4 H2O

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Example Problem 1 Fourth Step

• Here is what we have so far
• 5 e-1 8 H1 MnO4-1 ? Mn2 4 H2O
• FOURTH STEP Do a quick check to make sure that
  all atoms and charges are balanced.
• If they are, we go to the other half-reaction.
• This problem looks good to go, so on to the
  oxidation half-reaction.

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Example Problem 1 Oxidation

• Oxidation Fe2 ? Fe3 Fe goes from 2 to 3
  
• It is not unusual at all not to have to add any
  species nor to find Oxygen or Hydrogen on a
  half-reaction.
• We just happily go on our merry way and balance
  the charge, which in this case is by adding one
  electron to the product side to give us.
• Fe2 ? Fe3 e-1

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Example Problem 1

• Here are our two half-reactions
• 5 e-1 8 H1 MnO4-1 ? Mn2 4 H2O
• Fe2 ? Fe3
  e-1
• NOTE Notice that the electrons are always going
  to be found on opposite sides of the arrows.
  This makes sense if you think about it.
  Oxidation is loss of electrons, so we would
  expect to find electrons on the product side.
  Reduction is a gain of electrons, so they are
  picking up the electrons lost by the oxidized
  species.

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Example Problem 1

• We now have our two half-reactions and we need to
  put them together. First we need to analyze
  them, but putting them next to each other.
• REDUCTION 5 e-1 8 H1 MnO4-1 ? Mn2 4 H2O
• OXIDATION Fe2 ? Fe3 e-1
• Lets think about this. Does it seem right that
  we are only losing one electron but gaining 5 in
  this reaction ?
• Of course, not.
• We need to balance the charge. We do this by
  simply finding the lowest common multiple and
  make them equal. In this case the lowest common
  multiple is 5.

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Example Problem 1

• The Reduction half-reaction gets multiplied by 1
• The Oxidation half-reaction gets multiplied by 5.
  
• Here is what we have now.
• REDUCTION 5 e-1 8 H1 MnO4-1 ? Mn2 4
  H2O
• OXIDATION 5 Fe2 ? 5 Fe3 5
  e-1
• NOTE Always check to see if the number of
  electrons being oxidized is equaling the number
  of electrons being reduced. THEY MUST EQUAL !!

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Example Problem 1 Final Step

• FINAL STEP Combine the two half-reactions and
  cancel out anything that is found on both sides.
  The electrons will always be canceled out and
  often water will be as well. We now get the
  following.
• 8 H1 MnO4-1 5 Fe2 ? Mn2 5 Fe3 4 H2O
• One last check to verify everything is balanced
  and we go out and celebrate our victory over
  redox.

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Summary of Redox Steps for Acidic Medium

• 1. Determine the oxidation states of each atom.
• 2. Write out the oxidized and reduced species
  and set up their respective half-reactions
• 3. Balance the atoms (other than O and H).
• 4. Balance the Oxygen by adding water as needed.
• 5. Balance the Hydrogen by adding H1 as needed.
• 6. Add the number of electrons needed to balance
  the equation according to charge.
• 7. Combine the two half reactions

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Example Problem 2 (basic)

• Know we will look at a problem when redox takes
  place in a basic solution.
• Here is our equation
• NO2-1 (aq) Al (s) ? NH3 (g) AlO2-1
  (aq)
• Just like before, first we need to determine the
  oxidation states of each atom in the problem.
• Reactants N ? 3 O ? - 2 Al ? 0
• Products N ? - 3 H ? 1 Al ?
  3 O ? - 2

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Example Problem 2

• Once again, we split this into a reduction and an
  oxidation.
• Before we go any further, we need to talk about
  what are the different steps when balancing a
  reaction in basic solution as opposed to when it
  is in acidic solution.
• In an acidic medium, we work the problem, by
  adding water and hydrogen ions to make it
  balanced.
• If you are thinking that for basic solution, we
  will be adding water and hydroxide ions, you
  would be CORRECT.
• HOWEVER -- This is not as easy as it sounds. So
  a slightly different path is used to get this
  done.
• What is this you ask ?!! This might sound
  strange BUT

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Example Problem 2

• when balancing a redox reaction in basic
  medium, we start by pretending it is an acidic
  solution. Of course, we will need to
  compensate for this later in the problem, but
  this method is much simpler.
• From our reaction we can see that the Nitrogen is
  the species being reduced.
• Reduction NO2-1 ? NH3 N goes from 3 ? - 3

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Example Problem 2

• FIRST STEP The Nitrogen atoms are balanced so
  we start off by adding water to balance the
  oxygen atoms. Just like we did when we were
  balancing in acidic medium.
• NO2-1 ? NH3 2 H2O
• Remember that when we balance any type of
  equation, we only balance one atom at a time and
  if while doing that, we UNBALANCE something else,
  that is fine. We will correct that later.

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Example Problem 2

• Here is what we have so far.
• NO2-1 ? NH3 2 H2O
• SECOND STEP Add Hydrogen Ions to the other side
  to balance the Hydrogen. We need to add 6 H1
  to the reactant side to balance the 7 Hydrogen
  atoms on the reactant side. Make sure you take
  the Hydrogen atom in the HNO2-1 species.
• 6 H1 HNO2-1 ? NH3 2 H2O

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Example Problem 2

• Look at where we are now
• 6 H1 HNO2-1 ? NH3 2 H2O
• There are Hydrogen ions present, which means an
  acidic solution. But we were told that the
  reaction is taking place in basic solution.
• Remember that we said that we will have to
  account for the addition of the Hydrogen Ions at
  a later time.
• WELL THAT TIME IS NOW !!

