Experiment No. 2 OXIDATION – REDUCTION REACTIONS Report by: Group #2 Tricia Ojon Katrina Mae Lee
What is redox? Redox (shorthand for reductionoxidation reaction) describes all chemical reactions in which atoms have their oxidation number/oxidation state changed
Terminology: The
term redox comes from the two concepts of reduction and oxidation. Oxidation describes the loss of electrons/hydrogen, or gain of oxygen / increase in oxidation state by a molecule, atom or ion Reduction describes the gain of electrons / hydrogen, or loss of oxygen / decrease in oxidation state by a molecule, atom or ion
Oxidizing agents / oxidants / oxidizers Substances that have the ability to
oxidize other substances and are said to be oxidative The oxidant removes electrons from another substance, and is, thus, reduced itself. And, because it "accepts" electrons, it is also called an electron acceptor
Reducing agents/reductants/reducer s Substances that have the
ability to reduce other substances and are said to be reductive The reductant transfers electrons to another substance, and is, thus, oxidized itself. And, because it "donates" electrons it is also called an electron donor
Redox pair The pair of an oxidizing and reducing agent that are involved in a particular reaction
Oxidation state An indicator of the degree of oxidation of an atom in a chemical compound The hypothetical charge that an atom would have if all bonds to atoms of different elements were 100% ionic and is used for purposes such as predicting ratio and balancing reactions Represented by integers, which can be positive, negative, or zero (0).
Oxidation number The charge that a central atom in a coordination compound would have if all the ligands were removed along with the electron pairs that were shared with the central atom Represented using Roman numerals
OXIDATION OXIDATION STATE NUMBER Electrons are Every electron distributed based belongs to the on ligand, regardless electronegativity of of atoms electronegativity Written using Arabic numerals Ex.: O-2
Written using Roman numerals Ex.: FeIII or iron(III)
Assigning oxidation states: The
oxidation state in the free or uncombined or elemental state such as Fe, H2, O2, is zero (0). Fluorine has an oxidation state of −1 in all its compounds since it is the most electronegative Alkali metals (Group I) have an oxidation state of +1 in virtually all of their compounds Hydrogen has an oxidation state of +1 except when bonded to less electronegative elements such as Na, Al, and B, as in NaH, NaBH4, LiAlH4, where each H has an oxidation state of -1. Alkaline earth metals (Group II) have an oxidation state of +2 in virtually all of their compounds.
Halogens,
other than fluorine have an oxidation state of −1 except when they are bonded to oxygen, nitrogen or with another halogen Oxygen has an oxidation state of −2 except where it is −1 in peroxides, +2 in oxygen difluoride, OF2,+1 in O2F2. For monoatomic ions, the oxidation number is the valence of the ion, e.g., +2 for Ca2+ The sum of the oxidation numbers in a polyatomic ion is equal to the ion charge The sum of the oxidation numbers in a compound is zero
Balancing Redox reactions: Assign
oxidation states to reactants and products
K2Cr2O7 + H2SO4 + H2O2 → K2SO4 + Cr2(SO4)3 + O2 + H2O +1 +6 -2 +1 +6 -2 +1 -2
+1 -1
+1 +6-2
+3 +6-2
0
K2Cr2O7 + H2SO4 + H2O2 → K2SO4 + Cr2(SO4)3 + O2 + H2O
Determine
which elements show an increase or decrease in oxidation state
+1 +6 -2 +1 +6 -2 +1 -2
+1 -1
+1 +6-2
+3 +6-2
0
K2Cr2O7 + H2SO4 + H2O2 → K2SO4 + Cr2(SO4)3 + O2 + H2O
Determine
the total change in oxidation number of the elements oxidized and reduced and determine the no. of e- gained/lost per molecule or unit of the compound Decrease from +6 to +3 ; Gained 6 e- per molecule +6 -1 +3 0
K2Cr2O7 + H2SO4 + H2O2 → K2SO4 + Cr2(SO4)3 + O2 + H2O Increase from -1 to 0 ; Lost 2 e- per molecule
Balance
the number of electrons gained and lost. The numbers used become the coefficients.
