Transcript Balancing Chemical Equations

Balancing Chemical Equations
A chemical reaction is a process by which one set of chemicals is
transformed into a new set of chemicals.
A chemical equation uses standard chemical symbols to describe the
changes occurring during a reaction
General form: reactants  products
Example: Water
Skeletal equation
H2 + O2  H2O
Hydrogen and oxygen combine to form water. But the equation is not
balanced.
There are 2 oxygen atoms on the reactant side of the equation, but only
one on the product side. If we place a 2 in front of the water on the
product side, we will balance the oxygens.
H2 + O2  2H2O
We now need to balance the hydrogens
 Place a 2 before the hydrogens on the reactant side of the equation
2H2 + O2  2H2O
Balanced
Note: Diatomic molecules
Equation for formation of water included hydrogens and oxygens with
subscript 2.
Other elements which occur in this way are Nitrogen, Fluorine, Chlorine,
Bromine and Iodine.
These elements occur naturally as diatomic (meaning 2 atoms)
molecules
The stoichiometric coefficients multiplying the chemical formulas tell you
the relative numbers of moles of each substance that reacts or is
produced in a chemical reaction.
Therefore, we can conclude from the balanced equation for water that 2
moles of hydrogen and one mol of oxygen combine to form water.
A chemical equation may also tell you what physical state the reactants
and products are in. The following state symbols are used:
• s (solid)
• l (liquid)
• g (gas)
• aq (aqueous solution)
These are found as subscripts after each reactant and product.
Example
Balance the following equation:
Na(s) + H2O(l)  NaOH(aq) + H2(g)
Step 1: How many atoms of each element are present on the reactant
and product side?
Reactants
Products
Na
H
O
Na
H
O
1
2
1
1
3
1
Let us balance the hydrogen atoms first
Na + 2H2O  2NaOH + H2
Na
H
O
Na
H
O
1
4
2
2
4
2
Next we balance the sodium atoms
2Na + 2H2O  2NaOH + H2
Na
H
O
Na
H
O
2
4
2
2
4
2
Step 2: Add state symbols
2Na(s) + 2H2O(l)  2NaOH(aq) + H2(g)
Questions
Balance the following equations:
(a) P2O5(s) + H2O(l)  H3PO4(aq)
Answer: P2O5(s) + 3H2O(l)  2H3PO4(aq)
(b) NO2(g) + H2(g)  NH3(g) + H2O(g)
Answer: NO2(g) + 3 ½ H2(g)  NH3(g) + 2H2O(g)
(c) SO2(g) + O2(g)  SO3(g)
Answer: 2SO2(g) + O2(g)  2SO3(g)
Balancing redox reactions
• Redox (meaning reduction-oxidation) reactions involves transfer of
electrons
• Involves two processes
oxidation
reduction
OIL RIG
is
is
loss
gain
Whenever oxidation occurs, reduction must also occur
Can split a redox reaction into two half reactions:
The oxidation reaction which involves a loss of electrons
The reduction reaction which involves a gain of electrons
To keep track of electrons in redox reactions, use oxidation numbers
Oxidation number:
the charge an atom has or appears to have
when electrons are distributed according to
certain rules
Rules for assigning oxidation numbers (O.N.)
1.
A free element (or uncombined element) is assigned O.N. of zero
2.
The sum of O.N.s of atoms in a neutral compound must equal zero
3.
The charge of a monoatomic ion is same as its oxidation number,
e.g. Li+ has O.N. of +1
Fe+3 has O.N. of +3
4.
In their compounds, group 1 elements have an O.N. of +1, group 2
elements have an O.N. of +2.
5.
The sum of O.N. of atoms in polyatomic ion equals the charge on
the ion
6.
The O.N. of halogens is usually -1.
Exception: when they are bonded to a more electonegative
element
7.
The O.N. of oxygen in most compounds is -2.
Exceptions: H2O2 and peroxide ion, O2-2 (O.N. of -1), and the
compound OF2 (O.N. +2)
8.
