Redox Reactions
Half Reactions
Redox reactions can be split into two halves, oxidation
and reduction.
Redox reactions take place when a species (molecule,
ion or atom) that can give up an electron (reducing
agent) comes into contact with a species that can
accept an electron (oxidizing agent).
e.g. 2 Fe + 3 Cl2 → 2 FeCl3
Each iron is oxidized by losing 3 e- . This is the the
oxidation half of the reaction:
Active metals can be oxidized by many oxidizing
agents. An example is when iron rusts:
Fe → Fe3+ + 3 e-
4 Fe + 3 O2 → 2 Fe2O3
oxidation
At the same time each Cl atom in Cl2 is reduced by
accepting one e- becoming Cl- . This is the reduction
half:
Iron is oxidized by oxygen (also iron reduces oxygen).
Cl2 + 2 e- → 2 Cl-
reduction
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Balancing Redox Equations
Using Half Reactions
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Balancing Redox Equations
Using Half Reactions
Fe (s) + CuSO4 (aq) → Cu (s) + Fe2(SO4)3 (aq)
The following reaction is the unbalanced reaction of
what happens when you put an iron nail into a
copper(II) sulfate solution:
Steps for balancing using half reactions:
1. Net Ionic Equations
Fe (s) + CuSO4 (aq) → Cu (s) + Fe2(SO4)3 (aq)
2. Redox Half Reactions
Iron atoms are oxidized as they lose electrons to the
Cu2+ ions.
3. Balance the atoms and charges
4. Adjust coefficients (so that e- lost cancel e- gained)
5. Combine half reactions and return spectator ions
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Balancing Redox Equations
Using Half Reactions
Balancing Redox Equations
Using Half Reactions
Fe (s) + CuSO4 (aq) → Cu (s) + Fe2(SO4)3 (aq)
3. Balance the atoms and charges
Steps for balancing using half reactions:
2 Fe → 2 Fe3+ + 6 e-
1. Net Ionic Equations
Cu2+ + 2 e- → Cu
Fe + Cu2+ + SO42- → Cu + 2 Fe3+ + 3 SO42-
4. Adjust coefficients
Fe + Cu2+ → Cu + 2 Fe3+
2 Fe → 2 Fe3+ + 6 e-
2. Redox Half Reactions
Fe → 2 Fe3+ + 6 e-
oxidation
Cu2+ + 2 e- → Cu
reduction
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3(Cu2+ + 2 e- → Cu) to give: 3 Cu2+ + 6 e- → 3 Cu
5. Combine half reactions and return spectator ions
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Balancing Redox Equations
Using Half Reactions
Balancing Redox Equations
Half-Reactions Method - Acidic
Before balancing split the original equation into two half
reactions, one “reduction” and the other “oxidation”.
In each half-reaction, follow these steps:
1. Balance all elements except “H” and “O”.
2. Balance the “O’s” by adding water, H2O.
5. Combine half reactions
2 Fe + 3 Cu2+ → 3 Cu + 2 Fe3+
3. Balance the “H’s” by adding hydrogen ions, H+.
4. Balance the electric charge by adding electrons, e-.
5. Multiply the two equations by appropriate coefficients
to make the # of electrons in the equations equal.
6. Re-combine the two equations, canceling if needed.
... and return spectator ions.
2 Fe (s) + 3 CuSO4 (aq) → 3 Cu (s) + Fe2(SO4)3 (aq)
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Balancing with Half-Reactions - Example
What if the solution was basic?
Fe2+ + Cr2O72- -> Fe3+ + Cr3+
Notice that the method has assumed the solution was
acidic H+ was added to balance the equation.
Fe2+
->
Fe3+
Cr2O7 ->
2-
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Fe2+ -> Fe3+ + e-
Cr2O72- -> 2 Cr3+
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For this reason, add enough OH- ions to both sides of
the equation to neutralize the H+ in the overall
reaction.
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Cr2O72- -> 2 Cr3+ + 7 H2O
6 Fe2+ -> 6 Fe3+ + 6 e-
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The [H+] in a basic solution is very small. The [OH-] is
much greater.
Cr3+
Cr2O72- + 14 H+ -> 2 Cr3+ + 7 H2O
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The hydrogen and hydroxide ions will combine to make
water, and you may have to do some canceling before
you’re done.
Cr2O72- + 14 H+ + 6 e- -> 2 Cr3+ + 7 H2O
Cr2O72- + 6 Fe2+ + 7 H2O -> 2 Cr3+ + 6 Fe3+ + 14 OH-
Cr2O72- + 6 Fe2+ + 14 H+ -> 2 Cr3+ + 6 Fe3+ + 7 H2O
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Balancing Redox Equations Practice
•
Balance in acidic solution:
H2C2O4 + MnO4- -> Mn2+ + CO2
5 H2C2O4 + 2 MnO4- + 6 H+ -> 2 Mn2+ + 10 CO2 + 8 H2O
•
Balance in basic solution:
CN- + MnO4- -> CNO- + MnO2
3 CN- + 2 MnO4- + H2O -> 3 CNO- + 2 MnO2 + 2 OH-
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