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Chapter 19: Redox Reactions
Redox Reactions
Redox is a contraction of the terms reduction and oxidation and these reactions are often called
oxidation – reduction reactions
Redox reactions occur when electrons are transferred (or partially transferred as in the formation of
polar molecules)
Electrons follow the Law of Conservation of Matter in that they are neither created nor destroyed
during redox reactions
Oxidation is the loss of electrons
Reduction is the gain of electrons
Mnemonic: The charge of any atom or ion that gains electrons will be reduced
Example: the formation of water from its elements
2 H2 (g) + O2 (g) → 2 H2O (g)
Note that on the left side of the reaction the oxidation numbers of both the H and the O is zero
[The oxidation number of any free element is zero]
On the product side of the reaction:
H has an oxidation state of +1
O has an oxidation state of –2
The reaction could be rewritten to show all the oxidation states:
2 H20 (g) + O02 (g) → 2 H2+ O−2
(g)
Note that it is very common to use + instead of +1 (also – instead of –1)
Half reactions
Chemists often split redox reactions by separating the oxidation and reductions halves
Example: 2 H20 (g) + O02 (g) → 2 H2+ O−2
(g)
Oxidation (loss of elecrtons):
2 H20 → 2 H2+1 + 4 e−
Reduction (gain of elecrtons):
O02 + 4 e− → 2 O−2
Note that:
The number of electrons gained = the number of electrons lost
The sum of the two half reactions is the overall reaction and that the electrons cancel out
Using half-reactions to balance equations using Fe + CuSO4 → Fe2(SO4)3 + Cu as an example
Write the net ionic equation for the reaction
0
3+
Fe0(s) + Cu2+
(aq) → Cu(s) + 2 Fe(aq)
Write the oxidation and reduction half-reactions
−
Fe0(s) → 2 Fe3+
(aq) + 6e
0
−
Cu2+
(aq) + 2e → Cu(s)
Balance atoms and charges in each of the half-reactions
−
2 Fe0(s) → 2 Fe3+
(aq) + 6e
0
−
Cu2+
(aq) + 2e → Cu(s)
Adjust coefficients so that the number of e– gained = number of e– lost
−
2 Fe0(s) → 2 Fe3+
(aq) + 6e
0
−
3 Cu2+
(aq) + 6e → 3 Cu(s)
Add the two half-reactions to obtain the overall reaction and return the spectator ions
2 Fe(s) + 3 CuSO4 (aq) → 3 Cu(s) + Fe2 (SO4 )3 (aq)
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Ion – electron method of balancing redox reactions
3+
−
using Cr2 O2−
7 (aq) + I(aq) → Cr(aq) + I2 (s) in acidic solution as an example
Cr2+6 O−2
7 (aq)
2−
+3
1−
+ I(aq)
→ Cr(aq)
+ I20 (s)
Step 1: write oxidation numbers
Step 2: connect changed ox. no.s
Step 3: show half-reactions
Step 4: balance e–
Step 5: distribute coefficients
Balancing using H+
3+
−
Cr2 O2−
7 (aq) + I(aq) → Cr(aq) + I2 (s)
7
O
Step 6: count the O atoms
0
3+
−
Cr2 O2−
7 (aq) + I(aq) → Cr(aq) + I2 (s) + 7 H2 O(𝑙)
Step 7: balance O using water
+
3+
−
14 H(aq)
+ Cr2 O2−
7 (aq) + I(aq) → Cr(aq) + I2 (s) + 7 H2 O(𝑙)
Step 8: balance H using H+
Balancing using OH– starting at step 6 for a different reaction:
Step 6: count the O atoms
−
−
N2 O (g) + 2 ClO−
(aq) → 2 NO2 (aq) + 2 Cl2 (aq)
3
O
4
Step 7: balance O by adding twice as many OH– as needed on the deficient side
−
−
−
N2 O (g) + 2 ClO−
(aq) + 2OH(aq) → 2 NO2 (aq) + 2 Cl2 (aq)
Step 8: now finish balance O (and H) by adding water on the other side
−
−
−
N2 O (g) + 2 ClO−
(aq) + 2OH(aq) → 2 NO2 (aq) + 2 Cl2 (aq) + H2 O(𝑙)
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