The O.N. of hydrogen is +1,
Exception: when bonded to a less electronegative element, i.e. a
metal, where it’s O.N. is -1, e.g. LiH, NaH
Examples
Assign oxidation numbers to all the elements in the following:
(a) Li2O
Lithium: Group 1 element  Oxidation number +1
Oxygen: Oxidation number -2
(b) HNO3
Hydrogen: +1
Oxygen: -2
HNO3 is neutral
 adding the O.N.s of the atoms together must equal zero
(+1) + N + 3(-2) = 0
 N = +5
(c) Cr2O72Sum of O.N.s equals charge on ion  here the sum equals -2
Oxygen: Oxidation number -2
2Cr + 7(-2) = -2
2Cr -14 = -2
Cr = +6
Questions
Assign oxidation numbers to each of the following:
(a) ZnCl2
Answer:
Zn: +2
Cl: -1
Answer:
O: -2
S: +6
Answer:
O: -2
Mn: +7
(b) SO3
(c) MnO4-
Defining Oxidising and Reducing agents
• Remember that for an oxidation to occur, a reduction
must also occur
• OIL RIG
• Oxidation can be defined (in terms of electon transfer) as
the loss of electrons from a species
• Reduction can be defined as the gain of electrons by a
species
• An oxidising agent (or electron acceptor) must itself be
reduced
• A reducing agent (or electron donor) must itself be
oxidised
• The action of both oxidising and reducing agents can be
represented in the form of a half-equation
Identifying Oxidising and Reducing Agents
Example
Formation of calcium oxide (CaO)
2Ca(s) + O2(g)  2CaO(s)
Ca and O2 (reactants): neutral atoms
Calcium oxide (product): ionic compound
Ca2+ O2ions
Can write the reaction as two half-reactions:
2Ca  2Ca2+ + 4eO2 + 4e-  2O2-
Here, calcium is oxidised; i.e. it loses electrons to oxygen
OIL
RIG
It reduces the oxygen
 it acts as a reducing agent
Oxygen is reduced; i.e. it gains electrons from calcium
It oxidises the calcium
 it acts as an oxidising agent
Example
Identify the oxidising and reducing agents in the following reaction:
PbO(s) + CO(g)  Pb(s) + CO2(g)
Step 1: Assign oxidation numbers:
PbO(s) + CO(g)  Pb(s) + CO2(g)
+2 -2
+2 -2
Step 2: Write half equations:
Pb+2 + 2e-
Pb0
0 +4
-2
Step 3: Identify oxidising and reducing
agents:
Pb increases O.N. from +2 to 0
PbO is reduced; acts as oxidising agent
C+2
C+4 + 2e-
C decreases O.N. from +2 to +4
CO is oxidised; acts as reducing agent
Questions
Identify the oxidising and reducing agents and write the half reactions in
the following:
(a) Cl2 + 2NaBr  2NaCl + Br2
Answer: 2Br--  Br2 + 2eCl2 + 2e-  2ClCl2 : oxidising agent
Br - : reducing agent
(b) Si + 2F2  SiF4
Answer: Si  Si4+ + 4eF2 + 2e-  2FF2 : oxidising agent
Si : reducing agent
(c) 4Fe +3O2  2Fe2O3
Answer: Fe  Fe3+ + 3eO2 + 4e-  2O2O2 : oxidising agent
Fe: reducing agent
Example
Balancing redox equations
Balance the equation showing the oxidation of Fe2+ ions to Fe3+ ions by
dichromate ions (Cr2O72-) in an acidic medium
Fe2+ + Cr2O72-  Fe3+ + Cr3+
Step 1: Identify oxidising and reducing agents and write half reactions
Fe2+ : +2
Cr2O72- : 2Cr + 7(-2) = -2
Fe3+ : +3
2Cr = 12
Cr3+ : +3
Cr = +6
Fe2+  Fe3+
Iron loses one electron, i.e. it is oxidised. It donates one electron to
dichromate, i.e. it is the reducing agent.
Half equation:
Oxidation reaction: Fe2+  Fe3+ + e-
Cr26+  Cr3+
Chromium gains three electrons, i.e. it is reduced. It gains its electrons
from iron, i.e. it is the oxidising agent.
