Curriculum Standards
• National: Content Standard B: Physical Science: Chemical
reactions
– A large number of important reactions involve the transfer
of either electrons (oxidation/reduction reactions) or
hydrogen ions (acid/base reactions) between reacting ions,
molecules, or atoms. (no related standard)
• Florida: SC.912.P.8.10 Describe oxidation-reduction reactions
in living and non-living systems.
• Florida: SC.912.P.8.8 Characterize types of reactions, for
example: redox, acid-base, synthesis, and single and double
replacement reactions.
Curriculum Objectives for 7.06H
• Objectives (includes higher-order thinking):
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Review 7.06 regular concepts
Define oxidation and reduction
Apply rules for oxidation numbers
Assign oxidation numbers to elements in a reaction
Create algebra expressions to solve for an unknown
oxidation number
– Identify oxidation and reduction half reactions
– Balance oxidation and reduction half reactions using
eight steps
Vocabulary (Literacy Skill)
• What is an oxidation and reduction (redox)
reaction? An exchange of ___________.
• Real life example:
– Rusting: This is a redox reaction in which oxygen
oxidizes (takes electrons from) iron to form iron (III)
oxide, otherwise known as rust.
4Fe (s) + 3O2 (g) → 2Fe2O3 (s)
– Iron is a very strong metal, whereas rust is a brittle
ionic compound that flakes off of old cars and nails.
What a difference a few electrons and bonds can
make!
Vocabulary (Literacy Skill)
• Oxidation---Lose electrons
• Reduction---Gain electrons
Way to remember:
“LEO the lion says GER”
Lose electrons is Oxidation (LEO)
Gain electrons is Reduction (GER)
Vocabulary (Literacy Skill)
• For the “agents” think of the opposite result.
• Oxidizing Agent: A reactant that causes
another substance to be oxidized (not being
oxidized itself). The oxidizing agent is the
reactant that is reduced (gain electrons).
• Reducing Agent: A reactant that causes
another substance to be reduced (not being
reduced itself). The reducing agent is the
reactant that is oxidized (lose electrons).
Scaffolding:
Breaking directions/calculations into
easier steps (Literacy Skill)
• This is a challenging eight-step calculation. However,
with this tutorial, you will rock this calculation!
• I will break the calculation into easier steps and discuss
the directions for each step.
• You will see examples for each step.
• Please take notes, pause, rewind, etc.
• You might want to watch this tutorial twice, complete
several practice problems yourself in the lesson, and
attend tutoring for this concept.
• Let’s get started. 
Learn and apply the rules for Oxidation
Numbers:
Neutral elements in their standard state (alone) have a charge of zero
(diatomic elements such as H2, N2, O2, F2, Cl2, Br2, I2; solid Na, etc.)
Oxygen has an oxidation number of -2 except in a peroxide which is -1
(H2O2 exception)
Ions alone (has a charge +/-) or in a compound will use their charge
from 3.05). NaCl (Na+1 and Cl-1)
For covalent compounds, pretend the compound is ionic with the
more electronegative element (top right) forming the negative ion
(anion). For example: Fluorine is always -1 in a compound, oxygen is
almost always -2, and hydrogen is +1 in covalent compounds.
The algebraic sum of all the oxidation numbers (multiplied by any
subscripts) must add up to the total charge of the compound or ion.
The element that gains electrons will be reduced and the element that
loses electrons will be oxidized.
Review of ion charges from 3.05
Balancing redox equations (8 total steps):
1. Identify the element that is oxidized and the
element that is reduced (example from the lesson)
What changes oxidation numbers? What gains
electrons and what loses electrons?
Oxygen stays the same oxidation number and
hydrogen stays the same (from the rules).
Therefore, Chromium and Nitrogen are going to
be our focus (one is oxidized and one is
reduced).
Cr2O72-(aq) + HNO2 (aq) → Cr3+ (aq) + NO3 - (aq)
2. Split the reaction into half reactions and
determine oxidation numbers
Break up into half reactions (similar
elements on each side).
Cr2O72-(aq) + HNO2 (aq) → Cr3+ (aq) + NO3 - (aq)
Cr2O72- → Cr3+
HNO2 → NO3 -
2. Split the reaction into half reactions and
determine oxidation numbers
• Identify oxidation numbers (use the rules and create an algebra expression
to solve for the unknown). Do the first half reaction.
Cr2O72- → Cr3+
We know oxygen is -2 and the overall charge on the left is -2 so create an
algebra expression to solve for the charge of Cr on the left. Cr is our
variable (we need to find out the oxidation number of Cr) and we have a
subscript of 2 so 2x is the variable on the left.
2x – 14 = -2
2x = 14 – 2
2x = 12
X = +6 (Cr is +6 on the left)
On the right, Cr is already shown as a +3.
Therefore, Cr goes from a +6 to a +3.
Electrons are NEGATIVE. Did we GAIN or lose electrons?
Is this first half reaction Oxidation or REDUCATION?
2. Split the reaction into half reactions and
determine oxidation numbers
• Now, do the second half reaction. Identify oxidation numbers (use the
rules and create an algebra expression to solve for the unknown).
HNO2 → NO3 –
We know oxygen is -2 and the overall charge on the left is zero. On the right,
the overall charge is -1. Hydrogen is +1.
