Electrochemistry


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Introduction

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

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)

Iron: Fe+2 - e- --> Fe+3 Multiply by 5
Manganese: Mn+7 + 5e- --> Mn+2 Multiply by 1
------------------------------------------------
Overall: 5Fe+2 + Mn+7 --> 5Fe+3 + Mn+2

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

H2C2O4 --> 2CO2 + 2H+ + 2e-
Cr2O7-2 + 14H+ + 6e- --> 2Cr+3 + 7H2O


Solution:
Multiply the first equation by 3 and add them algebraically so the electrons in the two half-reaction equations cancel.

3H2C2O4 --> 6CO2 + 6H+ + 6e-
Cr2O7-2 + 14H+ + 6e- --> 2Cr+3 + 7H2O

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 DeltaG by inserting n=2 and E0cell=0.75 in eq.: DeltaG = -n F E0
Insert DeltaG in eq.: ln K = -DeltaG/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: DeltaG = -n F E0 and calculate.
Step 3) Transfer result of DeltaG to equation: -DeltaG = R T ln K and calculate K.

Example 4

Balance and decide the overall cell potential (E0cell), DeltaG 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).
                &nbspFe+2 --> Fe+3 + e-                Multiply by 4
                &nbspO2 + 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 DeltaG by inserting n=4 (-n=-4) and E0cell=0.46 V in eq.: DeltaG = -n F E0
Insert DeltaG in eq.: -DeltaG = RT ln K (ln K = -DeltaG/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: DeltaG = -n F E0 or ln K = -DeltaG/RT and calculate.
Step 3) Transfer result of DeltaG to equation: -DeltaG = R T ln K and calculate K.



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