Half equation:
Reduction reaction: Cr26+ + 3e-  Cr3+
Fe2+ + Cr2O72-  Fe3+ + Cr3+
Step 2: Balance each kind of atom other that H and O
Fe2+  Fe3+ + eCr2O72- + 3e-  2Cr3+
Step 3: Balance oxygen atoms by using H2O
Fe2+  Fe3+ + e-
Cr2O72- + 3e-  2Cr3+ + 7H2O
Step 4: Balance H atoms by using H+ ions
Fe2+  Fe3+ + e-
14H+ + Cr2O72- + 3e-  2Cr3+ + 7H2O
Step 5: Use electrons as needed to obtain a charge that is balanced
Fe2+  Fe3+ + e+2
+3
-1
14H+ + Cr2O72- +3e-  2Cr3+ + 7H2O
+14
-2
-3
+6
0
+2
+2
+9
+6
Add three electrons to the reactant side to balance the charges
Fe2+  Fe3+ + e-
14H+ + Cr2O72- + 6e-  2Cr3+ + 7H2O
Step 6: Ensure the number of electrons gained equals the number of
electrons lost and add the 2 half reactions together
Fe2+  Fe3+ + e-
×6
14H+ + Cr2O72- + 6e-  2Cr3+ + 7H2O
×1
6Fe2+  6Fe3+ + 6e14H+ + Cr2O72- + 6e-  2Cr3+ + 7H2O
14H+ + Cr2O72- + 6Fe2+ + 6e-  2Cr3+ + 7H2O + 6Fe3+ + 6e14H+ + Cr2O72- + 6Fe2+  2Cr3+ +7H2O + 6Fe3+
Balanced
Check to ensure that all atoms and charges are balanced.
Question
Balance the equation for the redox reaction of MnO4- with Fe2+ to
produce Mn2+ and Fe3+ in an acidic medium.
Procedure:
• Step 1: Identify the oxidising and reducing agents, then write
the half equations
• Step 2: Balance each atom, except H and O
• Step 3: Balance the O atoms (using H2O)
• Step 4: Balance the H atoms (using H+)
• Step 5: Check the charges are balanced for each half
equation
• Step 6: Ensure the numbers of electrons are equal (i.e.
there are the same numbers lost and gained)
• Step 7: Add the two equations together
• Finally: check that all the charges and the numbers of atoms
balance!
Question
Balance the equation for the redox reaction of MnO4- with Fe2+ to
produce Mn2+ and Fe3+ in an acidic medium.
Answer
Step 1: Identify oxidising and reducing agents and write half reactions
MnO4-  Mn2+
Mn + 4(-2) = -1
 Mn = +7
Mn : +2
MnO4- gains 5 electrons. It acts as the oxidising agent as it is reduced.
MnO4- + 5e- Mn2+
Reduction reaction
Fe2+  Fe3+
Fe loses an electron. It acts as the reducing agent as it is oxidised.
Fe2+  Fe3+ + e-
Oxidation reaction
Step 2: Balance each kind of atom other than H and O.
Balanced in this case
Step 3: Balance O atoms by adding H2O
MnO4- + 5e-  Mn2+ + 4H2O
Fe2+  Fe3+ + eStep 4: Balance H atoms by using H+ ions
8H+ + MnO4- + 5e-  Mn2+ + 4H2O
Fe2+  Fe3+ + eStep 5: Use electrons as needed to obtain a charge that is balanced
Fe2+  Fe3+ + e+2
+2
8H+ + MnO4- + 5e-  Mn2+ + 4H2O
+2
Already balanced!
+2
Step 6: Ensure the number of electrons gained equals the number of
electrons lost and add the two half reactions together
Fe2+  Fe3+ + e-
×5
×
1
8H+ + MnO4- + 5e-  Mn2+ + 4H2O
5Fe2+  5Fe3+ + 5e-
8H+ + MnO4- + 5e-  Mn2+ + 4H2O
8H+ + MnO4- + 5Fe2+ + 5e-  Mn2+ + 5Fe3+ + 4H2O + 5e8H+ + MnO4- + 5Fe2+  Mn2+ + 5Fe3+ + 4H2O
Balanced
Check to ensure that all atoms and charges are balanced.
