Transcript Electrochemistry - OCVTS.org | Ocean County Vocational

Chapter 19
Oxidation - Reduction
Reactions
19.1 Oxidation and Reduction
Oxidation – Reduction
• Rules for assigning oxidation numbers
1. Most substances have the same oxidation
number as their individual charge
(the more electronegative element 1st)
2. All oxidation numbers in a compound must add
up to equal the total charge on the compound
3. All single elements have a oxidation number of
zero
4. All single ions have the same oxidation number
as their charge
5. Oxygen has a charge of -2, except with F (+2),
or peroxides (-1)
6. Hydrogen is usually +1, unless with a metal (-1)
Oxidation
Oxidation
HalfReaction
Fe → Fe+2 + 2e-
• Reactions in which
the atoms or ions of
an element
experience an
increase in oxidation
state
• A species whose
oxidation number
increases is oxidized
Reduction
Reduction
HalfReaction
• Reactions in which
the oxidation state of
an element
decreases.
• A species that
undergoes a
decrease in oxidation
Cu+2 + 2e- → Cu
state is reduced.
Cu+2 + 2e- → Cu
+
Overall Rxn
Fe → Fe+2 + 2e-
Fe + Cu+2 → Fe+2 + Cu
Oxidation - Reduction
+1
-1
0
0
+1
-1
AgCl(aq) + Na(s)  Ag(s) + NaCl(aq)
Charge
reduced
from +1 to 0,
so e- were
gained
The silver in
silver
chloride was
reduced
The sodium
was oxidized
OIL RIG
Charge
increased
from 0 to +1,
so e- were
lost
Oxidation is
loss,
reduction is
gain
Leo the Lion goes Ger
Lose
Electrons
Oxidation
Gain
Electrons
Reduction
Electron Loss Means Oxidation
Oxidation - Reduction
+1
-2
0
0
H2O(l)  H2(g) + O2(g)
Charge
reduced
from +1 to 0,
so e- were
gained
Charge
increased
from -2 to 0,
so e- were
lost
The
hydrogen in
water was
reduced
The oxygen
in water was
oxidized
Example
• Household Bleach removes stains through a redox reaction:
Stain molecules (s) + OCl- (aq) → colorless molecules (s) + Cl- (aq)
• Determine the oxidation numbers of oxygen & chlorine in
OCl- .
OCl-2 +1
Chapter 19
Oxidation - Reduction
Reactions
19.2 Balancing Redox
Equations
Example
• Balance the following reaction:
I- + MnO4- + H+ → MnO2 + I2 + H2O
2I- + MnO4- + 4H+ → MnO2 + I2 + 2H2O
But this reaction is balanced for mass not charge!
A half-reaction system has to be used to balance
for charge.
Half-Reaction Method
1. Write the formula equation then ionic equation
2. Assign oxidation numbers. Exclude anything with an
ox. # of zero, or that doesn’t change ox. #
3. Write the ½ rxn for oxidation
4. Balance the atoms
5. Balance the charge (w/ electrons)
6. Write the ½ rxn for reduction
7. Balance the atoms
8. Balance the charge (w/ electrons)
9. Use coefficients to ensure the # of e- lost in ox. equals
the # of e- gained in red.
10.Combine both ½ rxns and cancel (like Hess’s Law)
11.Combine ions to form initial compounds.
Example
Now try to balance the following reaction:
I- + MnO4- + H+ → MnO2 + I2 + H2O
Half-Reactions:
2MnO4- + 8H+ + 6e- → 2MnO2 + 4H2O
6I- → 3I2 + 6eOverall Balanced Equation:
6KI + 2KMnO4 + 8HCl → 2MnO2 + 3I2 + 4H2O + 8KCl
Chapter 19
Oxidation - Reduction
Reactions
19.3 Oxidizing and Reducing
Agents
Reducing Agent
Fe + Cu+2 → Fe+2 + Cu
Fe → Fe+2 + 2e-
• Substance that has
the potential to
cause another
substance to be
reduced.
• They lose electrons;
are oxidized.
Iron causes Copper to
become reduced, so it is
the Reducing Agent
Oxidizing Agent
Fe + Cu+2 → Fe+2 + Cu
Cu+2 + 2e- → Cu
• Substance that has
the potential to
cause another
substance to be
oxidized.
• They gain electrons;
are reduced.
Copper causes Iron to
become oxidized, so it is
the Oxidizing Agent
Disproportionation/
Autooxidation
• A process by which a substance acts as
both an oxidizing and reducing agent