Re-dox Reactions
Oxidationis the loss of electrons
Reduction is a gain of electrons
Leo goes Grr!
Lose Electrons Oxidation
Gain electrons Reduce
3.
“Redox Reactions”
Many reactionsinvolve (or seem to
involve) a transfer of electrons.
Such reactions always involve both:
Oxidation Loss of electrons
Na Na+
+ e-
Reduction Gain of electrons
Cl + e-
Cl-
4.
Determine a Redox
Reaction
To determine if what you have is a redox
reaction, first rewrite the equation to
include the oxidation number for each
element
The oxidation number is the charge an
element has or appears to have when
combining to form compounds
5.
Oxidizing/Reducing Agents
1. Theelement that was oxidized is part of the
reducing agent.
2. The element that was reduced is part of the
oxidizing agent.
6.
Assigning Oxidation
Numbers
1. Allpure elements and homogeneous molecules = 0
2. Elements in group IA = +1
3. Elements in group IIA = +2
4. Ag+
, Zn+2
, Al+3
(unless it is a pure metal)
5. In binary compounds the second element = anion
charge
6. Oxygen is almost always = -2 *unless it is O2(g)
7. Hydrogen is almost always = +1 *unless it is H2(g)
8. The total charge of a compound is always = 0
7.
A Sample Problem
Whatare the oxidation numbers of the
elements in Na2SO4?
Na = +1 (times 2 atoms) = +2
O = -2 (times 4 atoms) = -8
+2 + -8 = -6
If the compound must = 0, then S must = +6
8.
Give the oxidationnumber for:
1. S in Na2SO3
2. Mn in KMnO4
3. N in Ca(NO3)2
4. C in Na2CO3
5. N in NO2
-
6. S in SO4
-2
7. S in H2S2O7
8. Fe in Fe(C2H3O2)2
+4
+7
+5
+4
+3
+6
+6
+2
9.
Ammonia plus Oxygen
NH3+ O2 NO + H2O
Is this equation balanced?
No, it does not matter
Step 1: Rewrite the equation with the
charges
N3-
H+
3 + O0
2 N2+
O2-
+ H+
2O2-
10.
Ammonia plus Oxygen
Step 2: Determine the changes in the
charges of each element
N3-
H+
3 + O0
2 N2+
O2-
+ H+
2O2-
Nitrogen changes from 3- to 2+
it lost an electron and was oxidized
Oxygen changes from 0 to 2-
it gains electrons and so is reduced
11.
Last Step
Step3: If there is a species oxidized and
one reduced, then it is a redox equation
12.
MASS RELATIONSHIPS IN
CHEMICALREACTIONS
MOLE METHOD- a balanced chemical
equation is used to establish ratios
among its reactants. Calculating the
amount of product formed or reactant in
the reaction involves the use the use of
moles as the stoichiometric coefficients in
chemical equations.
13.
A. MOLE-TO-MOLE
CALCULATION
Thistype of calculation is
needed to find out the amount
of moles that have been
consumed or have in a
chemical reaction, given the
number of moles of another
reactant or product.
14.
EXAMPLE
Vinegar isderived from many
substances, including alcohol
(CH3,CH2OH),which is then
converted to acetic acid (CH3COOH)
by acetobacter bacteria this is
described in the equation below.
CH3COOH+H2O CH3CH2+H2O
15.
1. Find thequantity of moles and
identify the useful unit conversions
and/or molar ratios.
1 mol of CH3CH2OH=1 mol of O2=
1 mol of CH3COOH=1mol H2O
If 10 moles of CH3CH2OH are used in the
production of CH3COOH, how many moles of acetic
acid will be produced?
16.
2. Use thesedata for unit conversion.
mol of = 10 mol x
=10 mol
Based on the computation, 10 mol can be
produced from 10 mol of
Mathematically, the ratio between alcohol and acetic
acid can be shown as follows