Question
Balance the following equation, which takes place in an acidic medium:
MnO4- + SO2  Mn2+ + SO42Step 1: Identify oxidising and reducing agents and write half reactions
MnO4-: Mn +4(-2) = -1
Mn - 8 = -1
Mn = +7
Mn2+ : +2
MnO4- gains 5 electrons: it acts as the oxidising agent as it is reduced
MnO4- + 5e-  Mn2+
SO2: S + 2(-2) = 0
S-4=0
S = +4
Reduction reaction
SO42-: S + 4(-2) = -2
S - 8 = -2
S = +6
SO2 loses 2 electrons: it acts as the reducing agent as it is oxidised
SO2  SO42- + 2e-
Oxidation reaction
MnO4- + SO2  Mn2+ + SO42Step 2: Balance each kind of atom other than H and O
MnO4- + 5e-  Mn2+
Balanced in this case
SO2  SO42- + 2eStep 3: Balance O atoms by using H2O
MnO4- + 5e-  Mn2+ + 4H2O
2H2O + SO2  SO42- + 2eStep 4: Balance H atoms by using H+ ions
8H+ + MnO4- + 5e-  Mn2+ + 4H2O
2H2O + SO2  SO42- + 2e- + 4H+
Step 5: Use electrons as needed to obtain a charge that is balanced
8H+ + MnO4- + 5e-  Mn2+ + 4H2O
+8
-1
+2
-5
+2
0
+2
Already balanced!
2H2O + SO2  SO42- + 2e- + 4H+
0
0
0
-2
-2
+4
0
Already balanced
Step 6: Ensure number of electrons gained equals number of electrons
lost and add the two half reactions together
8H+ + MnO4- + 5e-  Mn2+ + 4H2O
×2
2H2O + SO2  SO42- + 2e- + 4H+
×5
16H+ + 2MnO4- + 10e-  2Mn2+ + 8H2O
10H2O + 5SO2  5SO42- + 10e- + 20H+
16H+ + 2MnO4- + 10H2O + 5SO2  2Mn2+ + 8H2O + 5SO42- + 20H+
2MnO4- + 2H2O + 5SO2  2Mn2+ + 5SO42- + 4H+
Balanced
Check to ensure that all atoms and charges are balanced.
Questions
Write balanced equations to represent the following reactions in an
acidic solution:
(a) I- + SO42-  I2 + S
Answer: 6I- + 8H+ + SO42-  3I2 + S + 4H2O
(b) MnO4- + H2C2O4  Mn2+ + CO2
Answer: 2MnO4- + 5H2C2O4 + 6H+  2Mn2+ + 10CO2 + 8H2O
(c) ClO3- + Cl-  Cl2 + ClO2
Answer: 2Cl- + 2ClO3- + 4H+  Cl2 + 2ClO2 + 2H2O
Example
In basic solution
Write a balanced equation to represent the oxidation of iodide ion (I -) by
permanganate ion (MnO4-) in basic solution to yield molecular iodine (I2)
and manganese (IV) oxide (MnO2)
I- + MnO4-
Skeletal equation:
I2 + MnO2
Step 1: Identify oxidising and reducing agents and write half equations
I-  I2
O.N. of I- is -1
O.N of I2 is 0
I- loses an electron: it acts as the reducing agent as it is oxidised
I-  I2 + e MnO4-  MnO2
Oxidation reaction
MnO4-: Mn + 4(-2) = -1
Mn = +7
MnO2 : Mn + 2(-2) = 0
Mn = +4
MnO4- gains 3 electrons: it acts as the oxidising agent as it is reduced
MnO4- + 3e-  MnO2
Reduction reaction
Step 2: Balance each kind of atom other than H and O
2I-  I2 + eMnO4- + 3e-  MnO2
Step 3: Balance the O atoms by using H2O
2I-  I2 + eMnO4- + 3e-  MnO2 + 2H2O
Step 4: Balance H atoms by using H+ ions
2I-  I2 + e4H+ + MnO4- + 3e-  MnO2 + 2H2O
Step 5: Since the reaction occurs in a basic medium, for each H+ ion we
add an equal number of OH- ions to both sides of the equation. Where
H+ and OH- ions appear on the same side of the equation, they may be
combined to form H2O.