Gained 6 e- per molecule X 1 = 6 e- gained +6 -1 +3
0
K2Cr2O7 + H2SO4 + 3H2O2 → K2SO4 + Cr2(SO4)3 + 3O2 + H2O Lost 2 e- per molecule X 3 = 6 e- lost
Balance
the rest of the equation
K2Cr2O7 + 4H2SO4 + 3H2O2 → K2SO4 + Cr2(SO4)3 + 3O2 + 7H2O
Other way of balancing: Separate into two half-reactions: the oxidation portion and reduction portion. Each half-reaction is balanced separately and then the equations are added together to give a balanced overall reaction. We want the net charge and number of ions to be equal on both sides of the final balanced equation. Balance the atoms Next, balance the charges in each halfreaction so that the reduction half-reaction consumes the same number of electrons as the oxidation half-reaction supplies. Add the two half-reactions
Balancing in Acidic/Basic Medium: Divide the equation into an oxidation half-reaction
and a reduction half-reaction Balance these : Balance the elements other than H and O Balance the O by adding H2O Balance the H by adding H+ Balance the charge by adding e Multiply each half-reaction by an integer such that the number of e- lost in one part equals the number gained in the other part Combine the half-reactions and cancel the e For Basic:**Add OH- to each side until all H+ is gone and then cancel again**
Experimental Results: In
test tube A, the reaction produced very visible results such as the change in color of the aqueous solution from aqua blue (Cu(NO3)2) to colorless (Mg(NO3)2), and dissolving (Mg) and formation (Cu) of solid substances. The oxidizing agent in the reaction is Cu(NO3)2 while the reducing agent is Mg Cu(NO3)2 + Mg → Mg(NO3)2 + Cu
In
test tube B, the color of the solution changed from yellow orange to aqua blue. The oxidizing agent is K2Cr2O7 while the reducing agent is H2O2. K2Cr2O7 + 4H2SO4 + 3H2O2 → K2SO4 + Cr2(SO4)3 + 3O2 + 7H2O
In
test tube C, the color of the solution changed from reddish brown to being colorless. The oxidizing agent in this reaction is I2 while the reducing agent is Na2S2O3.
2Na2S2O3
+ I2 → Na2S4O6 + 2NaI
In
test tube D, there were several changes in the color of the substance upon addition of the last reagent. There was a slow change from deep purple to red, then orange, and lastly to being colorless. The oxidizing agent in this reaction is KMnO4 while the reducing agent is Na2C2O4 2KMnO4 + 8H2SO4 + 5Na2C2O4 → 2MnSO4 + 10CO2 + 8H2O + 5Na2SO4 + K2SO4
Visible indications of redox: When
transitional metal complexes are involved, a change in color often occurs The
d orbital of a transitional metal ion are split when it is surrounded by ligands. This is because the ligand outer electrons repel some of the d electrons more than the others. This allows an energy state transition from one d orbital to another in response to the absorption of a photon of light
As the light is absorbed, the solution takes on a color. Changes of oxidation state therefore change the color of the light absorbed, and so the color of the solution visible So you can observe the change in oxidation state of a transition metal in redox reaction by observing the change in color
Other
visible indications that a redox reaction took place are effervescence or the evolution of gas and the conversion of ions into metal (evident in the metal precipitate)
Questions and Answers:
Do all reactions involve redox? Why?
Not all reactions involve oxidation-reduction because some do not experience any change in the oxidation states of the substances. Non-redox reactions, which do not involve changes in formal charge, are known as metathesis reactions.
Why should H2O2 be freshly prepared?
Hydrogen peroxide or H2O2 needs to be freshly prepared because it decomposes exothermically into water and oxygen gas spontaneously. If not used in the freshly prepared state, it would not be useful for the characteristic reactions of H2O2.
Applications/ Importance
Redox
reactions have a large number of practical applications, including
the storage and later use of electric power the collection of hydrogen and oxygen from water the deposition of a thin layer of one metal on top of another an indicator not only of a system's capacity for cycling waste, but indeed of chemically supporting fish, plant, and invertebrate life oxidation of cyanide wastes and reduction of chromate waste
There
are both oxidation (e.g. biological conversion of ammonia to nitrites to nitrates) and reduction (ridding systems of nitrate aka denitrification, bio-phosphate PO3) that must occur readily in a truly closed system to support (macro-) life
APPENDIX
ELEMENT COMPOUND / COLOR (Aq. ION Sol’n) Cu Cu(NO3)2 Aqua blue Mg Cr
I
O
OXIDATION STATE +2
Cu
Brick red
0
Mg(NO3)2
Colorless
+2
Mg
Silvery white
0
K2Cr2O7
Yellow orange
+6
Chromic ion
Green
+3
I2
Reddish brown
0
KI
Colorless
-1
H2O
Colorless
-2
H2O2
Colorless
-1
S
Mn
C
Na2S2O3
Colorless
+2
Na2S4O6
Colorless
+2.5
KMnO4
Deep purple
+7
MnSO4
Colorless
+2
MnO2
Brown precipitate
+4
CO2
Colorless gas
+4
Na2C2O4
Colorless
+3
END
