Chemistry - Chapter 7 - Oxidation and Reduction

It is a type of bond where a hydrogen atom binds two strongly electronegative atoms. Example: H2O

 The metallic bond is the force of attraction between two metal atomd . Example: Gold, Aluminium

It is the weakest attraction between molecules or atoms.

It is a weak attractive intermolecular or interatomic force.

This is a reaction where one reactant is Oxidised and the other is Reduced. 

Oxidation = the loss of an electron 

Example : 

Cu -----> Cu2 + 2e- 

Reduction = the gaining of an electron

Example: 

S + 2e- -----> S2- 

Look for any H's, O's, or F's in the equation 

1) Any element just by itself not in a compound, their oxidation number is 0 

ex: H2's oxidation number is 0 

2) The oxidation number of H is always +1, unless its just by itself (see #1) 

3) The oxidation number of O is always -2, unless its just by itself (see #1) 

4) The oxidation number of F is always -1, unless its just by itself (see#1) 

5) The sum of oxidation numbers in a compound add up to 0 (when multiplied by the subscripts!!!) (see above example) 

6) The sum of oxidation numbers for a compound is the charge (for example, PO4 has a charge of (-3) SOO...  

...oxidation no. of O = -2. (there are 4 O's = -8 charge on that side ) P must have an oxidation number of 5. (-8+5= -3), and -3 is the total charge of the compound. 

1. Add H2O

2. Add H+ ions

3. Add electrons (e-)

Example: 

I- + 3H20   ------> IO3- + 6H + 6e-

1. Add OH-

2. Add H+ ions

3. Add electrons (e-)

Example: 

S2- + 4OH- ---> SO4 2- + 4H+  + 8e- 

Chlorine 

(H = +1, Cl = +7, O = -2) 

HClO4   = +7  

ClO-  3  = +5

ClO-     = +1

Cl2       = 0

Cl-       = -1

HCl       = -1

Sulfur

SO4 -2   = +6

SO3         = +6

SO3 -2  = +4

SO2      = +4

S         = 0

H2S      = -2

Mg 2+ (aq) + 2e- ---> Mg(s)    = -2.37V ....1

Ag+ + e- ----> Ag(s)            = +0.80V  .....2

1. The most reactive has to be flipped

Mg (s) ----> Mg 2+ (ag) + 2e-      = +2.37V .....1a2. Make the no. of electrons the same as the first one

2Ag+ + 2e- ----> 2Ag(s)            = 0+.80V .... 2a

3. Write the equation (add both reactions together)

2Ag+ + 2e- + Mg(s) ---> 2Ag(s) + Mg2+(ag) +2e-   (1a+2a)

                        = +2.37V +0.80V = +3.17V (fast reaction)