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Example Problem 2

• 6 H1 HNO2-1 ? NH3 2 H2O
• What we do is to add Hydroxide ions to any
  Hydrogen ions to form water.
• Of course, if we add OH-1 to the one side, then
  we have to add a similar number of OH-1 ions to
  the other side.
• Here is what we now have.
• 6 H2O HNO2-1 ? NH3 2 H2O 6 OH-1

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Example Problem 2

• 6 H2O HNO2-1 ? NH3 2 H2O 6 OH-1
• This should look a bit better. Now we have a
  solution with hydroxide ions, which makes the
  statement of this reaction being in a basic
  solution a bit easier to see.
• We are not quite done yet. We still have to add
  the electrons to balance the charge. Also,
  remember that we have only done the reduction
  half-reaction. We still have to do the oxidation
  half-reaction.

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Example Problem 2

• 6 H2O HNO2-1 ? NH3 2 H2O 6 OH-1
• The total of the charges on the reactant side is
  -1 while on the product side it is -6. 5
  electrons need to be added to the reactant side.
• 5 e-1 6 H2O HNO2-1 ? NH3 2 H2O 6 OH-1
• Now we are done with the reduction half-reaction.

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Example Problem 2

• Oxidation Al 0 ? 3 Al ? AlO2-1
• We need to repeat the same steps over again, that
  we did for the reduction half-reaction.
• First we need to determine the oxidation states
  of each atom in the problem.
• Reactants Al ? 0
• Products Al ? 3 O ? - 2
• We see that the Al is being reduced

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Example Problem 2

• Here is our reaction Al ? AlO2-1
• Since the Al is balanced, we balance the Oxygen
  with water to give us.
• 2 H2O Al ? AlO2-1
• We add Hydrogen atoms to balance out those that
  we added when we inserted H2O
• 2 H2O Al ? AlO2-1 4 H1

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Example Problem 2

• 2 H2O Al ? AlO2-1 4 H1
• Now we need to do our step to compensate for the
  basic medium. Lets add the OH-1 on both sides
  as needed. The H1 becomes H2O
• 4 OH-1 2 H2O Al ? AlO2-1 4 H2O
• And finally, balance according to charge.
• 4 OH-1 2 H2O Al ? AlO2-1 4 H2O 3 e-1

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Example Problem 2

• Now we need to bring the two half-reactions
  together.
• 5 e-1 6 H2O HNO2-1 ? NH3 2 H2O 6 OH-1
• 4 OH-1 2 H2O Al ? AlO2-1 4 H2O 3
  e-1
• We need to make the electrons being oxidized and
  the electrons being reduced equal to each other.
  So we multiply the oxidation half-reaction by 5
  and the reduction half-reaction by 3.
• The next slide has the updated equations

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Example Problem 2

• 15 e-1 18 H2O 3 HNO2-1 ? 3 NH3 6 H2O
  18 OH-1
• 20 OH-1 10 H2O 5 Al ? 5 AlO2-1 20
  H2O 15 e-1
• Now we can start cancelling out species.
• Electrons First 15 e-1 from each side
• 18 H2O 3 HNO2-1 ? 3 NH3 6 H2O 18 OH-1
• 20 OH-1 10 H2O 5 Al ? 5 AlO2-1 20 H2O

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Example Problem 2

• 18 H2O 3 HNO2-1 ? 3 NH3 6 H2O 18 OH-1
• 20 OH-1 10 H2O 5 Al ? 5 AlO2-1 20 H2O
• There are 28 water molecules on the reactant side
  and 26 on the right. As you can see, we can
  combine from the two equations when the same
  species is on the same side. This leaves 2
  water molecules on the reactant side
• 3 HNO2-1 ? 3 NH3 18 OH-1
• 20 OH-1 2 H2O 5 Al ? 5 AlO2-1

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Example Problem 2

• 3 HNO2-1 ? 3 NH3 18 OH-1
• 20 OH-1 2 H2O 5 Al ? 5 AlO2-1
• There are hydroxide ions on each side 20 ? 18
• 3 HNO2-1 ? 3 NH3
• 2 OH-1 2 H2O 5 Al ? 5 AlO2-1
• We can now combine the two equations

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Example Problem 2

• 3 HNO2-1 ? 3 NH3
• 2 OH-1 2 H2O 5 Al ? 5 AlO2-1
• This is what we get.
• 3 HNO2-1 2 OH-1 2 H2O 5 Al ? 5 AlO2-1 3
  NH3
• We do our final check to make sure everything is
  balanced.
• There are a few options at your disposal for
  balancing redox equations.

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Example Problem 2

• OPTIONS
• You can combine the half-reactions before you
  start to cancel out on each side.
• Many people find this easier to do.
• We had the HNO2-1 ion present. When we went to
  work that half-reaction, we left it as HNO2-1.
  We could have removed the H and used NO2-1 to
  work the problem through. Sometimes you get
  another equally valid balanced equation. Both
  are correct and would get full credit.

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Summary of Redox Steps for Basic Medium

• 1. Determine the oxidation states of each atom.
• 2. Write out the oxidized and reduced species
  and set up their respective half-reactions
• 3. Balance the atoms (other than O and H).
• 4. Balance the Oxygen by adding water as needed.
• 5. Balance the Hydrogen by adding H1 as needed.
• 6. Add Hydroxide Ion to each side to make the
  H1 ion become water and add free OH-1 ions
  to the other side.
• 7. Add the number of electrons needed to balance
  the equation according to charge.
• 8. Combine the two half reactions