Electrochemistry is the study of interchange of chemical and
electrical energy. Oxidation/Reduction (redox) involves the exchange
of electrons from one chemical species to another. Normally, this is
done when the two chemicals contact each other (bump into each
other). Separating the chemical species such that the electrons
transfer via an external circuit, we can measure the electrochemical
effects.
Redox
Redox, reduction-oxidation reactions are reactions that involves
transfer of electrons. In the following example we will see that a
redox reaction involves both reduction and oxidation: Example:
2Br- --> Br2 + 2e- : Oxidation
Cl2 + 2e- --> 2Cl- : Reduction
-------------------------------------------
2Br- + Cl2 = Br2 + 2Cl-
: Redox reaction
Reduction: the donation of electrons to a species. Oxidation: the removal of electrons from a species.
A substance which causes another to get oxidized is called an
oxidizing agent (or oxidant) and will itself get reduced.
A substance which causes another to get reduced is called a reducing
agent (or reductant) and will itself get oxidized.
Mnemonic: Oxidation Is Loss, Reduction
Is Gain -- (OIL RIG)
Guidelines for determining Oxidation States
An oxidation state change indicates how many electrons transferred
per species.
1) Oxygen in compounds is assigned an oxidation state of -2.
(Exception: peroxides, e.g. H2O2) 2) Hydrogen in compounds is assigned an oxidation state of
+1. (Exception: hydrides, e.g. NaH, KH ...) 3) Free elements such as e.g. O2 and Na are
assigned an oxidation state of zero. 4) The sum of the oxidation states of all the atoms in a
species must be equal to the net charge on the species. 5) The alkali metals (Li, Na, K, Rb, and Cs) in compounds are
always assigned an oxidation state of +1. 6) The alkaline earth metals(Be, Mg, Ca, Sr, Ba, and Ra) and
also Zn and Cd in compounds are always assigned an oxidation state
of +2.
7) In an acid solution, use H+ and H2O
to balance charges and other atoms. In a basic solution, use OH-
and H2O to balance charges and other atoms.
By using these guidelines we can figure out oxidation states for all
elements involved in a redox reaction.
CHEMIX
School
- Electrochemistry
Examples from program - problems and solutions
Example 1
Balance: Fe+2 + MnO4-
+ H+ --> Fe+3 + Mn+2 + H2O
Solution:
The oxidation state of oxygen is -2 (Guideline 1). By
knowing this we can decide oxidation state of manganese in MnO4-
by the use of guideline 4 (net charge).
The oxidation state of manganese must be +7 because : 4*(-2)+7 = -1
(net charge)
The other species in the equation can now be balanced by inspection.
5Fe+2 + MnO4- + H+
--> 5Fe+3 + Mn+2 + H2O
5Fe+2 + MnO4- + 8H+
--> 5Fe+3 + Mn+2 + 4H2O
CHEMIX solution:
Insert: Fe+2 + MnO4- + H+
> Fe+3 + Mn+2 + H2O
in one of the half-reaction fields and calculate.
NOTE: In this case only one of the two fields should contain an
equation.
Example 2
Balance the equation from the following two half reactions
Add the two equations. Cancel the electrons and remove right side H+:
Overall equation: 3H2C2O4 + Cr2O7-2
+ 8H+ --> 6CO2 + 2Cr+3 + 7H2O
CHEMIX solution:
Insert the two half cell reactions:
H2C2O4 > 2CO2 + 2H+
+ 2e-
Cr2O7-2 + 14H+ + 6e-
> 2Cr+3 + 7H2O
and calculate.
Example 3
What is the equilibrium constant for the reaction of metallic Cu
with bromine to form Cu+2 and Br- at 25oC
?
Br2 + 2e- --> 2Br-
E0 = 1.09 V
Cu --> Cu+2 + 2e-
E0 = -0.34 V
---------------------------------
Cu + Br2 --> Cu+2 + 2Br-
Solution:
The overall cell voltage : E0cell = 1.09 V
-0.34 V = 0.75 V
Calculate G by inserting n=2 and E0cell=0.75
in eq.: G = -n F E0
Insert G in eq.: ln K = -G/RT and calculate.
ln K = 58.36 --> K = e58.38 = 2.27E25
CHEMIX solution: Step 1) Insert first and second half-reaction
Br2 + 2e- > 2Br-
E0 = 1.09
Cu > Cu+2 + 2e-
E0 = -0.34
and calculate equation and overall cell voltage. Step 2) Insert n=2 (-n=-2) and E0cell=0.75
in
equation: G = -n F E0 and
calculate. Step 3) Transfer result of G to
equation: -G = R T ln K and calculate K.
Example 4
Balance and decide the overall cell potential (E0cell),
G and K. : Fe+2+
O2 + H+ --> Fe+3 + H2O
knowing that:
Fe+3 + e- --> Fe+2
E0
= 0.77 V
O2 + 4H+ + 4e- --> 2 H2O
E0 = 1.23 V
Solution:
Turn upper half-reaction according to species in unbalanced equation
and change sign of E0 (0.77 V --> -0.77 V).
 Fe+2
--> Fe+3 + e-
Multiply by 4
 O2
+ 4H+ + 4e- --> 2 H2O
--------------------------------------------
Overall: 4Fe+2+ O2 + 4H+
--> 4Fe+3 + 2H2O
The overall cell voltage can be summed from the half-cell potentials
of the oxidation and of the reduction reactions.
E0cell = -0.77 V + 1.23 V = 0.46 V
Calculate G by inserting n=4 (-n=-4) and
E0cell=0.46 V in eq.: G = -n F E0
Insert G in eq.: -G = RT ln K (ln K = -G/RT) and calculate.
ln K = 71.62 --> K = e71.62 = 1.27E31
CHEMIX solution: Step 1) Insert first and second half-reaction (remember to
turn first half-reaction)
Fe+3 + e- > Fe+2
O2 + 4H+ + 4e- > 2 H2O
and calculate equation and overall cell voltage. Step 2) Insert n=4 (-n=-4) and E0cell
in equation: G = -n F E0 or ln K
= -G/RT and calculate. Step 3) Transfer result of G to
equation: -G = R T ln K and calculate K.