Reduction Reactions (Redox
midation
-
-
-
Transfer of
electrons/electron
Oxygen
Is
Losing
For covalent
compounds
-
density from one atom to
Reduction Is Gaining
,
another
(elections)
:
transfer
-no actual
find agents via election density
-
-
elections
of
treated
there is
as if
transfer
atom with
to
greater electronegativity
-
-
-
-
-
Oxidizing
Reducing agent
Agents
Only
be
can
can
Even
contains atoms
being
contains
being oxidized
-
-
atoms
single atoms
specific
one
Agents
-
-
agent
,
,
is
have multiple atoms,
one
though other atoms in agent
-
oxidized/reduced
of
which
actually stimulates
don't affect reation
For union : actual
they
charge
reaction
are
still
of
ion
mentioned
↳ number after charge (e g 2t)
reducing
pagent ↑
polyatomic ions
molecules
atom/element
Oxidation Numbers/States
oxidizing
reduced
.
.
agent
-
8
+4
2
F 2 Fe 30
-
+
W
↓
reduced
W
oxidized
For an evalent bond
:
↳
-
number
after
hypothetical charge
if
reaction happened
charge le g + 2)
.
.
then
...
1
Naturally
.
occuring elements oxidation number O
2
. Simples mon atomic ion Oxidation number
. Sum of alloxidation
3
(polyatomic) ion
In
4 Oxidation
.
oxygen
in peroxides (e g H202)
=
.
.
when uncombined (8)
S
. Oxidation number of
when uncombined (H)
=
6 Group I metals
E
.
.
Group &metals
Tonic
-
rules
>
2
Lovalent > rules
-
Fez8 3
+3
=
↓
2
2
-
+
0
charge
=
on
ion
-2
rule
(very rare)
rule
to
pure element refer
,
,
Be
Rb (s
Fri)
My Ca Sr Ba Ra)
Li Nu K
,
,
,
,
,
=
,
,
,
1
+
=
+ 2
=
-2
,
maybe
3, 4, 5
Fez8 3 + 3(8
-
=
1
3 <
and
molecule
+
=
-1
0
numbers
to
pure element refer
,
=
a
(very rare)
hydrogen
.
=
-1
0
=
in hydrides (e g. Nut)
7
of all oxidation
sum
,
of
number
of atoms in
numbers
(S2-2)
charge on ion
=
,
3(8
↓ +2
-
Fez83 + 320
-
>
2Fe + 316
I
2
-
>
-
>
2
2Fe + 316
I
2
-
+
4
-
2
[Fe + 310 >
Oxygen
Fe = 0
(naturally occurring)
Balance the rest
accordingly
f
+3
-
reducing
oxidizing
agent
agent
-
+3
1
oxidized
+ 2
not
+ 1
2
+6
-
2
+
-
2
Y
reduced
+ 1
2
-
W
agent
-
Y
reduced
-
Oxidizing
agent
O
-
+ 2
f
3 +
reducing
+2
I
oxidized
+5
-
2
+2
+6
-
2
redox
-
Redox Equations
Balancing
Oxidation Number Method :
DAssign Oxidation Number to all
+ +5
-
+ 1
2
-
2
+ 1
② Identify atoms that
+1 + 5
-
+ 1
2
-
2
-
2
+2
-
Atoms
28
change oxidation number
+ 1
-
+2
2
-
28
=>
=>
-
&
J
Y
③ Balance
+1 + 5
-
+ 1
2
-
atoms that
2
+ 1
-
change oxidation
+2
2
-
28
already
>
-
-
&
number
Ja
balanced
① Determine total contribution of oxidation numbers by each type of atom
loxidation number) x (# of atoms) x (# of molecules)
+1 + 5
-
+ 1
2
-
2
+ 1
-
2
-
28
=>
&
&
+2
1
=>
Sx(x)
=
Ja
2x(x) = 2
5
-
2x(x)
=
-
2
8x(x)
=
g
⑤ Use LCM
+1 + 5
-
of
+ 1
2
gained and
elections
2
-
+ 1
-
+2
2
-
last
make them
equal
28
=>
=>
to
=>
-
⑮
↓+ S
↳
5
X
=
=
3
2
+ x
-
2
X
3
+ 1
2
of
2
2
+ 1
-
=
8
-
2
and
-
x
lost
3
+2
2
3
2
+
=
2
gained
LCM
+1 + 5
-
J
=
-
6
28
2
3
⑤
>
=>
+ S
2e- xB]
-
2
=
Ge lost
J
2
-
⑧ If occurring in acidic/basic solutions balance O
like
half-reaction
in
method
⑦ Balance remaining electrons via inspection
