Transcript OXIDATION REDUCTION REACTIONS

OXIDATION REDUCTION
REACTIONS
Rules for Assigning Oxidation States
The oxidation number corresponds to the number of electrons, e-, that an atom
loses, gains, or appears to use when joining with other atoms in compounds.
When determining the Oxidation State of an atom there are seven guidelines to
follow:
1. The Oxidation State of an individual atom is 0.
2. The total Oxidation State of all atoms in: a neutral species is 0 and in an ion
is equal to the ion charge.
3. Group 1 metals have an Oxidation State of +1 and group 2 an Oxidation State
of +2
4. The Oxidation State of fluorine is -1, when in compounds
5. Hydrogen generally has an Oxidation State of +1 in compounds
6. Oxygen generally has an Oxidation State of -2 in compounds.(EXCEPTION:
the oxygen in peroxide (O22 ) is -1
7. In binary metal compounds, group 17 elements have an Oxidation State of 1, group 16 of -2, and group 15 of -3.
8. (Note: The sum of the oxidation states is equal to zero for neutral
compounds and equal to the charge for polyatomic ion species.)
Example 1: Determine the oxidation state of the bold element in each of the following:
1. Na3PO3
2. H2PO4Answers to Example 1:
1. The oxidation numbers of Na and O are +1 and -2. Since sodium phosphite is neutral,
the sum of the oxidation numbers must be zero.. Letting x be the oxidation number of
phosphorus then, 0= 3(+1) + x + 3(-2). x=oxidation number of P= +3.
2. Hydrogen and oxygen have oxidation numbers of +1 and -2. The ion has a charge of -1,
so the sum of the oxidation numbers must be -1. Letting y be the oxidation number of
phosphorus, -1= y + 2(+1) +4(-2), y= oxidation number of P= +5.
Sample Problems: Determine the oxidation states:
1. Fe(s) + O2(g) → Fe2O3(g)
2. Fe2+
3. Ag(s) + H2S → Ag2S(g) + H2(g)
Solutions
1. Fe and O2 are free elements, therefore they have an O.S. of "0" according to Rule #1.
The product has a total O.S. equal to "0" and following Rule #6, O3 has an O.S. of -2,
which means Fe2 has an O.S. of +2.
2. The O.S. of Fe corresponds to its charge, therefore the O.S. is +2.
3. Ag has an O.S. of 0, H2 has an O.S. of +1 according to Rule #5 and S has an O.S. of -2
according to Rule #7.
OXIDATION REDUCTION REACTIONS
Redox reactions are comprised of two parts, a reduced half and an oxidized half, that
always occur together. The reduced half gains electrons and the oxidation number
decreases, while the oxidized half losses electrons and the oxidation number increases.
OXIDATION
REDUCTION
•Losses electrons
•Oxidation number increases.
•Electrons are shown on the
product side of the half reaction
• The species oxidized is called the
reducing agent
•Gain of electrons
•Oxidation number decreases
•Electrons are shown on the
reactant side of the half reaction
•The species reduced is called
the oxidizing agent
Simple ways to remember this are the mnemonic devices OIL RIG meaning
"oxidation is loss" and "reduction is gain" or LEO says GER meaning "loss of e- =
oxidation" and "gain of e- = reduced."
There is no net change in the number of electrons in a redox
reaction. Those given off in the oxidation half reaction are taken on
by another species in the reduction half reaction.
The ion or molecule that accepts electrons is called the oxidizing
agent; by accepting electrons it brings about the oxidation of
another species.
Conversely, the species that donates electrons is called the
reducing agent; when reaction occurs it reduces the other species.
Note: the oxidizing and reducing agents can be the same element or
compound This will be further discussed under Types of Redox
Reactions: Disproportionation).
Example 2: Determine which element is oxidized and which element is
reduced in the following reactions (be sure to include the oxidation state
of each):
Zn + 2H+ → Zn2+ + H2
2Al + 3Cu2+→2Al3+ +3Cu
CO32- + 2H+→ CO2 + H2O
Answers to Example 2:
Zn is oxidized (Oxidation number: 0 → +2); H+ is reduced (Oxidation
number: +1 → 0)
Al is oxidized (Oxidation number: 0 → +3); Cu2+ is reduced (+2 → 0)
This is not a redox type because each element has the same oxidation
number in both reactants and products: O= -2, H= +1, C= +4.
(For a more in depth look see oxidation numbers).
Oxidizing and Reducing Agents
An atom is oxidized when it oxidation number increases, the reducing
agent, and an atom is reduced when its oxidation number decreases, the
oxidizing agent. In other words, what is oxidized is the reducing agent
and what is reduced is the oxidizing agent. (Note: the oxidizing and
reducing agents can be the same element or compound).
Example 3. Using the equations from the previous
examples determine what is oxidized?
Zn + 2H+ → Zn2+ + H2
Answer:
The O.S. of H goes from +1 to 0 and the O.S. of Zn goes
from 0 to 2+. Hence, Zn is oxidized and acts as the
reducing agent.
Example 4. What is reduced?
