Redox Titration Practical Report - College Assignment by Hengky Kusniar
REDOX TITRATION EXPERIMENT
Principle of Redox Titration
The qualitative analysis
of substances can be done by using a titration technique. Titration is a method
for determining the level of a substance by using another substance already
known to its concentration. Titration is also known as volumetric analysis, in
which the substance to be analyzed is allowed to react with other substances
whose concentration is known and flowed from the burette in solution form.
Titration is differentiated by the type of reaction involved in the titration
process. Titration is divided into 4, namely: acid base titration, redox
titration, complexity titration and sedimentation titration.
In
this paper will be focused oh redox titration. Redox titration is an analytical
method based on the occurrence of a reduction oxidation reaction between the
analyte and the titrant. Analytes containing reductor species are titrated with
titrant in the form of standard solutions of the oxidizer or conversly. The
concept of redox reactions is the concept of oxidation-reduction reactions
based on their changing oxidation number (involving electron catch and
release). The substance that oxidised is known as reducing agent. On the contrary, the
substance that reduces is known as an oxidising agent in the redox reaction. We can easily
understand the redox reaction with the help of electron concept. A substance
which accepts electrons to form anion show reduction reaction and vice-versa.
During a redox reaction, we can check the change in the oxidation number of
substances. In an oxidation reaction, the oxidation number increases whereas in
a reduction reaction, the oxidation number reduces.
There
are two ways to equalize the equations of redox reactions namely the method of
oxidation number and the half-reaction method (electron ion method) Redox
reactions can be used in volumetric analysis when eligible.
Redox reaction is a combination of two reactions;
reduction and oxidation. The old definition of reduction is the removal of
oxygen from a compound and oxidation is the combination of a substance with
oxygen. But we can just focus on the current definition which says that
reduction is a process of a substance accepting electrons to form anion, and oxidation is a process of a substance losing
electrons. In redox reaction, each oxidation and reduction reaction is called
half reaction.
·
Example for oxidation
The positive charge represents electron
deficiency, one positive charge means deficient by one electron.
·
Example for reduction
The negative charge represents electron
richness, one negative charge means rich by one electron.
In
a reaction, if there is an atom undergoing oxidation, there is probably another
atom undergoing reduction. When there is an atom that donates electrons, there
is always an atom that electrons. Electron transfer happens from one atom to
another. To keep track of
electron transfer in redox reaction, we use oxidation number or oxidation state
(OS). There are some rules for assigning OS:
1.
The
sum of the oxidation numbers of all the atoms in a molecule or ion must be
equal in sign and value to the charge on the molecule or iron.
2.
In binary compounds
(those consisting of only two different elements), the element with greater electronegativity is assigned
a negative OS equal to its charge as a simple monoatomic ion.
3.
When
it is bonded directly to a non-metal atom, the hydrogen atom has an OS of +1.
(When bonded to a metal atom, hydrogen has an OS of -1.)
4. Except for substances termed peroxides or superoxides, the OS of oxygen in its compounds is -2. In peroxides, oxygen has an oxidation number of -1, and in superoxides, it has an OS of -1/2.
Reaction of Redox
Titration
Determine which element is oxidized and which
element is reduced in the following reactions (be sure to include the oxidation
state of each):
- Zn
+ 2H+ → Zn2+ + H2
- 2Al
+ 3Cu2+→2Al3+ +3Cu
- CO32- +
2H+→ CO2 + H2O
Solutions
- Zn is
oxidized (Oxidation number: 0 → +2); H+ is reduced
(Oxidation number: +1 → 0)
- Al is
oxidized (Oxidation number: 0 → +3); Cu2+ is reduced (+2 →
0)
- This is
not a redox reaction because each element has the same oxidation
number in both reactants and products: O= -2, H= +1, C= +4.
(For further discussion, see the article
on oxidation numbers).
An atom is oxidized if its oxidation number increases, the reducing agent, and an atom is reduced if its oxidation number decreases, the oxidizing agent. The atom that is oxidized is the reducing agent, and the atom that is reduced is the oxidizing agent. (Note: the oxidizing and reducing agents can be the same element or compound).
Steps and Example of Redox
Titration
A.
Examples
of Analytical Reduction and Oxidation
A 25.00-mL sample of a liquid bleach was diluted to
1000 mL in a volumetric flask. A 25-mL portion of the diluted sample was
transferred by pipet into an Erlenmeyer flask containing an excess of KI,
reducing the OCl– to Cl–, and producing I3–.
The liberated I3– was determined by titrating with
0.09892 M Na2S2O3, requiring 8.96 mL to reach
the starch indicator end point. Report the %w/v NaOCl in the sample of bleach.
B.
Steps of
Analytical Reduction and Oxidation reaction
To determine
the stoichiometry between the analyte, NaOCl, and the titrant, Na2S2O3,
we need to consider both the reaction between OCl– and I–,
and the titration of I3– with Na2S2O3.
·
First, in reducing OCl–
to Cl–, the oxidation state of chlorine changes from +1 to –1,
requiring two electrons. The oxidation of three I– to form I3–
releases two electrons as the oxidation state of each iodine changes from –1 in
I– to –⅓ in I3–. A conservation of electrons,
therefore, requires that each mole of OCl– produces one mole of I3–.
·
Second, in the titration reaction, I3–.
is reduced to I– and S2O32– is
oxidized to S4O62–. Reducing I3–
to 3I– requires two elections as each iodine changes from an
oxidation state of –⅓ to –1. In oxidizing S2O32–
to S4O62–, each sulfur changes its oxidation
state from +2 to +2.5, releasing one electron for each S2O32–.
A conservation of electrons, therefore, requires that each mole of I3–
reacts with two moles of S2O32–.
·
Finally, because each mole of OCl–
produces one mole of I3–, and each mole of I3–
reacts with two moles of S2O32–, we know that
every mole of NaOCl in the sample ultimately results in the consumption of two
moles of Na2S2O3.
The balanced
reactions for this analysis are:
OCl−(aq)+3I−(aq)+2H+(aq)→I3-(aq)+Cl−(aq)+H2O(l)
I3- (aq)+2S2O32− (aq)→S4O62− (aq)+3I‑(aq)
The moles of
Na2S2O3 used in reaching the titration’s end
point is:
(0.09892MNa2S2O3)×(0.00896LNa2S2O3)=8.86×10−4molNa2S2O3
which means
the sample contains:
8.86×10−4molNa2S2O3×1molNaOCl/2mol
Na2S2O3×74.44gNaOCl = 0.03299gNaOCl
mol NaOCl
Thus, the %w/v NaOCl in the diluted sample is:
0.03299gNaOCl ×100 =1.32%w/vNaOCl
25.00mL
Because the bleach was diluted by a factor of 40 (25 mL to 1000 mL), the concentration of NaOCl in the bleach is 5.28% (w/v).
References
-
Auroracahya. June 12,
2012. Titrasi Redoks. Retrivied from https://auroracahya.wordpress.com/tag/titrasi-redoks/
-
“Redox
Titration” (2016). Retrieved from http://chemistry.tutorvista.com/analytical-chemistry/redox-titration
- Kiruthiga, B. “Redox
Titration Oxidation-Reduction Titration” [PDF document]. Retrivied from http://www.srmuniv.ac.in/sites/default/files/downloads/Redox_Titration.pdf
- David harvey, Feb 25 2017.Redox titrations. Retrieved from
- https://chem.libretexts.org/Textbook_Maps/Analytical_Chemistry_Textbook_Maps/
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