Oxidation - reduction reactions are reactions in that electrons are transferred from one ion or atom or module to another of such species in the particular reaction under consideration whether the reaction involves formation of ionic compounds or covalent compounds though it is less obvious in the later. They are as well utilized in titrimetric analysis as the acid - base reactions we treated in previous chapter.
Oxidation - reduction titration involves a transform in the oxidation state for both substances being measured and the titrant oxidation involves the loss of electrons whereas reduction involves gain of electrons.
Such redox reactions unlike the acid - base reactions are extremely slow and the process generally involves the employ of catalyst or slight heat.
How can I find out the oxidation number or state of an element? The oxidation number or oxidation state of an element in a compound symbolizes the amount of oxidation (for example the number of electrons gained) essential to change one atom of the element from the free elemental state that has zero value to that in the compound. Both reduction and oxidation occur at similar time for example simultaneously; therefore they are sometimes simply called as redox reaction.
There are certain guidelines that approximately indicate the oxidation states of elements, atoms and ions in compounds. Three of which are:
1) If you remember in periodic table, the ions shaped from the elements in all the groups carry either positive or negative charges as their oxidation numbers.
Therefore, Hydrogen is in group one of the periodic table, so it has +1 oxidation state. Oxygen is in group 6 so it has - 2 oxidation states. Carbon is in group four so it has + 4 oxidation state.
2) The sum of the oxidation states of every elements in a neutral compound adjoin to zero for example .SO2 = 0 whereas in the case of polyatomic ions such as SO42- the sum of the oxidation states of the elements equivalents the charge on the ion for example -2.
3) The oxidation state of an element in its natural state in Zero for example Cl2, Na, and P4 etc.
Oxidizing and reducing agents:
In acid - base reactions, the reactants are so identified through their names. That is, there must be an acid and there must a base. In alike manner, Redox reactions have 2 different participants. The reducing agent and the oxidizing agent correspondingly. There can't be one with no the other therefore, the term Redox.
How can we distinguish between the 2 Redox participants? The substance that admits electrons is termed e oxidizing agent and the substance that donates the electrons is the reducing agent. Take this instance.
Na → Na+ + e-
Two things are obvious from the equation:
1) Sodium donates 1 electron to form sodium ion, thus it is the reducing agent.
2) Na moves from oxidation state of Zero to +1, therefore Na is oxidized.
How does one classify and differentiate a reducing agent and an oxidizing agent in a chemical reaction? In a chemical redox reaction, 2 half equations are generally recognized. One symbolizes oxidation and the other reduction. The 2 balanced half equations should adjoin up to the redox equations. Therefore, calculating the oxidation states of every one element in the equation disclose that moved from what value to the other. If an element undergoes an amplify in the oxidation number, then it is the reducing agent for example the element has been oxidation and if a reduce in oxidizing number is noticed, the substance is an oxidizing agent and it has been decreased,
Let us take an instance. In the reaction,
1) Na(s) + 1/2 Cl2 (g) → NaCl(g)
2) Sn4+(aq) + 2Fe 2+ (aq) → Sn2+(aq) + 2F3+(aq)
Classify the reducing agent, the oxidizing agent, the oxidized specie and the reduced specie.
1) Na(s) + ½ Cl (g) → NaCl(s)
Break the equation into 2 halves
Na(s) → Na+
1/2 C12 (g) → C1-
2) Balance the charges and coefficients
Na (s) → Na+ + e-
1/2 Cl 2 + e- → Cl-
3) Write the oxidation numbers
Na(s) → Na+ + e-
1/2 Cl2 + e- → Cl-
4) Identify the species.
The reducing agent is Na because it donates electron;
The oxidizing agent is Cl2 because it accepts one electron Na is oxidized because it moves from 0 to + 1 that is an enhance. Cl2 is reduced because it moves from 0 to -1 i.e. a decrease
5) In alike manner, we will find out that Sn" is reduced, Sn4+ is the reducing agent and Fe2+ is oxidized. Fe2+ is the oxidizing agent
The subsequent other rules assist in providing assistance in assigning oxidation states to some specific cases.