Zn + 2H+ → Zn2+ + H2
Answer:
The O.S. of H goes from +1 to 0 and the O.S. of Zn goes
from 0 to 2+. Hence, H+ ion is reduced and acts as the
oxidizing agent.
2Al + 3Cu2+→2Al3+ +3Cu
Al goes from 0 to 3+ so it is oxidized so Al acts as the reducing
agent
Cu goes from 2+ to 0 so it is reduced so Cu2+ acts as the
oxidizing agent
CO32- + 2H+→ CO2 + H2O
As stated in a previous slide, C goes from 4+ to 4+ and H+ goes
from 1+ to 1+ so this is not a redox reaction b/c there is no gain
or loss of electrons.
BALANCING REDOX REACTIONS USING THE HALF REACTION METHOD
Half Reactions
Before one can balance an overall redox equation, one has to be
able to balance two half-equations, one for oxidation
(electron loss) and one for reduction (electron gain).
Collectively, oxidation and reduction are known as redox, or an
electron transfer reaction. After balancing the two halfequations one can determine the total net reaction.
Each equation is balanced by adjusting coefficients and adding
H2O, H+, and e- in this order:
1) Balance the number of atoms of each element.
2) Balance the number of electrons transferred.
3) Balance the total charge on reactants and products
(Note: If #1 and #2 are done correctly, #3 will follow. Thus, it
serves as a means of checking your work).
Balancing in an Acidic Solution
Balancing in acidic solution is similar to balancing in neutral
solutions however, instead of three steps to follow, there are six.
These rules are:
Write and balance the half reactions.
Balance oxygen, O, by adding with H2O
Balance hydrogen, H, by adding H+ (acidic)
Balance charge by adding electrons (you should be adding the
same number of electrons as H+ ions)
Multiply both half reactions by some integer to cancel out
electrons
Add the half reactions together and cancel out what appears on
both sides
Example 5:
Balance the redox reaction in acidic solution: MnO4- + I- --> Mn2+ + I2(s)
Write and balance the half reactions:
MnO4- + I- --> Mn2+ + I2(s)
O.S:
+7
-2
-1
+2
0
(Mn is reduced and I is oxidized)
Write and balance the half reactions.
Oxidation Rx:2I-(aq) --> I2(s)
Reduction Rx: MnO4- + 5e- --> Mn2+
Balance oxygen, O, by adding H2O
Reduction Rx: MnO4- + 5e- --> Mn2+ + 4H2O
Balance hydrogen, H, by adding H+
Reduction Rx: MnO4- + 5e- + 8H+ --> Mn2+ + 4H2O
Balance charge by adding electrons
Oxidation Rx: 2I-(aq) --> I2(s) + 2eReduction Rx: MnO4- + 5e- + 8H+ --> Mn2+ + 4H2O
Multiply both half reactions by some integer to cancel out electrons
(Oxidation Rx: 2I-(aq) --> I2(s) + 2e-) * 5
(Reduction Rx: MnO4- + 5e- + 8H+ --> Mn2+ + 4H2O) *2
Oxidation Rx: 10I-(aq) --> 5I2(s) + 10eReduction Rx: 2MnO4- + 10e- + 16H+ --> 2Mn2+ + 8H2O
Add the half reactions together and cancel out what appears on both sides:
10I-(aq) + 2MnO4-(aq) + 16H+(aq) --> 2Mn2+(aq) + 5I2(s) + 8H2O(l)
(Note: Don't forget the states of matter! Generally, anything with a charge is (aq) and H2O is
(l))
BALANCING REDOX REACTION THAT TAKE PLACE IN A BASIC SOLUTION
FOLLOWS THE SAME STEPS AS IN AN ACIDIC SOLUTION
HOWEVER THERE ARE A FEW MORE EXTRA STEPS
AFTER COMPLETING THE STEPS FOR AN ACID, THE FOLLOWING STEPS ARE
ADDED:
•ADD OH- TO BOTH SIDES TO BALANCE THE H+ PRESENT
•COMBINE THE H+ AND OH- TO FORM WATER
•CANCEL THE WATER MOLECULES APPEARING ON BOTH SIDES
Balance the above redox reaction in basic solution:
Balance the reaction in acidic solution
10I-(aq) + 2MnO4-(aq) + 16H+(aq) --> 2Mn2+(aq) + 5I2(s) + 8H2O(l)
Add the same amount of OH- ions as H+ ions to both sides of the
equation.
10I-(aq) + 2MnO4-(aq) + 16H+(aq) + 16OH- --> 2Mn2+(aq) + 5I2(s) + 8H2O(l) +
16OHOn one side, the OH- and H+ will react to form water (H2O) in a 1:1
ratio.
10I-(aq) + 2MnO4-(aq) + 16H2O --> 2Mn2+(aq) + 5I2(s) + 8H2O(l) + 16OHCancel out the water molecules appearing on both sides
10I-(aq) + 2MnO4-(aq) + 8H2O(l) --> 2Mn2+(aq) + 5I2(s) + 16OH-(aq)