Redox (Oxidation Reduction) Reactions
Redox (Oxidation-Reduction) Reactions
If there is an electron transfer between matters, these reactions are called oxidation reduction or redox reactions. If atom/compound or element accept electron this process is called reduction, on the contrary, if atom/compound or element donate electron this process is called oxidation. Look at following redox reaction examples;
Examples:
1. Mg → Mg+2 + 2e-
Mg atom loses two electrons and it is oxidized.
2. S-2 → S+6 + 8e-
S ion loses eight electrons and it is oxidized.
3. S + 2e- → S-2
S atom gains two electrons and it is reduced.
4. S+6 + 2e- → S+4
S+6 ion gains two electrons and it is reduced.
5. 2Al(s) + 3Cu+2(aq) → 2Al+3(aq) + 3Cu(s)
In this reaction, neutral reactant Al donate 3 electrons and oxidized and since it reduce Cu we call Al “reducing agent “, Cu initially has oxidation state of plus two and it gains two electrons and reduced, since it oxidize Al we call it “oxidizing agent “.This reaction is called redox or oxidation-reduction reaction.
Some Important Points about Oxidation State of Matters
1. Free elements have oxidation state 0. H2, Na, Cu has 0 oxidation state.
2. Oxidation state of mono atomic ion is equal to charge of ion. For example, Na+ has oxidation sate of +1, S-2 has oxidation state of -2.
3. Fluorine has oxidation sate of -1 in all compounds.
4. In general Hydrogen has oxidation sate of +1, but there are some exceptions that it has oxidation state of -1 in compounds like LiH, NaH, BaH2.
5. In general, oxygen has oxidation state -2, there are two exceptions in which it has oxidation state -1, like Na2O2, H2O2 and in compound OF2 O has oxidation state +2.
6. In a compound sum of oxidation states of elements is zero. For example;
In K2CO3 compound let me find oxidation state of C using known values.
K has +1 oxidation sate and O has oxidation state -2.
2.(+1) +(X)+3(-2)=0
X=+4
7. In polyatomic ion, sum of oxidation states of atoms is equal to charge of ion.
Example:
Find oxidation state of Cr in Cr2O7-2 compound.
O has oxidation state -2.
2X+7.(-2)=-2
X=+6
8. If a metal have more than one oxidation state, we find oxidation state of it by using known values in ion.
Example: Find oxidation states of Cu and N in compound CuNO3.
Cu can have +1 and +2 oxidation states in compounds. Nitrate NO3- has oxidation state -1, thus Cu must have oxidation state of +1.
We find oxidation state of N using compound as given below;
CuNO3
+1+X+3.(-2)=0
X=+5
N has oxidation state of +5 in this compound.
Example: Which ones of the following reactions are redox reaction?
I. 2SO2 + O2 → 2SO3
II. Mg + 2HCl → MgCl2 + H2
III. AgNO3 + KCl → AgCl +KNO3
Being a redox reaction; at least one reduction or one oxidation must take place. Now we examine given reactions whether oxidation states of elements are changed or not.
I . 2SO2 + O2→ 2SO3
In SO2 S has value
S+2(-2)=0
S=+4
In SO3 S has oxidation state;
S+3.(-2)=0
S=+6 Thus, I is redox reaction.
II. Mg + 2HCl → MgCl2 + H2
Mg in left hand side has 0 oxidation state, however, in product side it has value;
Mg+2(-1)=0
Mg=+2
And H has +1 value in compound HCl and 0 value in product side.
II is also redox reaction.
III. AgNO3 + KCl → AgCl +KNO3
Since oxidation states of species are not changed this reaction is not redox reaction.
Ag has +1 oxidation state , K has +1 oxidation state, Cl has -1 oxidation state and NO3 has -1 oxidation state in reactants and products sides.