4H+ + 4OH- + MnO4- + 3e-  MnO2 + 2H2O + 4OH-
4H2O + MnO4- + 3e-  MnO2 + 2H2O + 4OH2H2O + MnO4- + 3e-  MnO2 + 4OHStep 6: Use electrons as needed to obtain a charge that is balanced
2I-  I2 + e-2
0
-2
-1
-1
Add one electron to the product side to balance charges
2I-  I2 + 2e2H2O + MnO4- + 3e-  MnO2 + 4OH0
-1
-3
-4
0
-4
-4
Already balanced!
Step 7: Ensure the number of electrons gained equals the number of
electrons lost and add the two half reactions together
2I-  I2 + 2e2H2O + MnO4- + 3e-  MnO2 + 4OH-
×3
×2
6I-  3I2 + 6e4H2O + 2MnO4- + 6e-  2MnO2 + 8OH-
2MnO4- + 6I- + 4H2O  2MnO2 + 3I2 + 8OHBalanced
Check to ensure that all atoms and charges are balanced.
Question
Balance the following redox equation which occurs in basic solution
Mn2+ + H2O2  MnO2 + H2O
Step 1: Identify the oxidising and reducing agents and write half reactions
Mn2+  MnO2
Mn2+ : +2
MnO2: Mn +2(-2) = 0
Mn = +4
Mn2+ loses two electrons: it acts as the reducing agent as it is oxidised
Mn2+  MnO2 + 2eH2O2  H2O
Oxidation reaction
H2O2: 2(+1) + 2O = 0
O = -1
H2O: 2(+1) + O = 0
O = -2
H2O2 gains one electron: it acts as the oxidising agent as it is reduced
H2O2 + e-  H2O
Reduction reaction
Step 2: Balance each kind of atom other than H and O.
Mn2+  MnO2 + 2eAlready balanced in this case
H2O2 + e-  H2O
Step 3: Balance the O atoms by using H2O
2H2O + Mn2+  MnO2 + 2eH2O2 + e-  2H2O
Step 4: Balance the H atoms by using H+
2H2O + Mn2+  MnO2 + 2e- + 4H+
2H+ + H2O2 + e-  2H2O
Step 5: For each H+ ion, add equal no. of OH- to both sides of equation
2H2O + Mn2+ + 4OH-  MnO2 + 2e- + 4H+ + 4OH2H2O + Mn2+ + 4OH-  MnO2 + 2e- + 4H2O
Mn2+ + 4OH-  MnO2 + 2e- + 2H2O
2H+ + 2OH- + H2O2 + e-  2H2O + 2OH2H2O + H2O2 + e-  2H2O + 2OHH2O2 + e-  2OHStep 6: Use electrons as needed to obtain a charge that is balanced
Mn2+ + 4OH-  MnO2 + 2e- + 2H2O
+2
-4
0
-2
0
-2
-2
H2O2 + e-  2OH-1
0
-1
-2
-2
Add one electron to the reactant side to balance charges
H2O2 + 2e-  2OHStep 7: Ensure no. of electrons gained equals no. of electrons lost and
add the half reactions together
Mn2+ + 4OH-  MnO2 + 2e- + 2H2O
H2O2 + 2e-  2OHMn2+ + H2O2 + 2OH-  MnO2 + 2H2O
Questions
Write balanced equations to represent the following reactions in a basic
solution
(a) Fe(OH)2 + MnO4-  MnO2 + Fe(OH)3
Answer: 3Fe(OH)2 + MnO4- + 2H2O  MnO2 + 3Fe(OH)3 + OH-
(b) Bi(OH)3 + SnO22-  SnO32- + Bi
Answer: 2Bi(OH)3 + 3SnO22-  2Bi + 3H2O + 3SnO32-