Balancing oxidation-reduction reactions by the half-reaction method
R. Hart, 27 July 1993
Based on studies of students sketching and verbalizing their thinking. Ref: Chemistry, 3rd Ed, Steven S. Zumdhal, Heath & Company, 1993, 1210 pages.
I can visualize an atom as a dense central area surrounded by a cloud of orbiting sub-atomic particles. The central area I will call:
the nucleus
a proton
a neutron
an electron
The particles orbiting the nucleus of an atom I will call:
electrons
protons
neutrons
nuclei
I can imagine the electron cloud organized into restricted regions in which the electrons orbit the nucleus. I will call each of these regions a(n):
subshell
orbital
valence
The electrons in the outermost subshell of an atom determine an atom's capacity to react or unite (the number of bonds it can make) with other atoms. I will call these electrons:
valence electrons
core electrons
covalent electrons
ionic electrons
I can imagine the "Lewis structure" of molecules by seeing the symbol for each atom surrounded by dots for the:
valence electrons
bonding electrons
core electrons
electron cloud
I can imagine the structural formula for a molecule by seeing the symbol for each atom connected with a line for each ( ) pair of electrons.
bonding
valence
core
lone
I will call bonds holding atoms together in a molecule formed by shared electrons:
covalent bonds
ionic bonds
polar-covalent bonds
chemical bonds
I will call the simplest method of representing a molecule with symbols for the included elements a:
chemical formula
structural formula
Lewis structure
chemical equation
I will call an atom that has a positive or a negative charge (from the loss or gain of an electron):
an ion
a cation
an anion
a salt
an ionic solid
The reorganization of the atoms in a molecule I will call:
a chemical reaction
a chemical equation
a chemcial formula
a half-reaction
I can imagine a chemical equation with formulas for the reactants on the left and the ( ) on the right.
products
oxidants
spectator ions
solvent
I know that a chemical equation is not balanced if:
the number of atoms on each side are different
the number of atoms on each side are the same
the number of molecules on each side are different
the number of molecules on each side are the same
The relative number of reactant and product molecules is given by numbers I will call:
coefficients
subscripts
atomic numbers
molecular numbers
nuclear numbers
Before balancing I will describe the reaction with:
a net ionic equation
a molecular equation
a complete ionic equation
spectator ions included
A chemical reaction in which one or more valence electrons are transferred I will call:
an oxidation-reduction reaction
a redox reaction
an acid-base reaction
a precipitation reaction
The number of negatively charged valence electrons an atom has lost or obtained, I will call its:
oxidation state
oxidation number
subscript
coefficient
An increase in oxidation state by the removal of electrons I will call:
oxidation
reduction
ionization
dissolving
The atom that obtains or is assigned a new electron I will call the:
electron acceptor
reducing agent
electron donor
The whole compound or molecule that losses an electron I will call the:
reducing agent
oxidizing agent
electron acceptor
When two net ionic equations are written for a reaction, one for oxidation and one for reduction, I will call them:
half-reactions
Water is a covalent molecule: H2O. The oxidation state of oxygen is given as: (The 2 in H2O is generally written as a subscript).
-2
2-
-1
+2
1+
Water is a covalent molecule: H20. The oxidation state of hydrogen is given as:
+1
1+
-1
1-
+2
Ba(ClO4)2 + H2SO4 ---> (Step 1) will result in the following before any reaction:
Ba++ + ClO4-- + H+ + SO4--
2 BaClO4 + 2 HSO2
Ba + ClO4 + H + SO4
2 BaClO4 + H++ + 2 SO4-
Ba(ClO4)2 + H2SO4 ---> (Step 2) will yield the possible products of:
HCl04 + BaSO4(s)
Ba++ + ClO4 + H+ + SO4--
BaH2 + (ClO4)SO4
Ba(ClO4)2 + H2SO4 ---> 2 HClO4 + BaSO4(s) (Step 3) is a(n) ( ) type of reaction because:
precipitation, a solid was formed
acid-base, a strong acid was used
oxidation-reduction, electrons were transferred from one molecule to another by the cations
complex, a solid was formed and a strong acid was used
KMnO4 + FeCl2 ---> (Step 1) will result in the following in solution before any reaction:
K+ + MnO4- + Fe++ + Cl-
K+ + MnO4- + Fe++ + Cl--
K+ + Mn++ + O-- + Fe++ + Cl-
KMnO4 + FeCl2 ---> (Step 2) will result in these ions after reaction: Fe++ + MnO4- ---> Fe+ + Mn++. This is a(n) ( ) type of reaction because:
oxidation-reduction, electrons were transferred from one ion to another
precipitation, as 4 atoms of oxygen (2 molecules of oxygen) were released from each permanganate ion
acid-base, as this reaction takes place in an acid solution.
