Transcript OXIDATION-REDUCTION REACTIONS (REDOX)

OXIDATION-REDUCTION REACTIONS
(REDOX)
A Change in Charge
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REDOX BASICS
• transfer of electrons between two atoms
Na
F
Oxidation Numbers
• the ox. # of free elements is ZERO (including
diatomics like O2 and polyatomics like P4 or S8)
• the ox. # of a monatomic ion = its charge (like Na+1)
• Other elements:
o O = -2 (except in peroxide, each O = -1)
o H = +1 (except in metal hydrides, then H = -1)
o Halogens = -1 (except when central atom of
polyatomic ions, like ClO3-1)
Identify the oxidation number for the
underlined element
• H2O
• H2O2
• FeSO4
• FeSO4
• HClO
• HClO2
• HClO3
• HClO4
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Is it Oxidation or Reduction or neither?
1. Silver ions change to raw (elemental) silver
2. Hydrogen gas changes to hydrogen ions
3. Blue CrCl2 changes to green CrCl3
4. Yellow K2CrO4 changes to orange K2Cr2O7
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Oxidation-Reduction Reactions
Single Replacement:
Zn(s) + HCl(aq)  ZnCl2(aq) + H2(g)
Synthesis:
Mg(s) + O2(g)  MgO(s)
Decomposition:
Fe2O3(s)  Fe(s) + O2(g)
Combustion:
CH4(g) + O2(g)  CO2(g) + H2O(g)
These types of reactions can be balanced by inspection
Redox reactions in acid/base solutions
MnO4- + S2O3-2 + H+  Mn+2 + SO4-2 + H2O
Br2 + OH-  Br- + BrO3- + H2O
S8 + O2 + H2O  SO4-2 + H+
These will be balanced using the method of half-reactions
Balancing with Half-Reactions
1. Split the reaction into half reactions involving similar
species (ignore H+, OH-, and H2O)
2. Balance atoms of all elements EXCEPT O and H
3. Balance O by adding H2O to the side that has fewer O
atoms
4. Balance H by adding H+ to the side that has fewer H
atoms
a. If acidic solution: continue on to step 5
b. If basic solution: add enough OH- to the same side that you
added H+ to. Change H+ and OH- to H2O. Add same amount
of OH- to the other side.
5. Balance by CHARGE
a. Determine total charge on each side of each half-reaction
b. Add electrons to the more positive side
c. Use a multiplier to get electrons in each reaction to be equal
6. Add half reactions together and cancel any terms that are
the same on opposite sides.
MnO4- + S2O3-2 + H+  Mn+2 + SO4-2 + H2O
Br2 + OH-  Br- + BrO3- + H2O
Fe+2 + Cr2O7-2  Fe+3 + Cr+3
(in acidic solution)
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S8  S2O3-2 + S-2
(in basic solution)
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