Nitrogen is the variable.
Left calculation:
Right calculation:
+1 + x – 4 = 0
x – 6 = -1
X = +3 on left
x = +5 on right
Therefore, nitrogen goes from +3 to +5.
Electrons are NEGATIVE. The oxidation number became more
positive.
Did we gain or lose electrons?
Is this second half reaction oxidation or reduction?
3. Balance elements except hydrogen and
oxygen by using coefficients
Cr2O72− → 2Cr3+ (balanced Cr)
HNO2 → NO3 − (N is already balanced)
4. Balance any oxygen atoms by adding H2O to the
other side (you are allowed to do this because there is
plenty of water available in the aqueous solution).
Cr2O72− → 2Cr3++ 7H2O
This half reaction needs seven oxygen atoms on
the right, so we add seven H2O molecules to the
right.
HNO2 + H2O → NO3 −
You have three oxygen atoms on the right and
two on the left so this half reaction needs one
more oxygen atom on the left (add one H2O
molecule to the left).
5. Balance hydrogen by adding H+ to the other side
(you are allowed to do this because the solution is
acidic, so there are H+ ions available).
Cr2O72− + 14H+ → 2Cr3++ 7H2O
This half reaction needs 14 hydrogen atoms on
the left to balance the 14 hydrogen atoms in the
seven H2O molecules, so we add 14 H+ ions to
the left.
HNO2 + H2O → NO3 − + 3H+
This half reaction needs three hydrogen atoms on
the right to balance the three hydrogen atoms on
the left, so we add three H+ ions to the right.
6. Balance the total charges on each side by adding
electrons to the side with the higher charge to make
the charges equal.
Cr2O72− + 14H+ → 2Cr3++ 7H2O
-2 + 14 → +6
+12 → +6
Electrons are NEGATIVE.
Need 6 electrons on the left (6e-).
6e− + Cr2O72− + 14H+ → 2Cr3++ 7H2O
HNO2 + H2O → NO3 − + 3H+
0 = -1 + 3
0 = +2
Need 2 electrons on the right (2e-).
HNO2 + H2O → NO3 − + 3H+ + 2e−
7. Balance the electrons
Make the electrons equal in both half reactions.
6e− + Cr2O72− + 14H+ → 2Cr3++ 7H2O
HNO2 + H2O → NO3 − + 3H+ + 2e−
When multiplying a half reaction by a number, that number
gets multiplied by every coefficient in that half reaction.
The first half reaction has 6 electrons and the second half
reaction has 3 electrons so multiply EVERYTHING in the
second half reaction by 3.
• 6e− + Cr2O72− + 14H+ → 2Cr3++ 7H2O
3(HNO2 + H2O → NO3 − + 3H+ + 2e−) equals
3HNO2 + 3H2O → 3NO3 − + 9H+ + 6e−
8. Combine the two half reactions and
simplify
• Put both of the left half reactions together on the left and both of
the right half reactions on the right.
6e− + Cr2O72− + 14H+ → 2Cr3++ 7H2O
3HNO2 + 3H2O → 3NO3 − + 9H+ + 6e−
6e− + Cr2O72- + 3HNO2 + 3H2O + 14H+ → 2Cr3+ + 3NO3- + 9H+ + 7H2O + 6eNow, time to simplify to get the final answer.
• The electrons simplify (here they cancel each other out).
• The three H2O on the left cancel three of the seven H2O to yield
four H2O on the right of the final equation.
• The nine H+ on the right cancel nine of the 14 H+ on the left leaving
five H+ on the left of the final equation.
FINAL ANSWER (SHOW THE CHARGES)
Cr2O72- + 3HNO2 + 5H+ → 2Cr3+ + 3NO3- + 4H2O
Tips for the last step
– If you have equal amounts of the same substance on
both sides of the balanced equation (like you will for
the electrons), they cancel out and do not get written
in the final equation.
– If the same substance is on both sides of the balanced
equation with different coefficients, you can reduce
them by subtracting the lower number from both
(one side will cancel, the other will reduce).
– If the same substance appears more than once on the
same side of the equation, you can combine them by
adding the coefficients together.
Summary and review of eight-step calculation to
balance redox equations (Literacy Skill)
• 1. Identify the element that is oxidized and the element that is reduced
(example from the lesson).
• 2. Split the reaction into half reactions and determine oxidation numbers
• 3. Balance elements except hydrogen and oxygen by using coefficients
• 4. Balance any oxygen atoms by adding H2O to the other side (you are
allowed to do this because there is plenty of water available in the
aqueous solution).
• 5. Balance hydrogen by adding H+ to the other side (you are allowed to do
this because the solution is acidic, so there are H+ ions available).
• 6. Balance the total charges on each side by adding electrons to the side
with the higher charge to make the charges equal.
• 7. Balance the electrons
• 8. Combine the two half reactions and simplify
You are done! 
Woohoo!
• Now it is your turn to complete the practice problems in
the 7.06H lesson before submitting the 7.06H assessment.
• Feel free to watch this tutorial again and attend tutoring.
Remember, you may resubmit 7.06H until you master this
topic.
• Please review this tutorial for the module 7 honors exam
and the final exam.
• Have a wonderful day!