Sunday, 28 November 2010

Chemistry : Redox I

Definitions of reduction and oxidation:
In terms of
O
H
e-
Receive
Oxidation
Reduction
Reduction
Lose
Reduction
Oxidation
Oxidation
Example
2Mg + O2 → 2MgO, Mg undergoes oxidation
CuO + H2 → Cu + H2O or N2H4 + O2 → N2 + 2H2O,
Cu and O is reduced.
Li+ + K → K+ + Li
K is oxidized.

The chemical species that oxidizes itself and reduce another species is called reducing agent, the chemical species that reduces itself and oxidize another species is called oxidizing agent.
Since number of e- conserves, reduction and oxidation must go in pairs.
Assignment of oxidation number (O.N.)
1)       Algebraical sum of oxidation number of each atom in a compound is equal to its charge of the ion (or equal to zero if it is neutral)
e.g., H2 is zero, so H in H2 is zero. K in K+ is +1.
2)       In a compound the element with higher electronegativity is assigned the O.N. first.
e.g., forCCl4, Cl was first assigned -1, then C is +4.
3)       Generally max. possible O.N. = group number, while the min. possible O.N. = 8 – group number, zero for metal.
Common O.N. of some elements:
H: +1 for H+, 0 for H2, -1 for H- (hydride)
O: -2 for O2-, 0 for O2, -1 for O22- (peroxide ion), -1/2 for O2- (superoxide ion)
Halogen: -1 for X-, 0 for X2; can have other O.N., e.g.: +1/3/5/7 for ClO- to ClO4-
Metal: 0 in neutral form and group number in ion form
Vanadium: +2 (V2+), +3 (V3+), +4(VO32+), +5(VO3+)
Stocking system to name ions: we use –ate(O.N.) to present the ion. e.g., SO42- is sulphate (VI) and Cr2O72- is dichromate (VI).
In a reaction, total decrease of O.N. is equal to the increase in O.N., increase in O.N. implies oxidation and decrease in O.N. implies reduction.
e.g., 2Mg + O2 → 2MgO, O.N. of Mg is raised from 0 to +2, while O decreased from 0 to -2.
We can balance the ionic-half equation by the following method
1)       If it is monatomic ion, then Xn+ + n e- → X (exceptional: Fe3+ + e- → Fe2+)
2)       If it is polyatomic ion (oxides), then a double amount of hydrogen ion is added and balance the charge with appropriate number of e-. e.g., Cr2O72- + 14H+ + 6e- → Cr3+ + 7H2O.
3)       Balance two ionic-half equation by multiply each of them so that the number of e- is balanced and can be eliminated completely.
The following is a electrochemical series (E.C.S.), where oxidizing power increase down the group, reducing power decrease down the group. Also, the LHS will be oxidized form while RHS will be the reduced form. i.e., oxidation goes as reverse and reduction go as written.
Li+ + e- → Li
Note that reactivity of halides decrease down the group, while reactivity of metals increase down the group, so Li and K is the strongest reducing agents while F2 is the strongest oxidizing agents.
Also, dichromate and permanganate ion is an oxidizing agent only if they’re in acidic solution (i.e. presence of H+), so they’re called acidified permanganate or dichromate solution.
Order of other metals is the same as metal reactivity series from K to Pb except Ca and Na is interchanged. Xn+ + n e- → X
2H+ + 2e- → H2
H2SO4 + 2H+ + 2 e- → SO2 + 2H2O
Cu2+ + 2e- → Cu
O2 + 2H2O + 4e- → 4OH-
I2 + 2e- → 2I-
Fe3+ + e- → Fe2+
Ag+ + e- → Ag
Br2 + 2e- → 2Br-
Cr2O72- + 14H+ + 6e- → Cr3+ + 7H2O
Cl2 + 2e- → 2Cl-
MnO4- + 8H+ + 5e- → Mn2+ + 4H2O
F2 + 2e- → 2F-

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