+1 + 5
-
+ 1
2
-
2
+ 1
-
+2
2
3
2
-
2
28
3
2
=>
⑤
&
=>
-
+ S
2e- xB]
-
2
=
Ge lost
J
and
H
-
3
2 + )
=
+6
2
-
+ 1 +6
2
-
+3
2
+6
-
2
+
3
2
3
3x2
- x3
Entgains
2x6 = + 12
-
-
2x3 =
-
6
-
6-4
=
2
+5
-
+ 1
2
3
-
2
+
6
+6
=
-
4
loses 2
2x3(L(M)
+ 1
=
-
=
2
loses 6
+ 1
-
4
+ 2
2
2
-
I
2
3
grind
Ru
-
2x3 = loses6
-
2
Half-Reaction
D Assign
Method :
Oxidation Numbers
f
+5
-
2
-
to all
2 + 1
atoms
+ 3
11
-
[
② Identify Oxidation
and reaction
oxidation half-reaction :
Al > Al(8H)y-
reduction half-reaction :
N85- NH3
2
important for Step
-
+
-
I
3
+
(basic solution)
half-reactions
-
>
③ Balance non hydrogen and oxygen
atoms
oxidation half-reaction :
Al > Al(8H(4-
reduction half-reaction :
N85
#Balance oxggen
-) for entire molecule ,
half
each
in
reaction
-
->
oxidation half-reaction : Al
reduction half-reaction :
NH3
by adding
atoms
+
4H20
-
>
(1Al
already balanced
Has
to
each
1N
per
side,
per side
half-reaction
Al(8H(4-
N85- NH3
+
>
3H28
③ Balance hydrogen atoms by
adding Ht to each half-reaction
Al(OH(4- + 4H
oxidation half-reaction Al + 4H20
reduction half-reaction :
⑥ Add OH-ions
He added
to
+
>
-
:
NO3 + 9H*
-
>
NH3
+
3H28
to BOTH sides of each half-reaction
,
one
for each
of half reaction
ONE side
Only applies to basic solutions skip if solution is acidic
,
oxidation hulf-reaction : Al
reduction half-reaction :
+
4H20 + 48H-
N85 + 9H
+
+
>
40H
-
Al(OH(4 + 4H
>
NH3
+
+
+
40H
3428 + 90H
⑦ Lancel out
#20 molecules
excess
Only applies to basic solutions skip if solution is acidic
,
20 + 48H+
oxidation hulf-reaction : Al
Al
48H
+
-
AlL8H(4- +
>
Al(8H(4-
-
>
NH3
+
3
20
JOH- NH3
+
90H-
N85 + + /H
*
reduction half-reaction :
No5 +
JH
-
+
>
+
N85 + 6H20
L
>
-
⑧Add electrons to balance charge of each
oxidation half-reaction : Al
reduction half-reaction :
48H
+
-
-
>
>
NH3
+
+
90H
98H-
half-reaction
Al(8H(4 + ze-
N85 + 6H20 + 8e
-
>
NH3
+
90H-
⑨ Balance two half-reactions by multiplying each side by LCM of added eI
hulf-reaction
Al(8H(4 + 3e -]x8
[Al + 48H
oxidation
-
:
8 Al
+
328H
-
>
-
8 Al(8H(4- + 24e
reduction half-reaction : [NO3
+
6H20 + 8e
3N85
+
18H20 + 24e
-
-
>
NH3
>
3 NH3
+
90H]
+
270H
⑱ Add reactions together
8 Al + 328H+ 3N0z+ 10H 20 + 24e- 8 Al(8H(4 + 24e =3 NH3+ 270H-
⑪ Lancel out
8
,
optional add spectator ions
:
Alth 3 Noz+ 10 H20tXe =8 AllOH(4
-3 NAzt2/H
+
↓
8 Al + 58H + 3N8z + 18H20 - 8 A)(8H(4 + 3NH3
In +* -- > In
-
3
NHyt
+
oxidation : [In
reduction :
(Oz + 18 H
42n
5
+
-
g
+
8e
+
-
Noz- +18H"
-
>
(2
oxidation :
[CuS + 4H28
reduction :
[NO3- + 4H
+
Mn((N)g"
+
82
-
(n2
>
3
-
8
8H
>
+
-
-
+
-
>
+
+
8H
+
+
82 ]
-
+
+3
4
3(ut + 35042- 2 + 80 +
4
+
3(u* + 35042
Mn((N) ,
-
+ 8N8
+
4H2
3
Mn((N)6" - Mn((N)63- + 1e
82 4H + 487) - He 2H28
>
5842
+
NO + 24283x8
>
-
+
+j
-
+
223x4
NHyt + 3H20 + 22n
+ 32
+
3(u) + 12
48 + 8N85 + 3
+
+
NHy* + 3H202x/
-
+
3(u) + 8N0z
>
-
2t
In
>
-
-
Oxidation hulf-reaction ,
Oxidation loses
product
half-reaction ,
- reduction
elections, elections in
+
48H
reduction is
elections
going elections ,
in
reactant