Transcript Half Equations

Title: Lesson 2 Redox Equations
Learning Objectives:
– Deduce simple half-equations
– Combine half-equations to form full equations
– Conduct a series of redox reactions
– Use H+ and H2O to balance redox equations
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Fertilizers may cause health problems for babies because nitrates can change
into nitrites in water used for drinking.
a)
Define oxidation in terms of oxidation numbers.
b)
Deduce the oxidation states of nitrogen in the nitrate, NO3–, and nitrite, NO2–, ions.
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Half-equations
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Half equations show the changes to individual species in a redox reaction.
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Fe2O3 + 2 Al  2 Fe + Al2O3
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Fe3+ + 3 e-  Fe ….this is the reduction
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Al  Al3+ + 3 e- ….this is the oxidation
A wide variety of half equations can be found in the data booklet
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Combining half equations – example
Chlorine oxidizes iron(II) to iron(III) and is itself reduced to
chloride ions. Write a balanced equation for this reaction.
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Step 1: Write the half equations.
Eqn. A
Cl2(aq) + 2e-  2Cl-(aq)
chlorine has been reduced
Eqn. B
Fe2+(aq)  Fe3+(aq) + e-
iron(II) has been oxidized
Step 2: Eqn. A involves 2 electrons and Eqn. B involves 1
electron, so multiply both sides of Eqn. B by two.
2Fe2+(aq)  2Fe3+(aq) + 2e-
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Step 3: Add the half equations and cancel the electrons.
Cl2(aq) + 2Fe2+(aq)  2Cl-(aq) + 2Fe3+(aq)
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Combining half equations
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Combining Half Equations and Balancing Redox Reactions
– Example 1
Step
Example 1: Reaction of iodine with
copper
Write out half-equations side by side and the number
of electrons lost or gained…can be found in the data
booklet
½ I2 + e-  I- … 1 e- gained
Cu  Cu2+ + 2 e- …2 e- lost
Combine them to form a single reaction, multiplying
each half-equation in order to balance the electrons
gained/lost
I2 + 2 e- + Cu  2 I- + Cu2+ + 2 e-
Ensure all atoms other than O and H balance
I2 + 2 e- + Cu  2 I- + Cu2+ + 2 e- no change needed
Cancel out any electrons that are duplicated on both
sides
I2 + Cu  2 I- + Cu2+
Balance oxygen atoms by adding H2O to the side that
needs extra
I2 + Cu  2 I- + Cu2+
Balance the hydrogens by adding H+ to the side that
needs extra
I2 + Cu  2 I- + Cu2+
- no change needed
Count the total charge on each side and add enough
electrons to the more positive side to balance it
I2 + Cu  2 I- + Cu2+
- no change needed
- the iodine ½-equation is doubled to make
. 2 e- gained, the other remains unchanged
- 2 e- appear on each side so are cancelled out
- no change needed
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Combining Half Equations and Balancing Redox Reactions
– Example 2
Step
Example 2: Reaction of nickel with manganate
ions
Write out half-equations side by side
and the number of electrons lost or
gained
MnO4- +5e-  Mn2+ …5 e- gained by Mn, so 2 lots needed
Ni  Ni2+ + 2 e- … 2 e- lost by Ni, so 5 lots needed
Combine them to form a single reaction
2MnO4- + 5Ni + 10e- 2Mn2+ + 5Ni2+ + 10e•Ni ½-equation multiplied by 5
•MnO4- ½-equation multiplied by 2
Ensure all atoms other than O and H
balance
2MnO4- + 5Ni + 10e- 2Mn2+ + 5Ni2+ + 10e•No change needed
Cancel out any electrons that are
duplicated on both sides
2MnO4- + 5Ni  2Mn2+ + 5Ni2+
•10 e- cancelled
Balance oxygen atoms by adding H2O to
the side that needs extra
2MnO4- + 5Ni  2Mn2+ + 5Ni2+ + 8H2O
•8H2O on the right balance the 8 O on the left
Balance the hydrogens by adding H+ to
the side that needs extra
2MnO4- + 5Ni + 16H+  2Mn2+ + 5Ni2+ + 8H2O
•16H+ added on left to balance 16 H on the right
Count the total charge on each side and
add electrons to the more positive side
to balance it
2MnO4- + 5Ni + 16H+  2Mn2+ + 5Ni2+ + 8H2O
•Total charge balances so no change needed’
•This step only needed on half-equations
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Use the worked example to help
you solve the following problems
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Solutions
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What are oxidizing and reducing agents?
Oxidizing agents…
Reducing agents…
…oxidize other species
…reduce other species
…accept electrons
…donate electrons
…are themselves reduced
…are themselves oxidized
For example, in the reaction below:
2NaCl + F2  2NaF + Cl2
Fluorine:
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oxidizes Cl- (to chlorine gas)
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accepts electrons (from Cl-)
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is reduced (to F-)
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is an oxidizing agent
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Oxidising and Reducing Agents
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The reactant that accepts electrons is called an oxidizing agent (it brings about
oxidation of the other reactant).  oxidant
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The reactant that supplies the electrons is called the reducing agent (it brings
about the reduction and itself becomes oxidized)  reductant
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E.g. Iron is extracted from its ore:
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The reducing agent C brings about the reduction of Fe (+3 0), whilst its oxidized
to CO (0  +2).
The oxidising agent Fe2O3 brings about the oxidation of C, whilst itself is reduced
to Fe.
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Oxidizing and reducing agents
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Common oxidizing agents:
Common reducing agents:
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concentrated sulfuric acid
(H2SO4)
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hydrogen (H2)
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potassium manganate(VII)
(KMnO4)
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zinc (Zn)
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carbon (C)
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carbon monoxide (CO)
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lithium tetrahydridoaluminate(III), (LiAlH4)
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sodium tetrahydridoborate(III) (NaBH4)
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potassium dichromate(VI)
(K2Cr2O7)
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manganese(IV) oxide
(MnO2)

chlorine (Cl2)

hydrogen peroxide (H2O2)
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Spot the agent
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Microscale Redox Reactions
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Complete the experiment here, on microscale redox
reactions
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Rather than doing it on a plastic sheet, use a dropping tile
For each change you observe, you should write a balanced
redox equation to describe it
Once you finish, you should practice balancing the redox
equations on the following slide
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Produce balanced redox equations for the reactions of:
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Bromine with sodium
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Copper (II) oxide with hydrogen
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Cu2+ + 2e-  Cu
½ H 2  H + + e-
Aluminium reacting with chromate ions
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½ Br2 + e-  BrNa Na+ + e-
Al  Al3+ + 3eCr2O72- + 6e- 2Cr3+
Iron (II) chloride reacting with manganate ions
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Fe2+  Fe3+ + eMnO4- + 5 e- Mn2+
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Key Points
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Half-equations show the changes to each species in a redox reaction
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To combine half-equations into a full equation
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Multiply each such that the electron-transfers balance
Add H2O to balance O
Add H+ to balance H
Add e- to balance charge
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