1) Oxygen though is generally -2 its oxidation state is lower in peroxides and super oxides these as Na2O2 .
2) Hydrogen though usually is +1 but can be originating as -1 in its combination via electropositive metals these as NaH.
3) In several compounds through constituents atoms having natural negative oxidation states for example F2O. The more electronegative takes the negative value which in this instance means oxygen has a + 2 oxidation state.
Balancing redox equations:
Actually, this is the main problem you will encounter in redox reactions. Whereas in acid base reaction, the coefficients of the reacting species is all the important criteria, in redox reactions apart from the coefficient, the electrons transferred and the medium of the reaction are also very important.
The ion electron method
We talked previous of writing 2 half equations in order to differentiate a reducing agent from an oxidizing agent. The similar applies at this time. The 2 half equations are then added to generate the final balanced equation. In doing this balancing, we must comprehend that there are no specific laws but instead we have guidelines which are hereby suggested.
a) Write the whole net equation with no attempting to balance it
MnO4- + SO2 → Mn2+ + SO42-
b) Decide which elements undergo transforms in oxidation states. Write the partial equations for each of such elements and determine which of the reaction oxidation is and which reduction is.
MnO4- → Mn2+
+7 +2 reduction
SO2 → SO42-
+4 +6 oxidation
c) Attempt to balance every the atoms in the 2 half equations. This you will do by adding coefficients to balance all atoms except those of H and 0 atoms.
MnO4- → Mn2+
SO2 → SO42-
d) Add appropriate coefficient of H2O molecules to the side deficient in oxygen. Afterward we either add H+ or OH- to balance the H atoms for acid and alkaline medium respectively.
MnO4 + 8H+ → Mn2+ + 4H2O
SO2 + 2H2O → SO42- + 4H+
e) Balance the 2 half equations electrically via adding electrons to the appropriate side of the equation so that the charges on both sides of the ½ equations are equal (always add electrons to the side through the higher oxidation)
MnO4- + 8H+ → Mn2+ + 4H2O
+7 +2(so add5e- to the +7 side)
MnO4- + 8H+ + 5e- → Mn2+ + 4H2O
SO2 + 2H2O → SO42- + 4H+
0 -2 +4 so add + 2e-
SO2 + 2H2O → SO42- + 4H+ + 2e-
f) Balance the electrons transferred in both half equations
2[MnO4 + 8H+ + 5e- → Mn 2+ + 4H2O]
5 [SO2 + 2H2O → SO42- + 4H+ + 2e-]
Add 2 half equations together and eliminate those electrons, ions or water molecules that appear on both sides of the equation.
2MnO4 + 16H+ + 10e- → 2Mn2+ + 8H2O
5SO2 + 10H2O → 5SO42- + 20H+ + 105
2MnO4 + 5SO2 + 2H2O → 2Mn2+ + 5SO42- + 4H+
The oxidation state method:
As it was indicated whenever we considered the ion- electron process of balancing a redox reaction, I can only suggest some guidelines which might assist us in balancing a Redox reaction via the oxidation state process.
For instance, balance the redox reaction.
C + H2SO4 → CO2 + SO2 + H2O
Determine which elements undergo a change in oxidation state from the reactants to the products.
CO + H2SO4 → CO2 + SO2 + H2O
C= 0 to + 4
H= +1 to +1
S= + 6 to + 4
O = -2 to -2
Only C and S underwent transforms in oxidation states: C is oxidized and S is decreased.
2. Write the differences in the oxidation states for the elements.
For the increase, C moves from 0 to + 4 for example a difference of +4
For the decrease, S moves from + 6 to +4 for example a difference of +2
3. Balance the enhance and decrease of oxidation states by placing coefficients in front of the increase or decrease such that the total increase equals the total decrease.
For C: = increase, +4 x 1 = + 4
For S = decrease +2 x2 = + 4
4. Multiply the compounds or molecules by the coefficients wherever the elements concerned are incorporated
IC + 2H2SO4 → ICO2 + 2SO2 + 2H2O
5. Balance the coefficients of other reactants.
C + 2H2SO4 → CO2 + 2SO2 + 2H2O
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