Oxidation-Reduction Reactions
Part I: Oxidation States
Redox Reactions
Involve transfer of 1 or more e-
Convention:
Ionic charges: n- or n+
Oxidation states: –n or +n
Example:
Mg + Pb(NO3)2 Mg(NO3)2 + Pb
AKA: Single Replacement
Oxidation States (Numbers)
Used to keep track of e- Redox Rxns
Like imaginary charges, BUT…
E- transfer has not necessarily
occurred!!!
Let’s look at the ones you did for HW!
Oxidation-Reduction Reactions
Part II: Oxidizing & Reducing
Redox Rxns:
e- transfer can be a real transfer as in
the formation of ionic compounds:
2Na ( s ) 1 Cl 2( g ) 2NaCl ( s )
Oxidation
States: 0 0 +1 -1
But with covalent compounds…
The transfer is “on paper” (ions are not formed)
Example:
1 CH 4(g) 2 O2 ( g ) 1 CO2 ( g ) 2 H 2O( g )
Oxidation -4 Each H 0 +4
Each O Each H
States: is +1 is -2 is +1
Let’s just look at the carbon:
1 CH 4 1 CO2 8 e
-
-4 +4
Account for e- loss by
The carbon has lost 8 e- adding e- to the right
side of the eqn
Oxidation is when an atom loses e-…
We say that atom has been oxidized
It is also the reducing agent!
Oxidation is:
Increase in oxidation state
Atom becomes more positive
Also called an Electron Donor
Look at the oxygen:
8 e 2 O2 1 CO2 2 H 2O
-
0 -2
Account for e-
4 oxygen atoms go from
gain by adding e-
oxidation state 0 to oxidation
to left side of eqn
state -2:
4 * (-2) = gain 8 electrons
Reduction is when an atom gains e-…
It has been reduced
It is the oxidizing agent
So Reduction is:
Decrease in oxidation state
Atom becomes more negative
Also called an Electron Acceptor
Example:
Cu(s) is the reducing agent
Cu(s) Lost 2 electrons
0 +2
Cu(s) is oxidized
2
1 Cu (s) 2 Ag (aq) 2 Ag (s) 1 Cu (aq)
+1 0
Each Ag Gained 1 electron
Ag is reduced
Ag is the oxidizing agent
Oxidation-Reduction Reactions
Part III: Balancing Redox Reactions
Balancing is different for Redox
In addition to balancing atoms, we need
to balance e-
We will use the Half-reaction method
Acidic & Basic solutions…
Are handled differently but you start
balancing the same way
We will look at the acidic case today
Overview (Acidic Solution):
Write separate eqns for Oxidation and Reduction
Then:
Balance elements other than O or H
Balance Oxygen by adding H2O
Balance Hydrogen with H+
Balance the charge with electrons
If necessary, find Least Common Multiple of the
number of electrons for the 2 half reactions &
multiply by a constant or constants
Add the half reactions, cancel identical species
Check: Are elements & charges balanced?
Balancing: Acidic Redox Reactions
Example - Balance the following redox rxn in
acidic solution:
2 3
Cr2O C2 H5OH Cr CO2
7
Oxidation Reaction:
C2 H 5OH CO2
Oxidation
States: -2 +4
Reduction Reaction:
3
Cr2O Cr
2-
7
Oxidation
States: +6 +3
Oxidation Reaction:
3H 2O 1 C22H 55OH 2 CO22 12H 12e
C H OH CO
Oxidation +4
States: -2
hydrogen with H+ 2O
oxygen with H
charge
Balance the carbon with electrons
Reduction Reaction:
6 e 14H 1 Cr22O77-- 2 Cr 3 7 H 2O
-
Cr O 22
Oxidation
States: +6 +3
Balance all elements H+ from
Balance Hydrogen with aside2O
Balance the oxygen with H
oxygen and hydrogen
Balance the charge with electrons
Balance the electrons…
The reduction equation has 6 electrons
on the left
The oxidation equation has 12 electrons
on the right
Least common multiple: 12
So…
2 3
6 e 14 H 1 Cr2O 2 Cr 7 H 2O
7
Multiply the reduction equation by 2 to get:
2 3
12 e 28 H 2 Cr2O 4 Cr 14 H 2O
7
Add the half-reactions:
16 11
12e 28H 2Cr2O7 4Cr 3 14H 2O
2
1 2 H5OH 3H 2O 2CO2 12H 12e
C
16 H 2 Cr2O7 1 C2 H5OH 4 Cr 3 11 H 2O 2 CO2
2
Are elements & charges balanced?
16H(aq) 2Cr2O72(aq) C2 H5OH(l ) 4Cr(3 ) 11H2O(l ) 2CO2( g )
aq
16(1) 2(2) 0 12 4(3) 0 0 12
22 H 22 H
4 Cr Atoms Charges 4 Cr
Balanced!
15 O Balance! 15 O
2C 2C
For Basic solution…
Balance as above
To both sides, add a number of OH
One side will have both H & OH
Change those to an equal number of H2O
Cancel H2O on opposite sides
Check for atom/charge balance
Let’s revise our example for basic
solution:
5
2 2 2 3 3
16 16 aq ( 2HCr2O7 )2C22O57OH (l ) ) 45OH() 11H Oaq
CrH 5OH 2 H Cr( aq 4Cr( (
16OH H 2 O(H) aq ) 5Cr2O(7( aq()aq C2 H( aq) (lC 4Cr( aql) 11H 22O(ll)) 2CO2( g ) 16OH
22 Oaq) 3
Now…combine H+ & OH- on
How H2O on both sides
Cancel many H+ do we have? 16-So we will
The same side to form H2O
add 16 OH- to
each side