In the half-reaction Fe++ ---> Fe+++ + e- iron:
is oxidized
is reduced
is the oxidizing agent
gains one electron
In the half-reaction MnO4- + 5e- + 8H+ --> Mn++ + H2O the permangante ion:
is the oxidizing agent
is oxidized
is the reducing agent
losses two electrons and four oxygen atoms
To balance the reaction, Fe++ + MnO4- --> Fe+++ + Mn++, using half-reactions (Step 3) in which each oxidation releases one electron and each reduction gains five electrons, I will:
multiply the oxidation by five
divide the reduction by five
multiply the reduction by five
In the reaction Fe++ + MnO4- --> Fe+++ + Mn++ the oxidation states of iron and Mn++ are easy to see. The reactant oxidation state of Mn:
can be found to be +7 working with oxygen as -2
can be found to be +8 since each oxygen is -2
is +1 since 2 molecules of oxygen are released
is -1 since 2 molecules of oxygen are released
Each half-reaction must be balanced for the number of atoms and charges among the reactants and products. In an acid solution H+ and H20 can be added as needed. This can be done by:
the process of inspection (trial and error) balancing the atoms in the most complicated molecule first
consulting a table of standard reduction potentials for half-reactions
the process of inspection (trail and error) balancing the atoms in the simplest molecule first
consulting the IUPAC periodic table
I can work out oxidation states by remembering:
H20 and HONC (1,2,3,4)
the number of valence electrons for the elements in the second period of the periodic table
the oxidation states of the noble gases
all the common cations and anions
all the common polyatomic ions
compounds must be electrically charged in solution
I can work out the molecules in a reaction:
by sketching all the ions in the solution
by writing down all the compounds
by writing half-reactions
I can balance a redox reaction with all species identified using the following steps (1. Add half-reactions and cancel identical species, 2. Balance each half-reaction, 3. Check that elements and charges are balanced, 4. Equalize electrons transferred, 5. Write separate oxidation and reduction reactions) in the order:
5, 2, 4, 1, 3
5, 3, 2, 1, 4
3, 2, 1, 4, 5
1, 2, 3, 4, 5
I can balance elements and charges in each half-reaction with all species identified in an acid solution (1. all elements except hydrogen and oxygen, 2. charges, 3. hydrogen with the hydrogen ion, 4. oxygen with water ) in the order:
1, 4, 3, 2
3, 1, 4, 2
3, 2, 1, 4
1, 2, 3, 4
I can balance elements and charges in each half-reaction with all species identified in a basic solution (1. all elements except hydrogen and oxygen, 2. cancel water on each side of equation, 3. charges, 4. hydrogen with hydrogen ions, 5. hydrogen ions with hydroxide ions, 6. oxygen with water) in the order:
1, 6, 4, 3, 5, 2
5, 2, 4, 6, 3, 1
5, 1, 2, 6, 3, 4
1, 2, 3, 4, 5, 6
I can demonstrate (measure) the flow of electrons from the reducing agent to the oxidizing agent:
with a galvanic cell
across a salt bridge or porous disk
with platinum electrodes
I can measure the driving force on the electrons moving from the reducing agent to the oxidizing agent in the cell. This force I can call:
the cell potential
the electromotive force
a joule
a volt
a coulomb
The electrode in the reducing agent compartment I will call:
the anode
the cathode
The standard reduction potential for the half-reaction, Fe+++ + e- ---> Fe++ is +0.77 V. To use this in the reducing agent compartment, I must:
change the sign on +0.77 V to -0.77 V
reverse the half-reaction
add the +0.77 V to the oxidizing agent potential
subtract the +0.77 V from the oxidizing agent potential
To set up the reducing agent compartment, I can add:
FeCl2 and FeCl3 to get 1 M Fe++ and 1 M Fe+++
1 M FeCl2
1 M FeCl3
1 M FeCl2 and 1 M FeCl3 to get 1 M Fe++ and I M Fe+++
The standard reduction potential for the half-reaction, MnO4- + 5e- + 8H- --> Mn++ + 4H2O, is 1.51. That for Fe+++ + e- --> Fe++ is 0.77. The resulting cell potential I can expect is:
+0.74 V
+2.27 V
-2.27 V
-0.74 V
To set up the oxidizing agent compartment, I can add:
KMnO4 and MnCl2 to get 1 M MnO4- and 1 M Mn++
1 M KMnO4
1 M MnCl2
1 M MnSO4
1 M KMnO4 and 1 M MnCl2 to get 1 M MnO4- and 1 M Mn++
The reaction MnO4- + Fe++ --> Mn++ + Fe+++, in a galvanic cell will occur as:
an oxidation of Fe++ at the anode
a reduction of Mn+7 at the cathode
an oxidation of Fe+++ at the anode
an oxidation of Fe++ at the cathode
a reduction of MnO4- at the cathode
For the reaction, MnO4- + Fe++ --> Mn++ + Fe+++, I must use ( ) electrode(s) in the galvanic cell.
platinum
one iron and one platinum
iron
one iron and one manganese
The standard reduction potential for the half-reaction, MnO4- + 5e- + 8H+ --> Mn++ + 4H20, is 1.51 V. I can expect this potential in:
an acid solution
a basic solution
a neutral solution
a wide range of pH from acid to basic
In the chemical formula, MnCl2, the 2 is:
generally written as a subscript for 2 Cl
a coefficient for two MnCl molecules
generally written as a superscript for two charges on the molecule
In the half-reaction, MnO4- + 5e- + 8H+ --> Mn++ + H20, the 8 is:
a coefficient indicating the number of H+ relative to the other species
a subscript for 8 H+ in the solution
a superscript for 8 charges in the solution
In the half-reaction, MnO4- + 5e- + 8H+ --> Mn++ + H2O, the 4- is:
a subscript of 4 oxygens per Mn and a net negative charge of one
a superscript for four negative charges on the ion
an exponent of four MnO with a net negative charge of one
