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Redox reaction

Reduction	Loss Oxygen
	Gain Hydrogen
	Gain Electron
Oxidation	Gain Oxygen
	Loss Hydrogen
	Loss Electron

Following reaction is an example of redox reaction
The small green numbers written on the top are oxidation state

Oxidation state is " a measure of the number of electron an atom has gained or lost control of ".

Rules for oxidation state

1. The sum of oxidation state for a compound is zero.
   e.g. NaCl : Na has +1 oxidation state, Cl has -1 oxidation state.

2. The sum of oxidation state for ion is charge.
   e.g. CO₃^2-

3. Elements are always zero.
   e.g. Cl₂, H₂, Fe, Cu etc.....

4. Some elements have fixed oxidation number when they form a compound.
   • F is always 1^- (e.g. KF : oxidation of K is 1+ and oxidation of F is 1-)
   • O is always 2^- (Unless with F or exist as peroxide)
   • Cl is always 1^- (Unless with F or O)
   • H is always 1⁺ (Unless exist as metal hydride: e.g. NaH...... in this case oxidation od H is 1-)
   • Group 1 elements ( LI, Na, K, Pb, Cs ) are always 1+
   • Group2 elements (Be, Mg, Ca, Sr, Ba) are always 2+
   • Al is always 3+

Acid and Base

e.g. CH₃COOH (Acid) + H₂O (Base) « CH₃COO^- (Base) +H₃O⁺ (Acid)

Strength of an acid or a base

Note: Acids partially dissociate in water are weak acids. A good example is ethanoic acid (CH₃CH₂OOH)

Ionization of water

H₂O+H₂O « H₃O⁺ + OH^-
(H₃O⁺ = H⁺)

So it can be written as H₂O « H₃O⁺ + OH^-, let's express this in equilibrium equation.
K_c = [H⁺][OH^-]/[H₂O] rearrange this to
K_c[H₂O] = [H⁺][OH^-] call K_c[H₂O] as a new constant K_w
K_w = [H⁺][OH^-]

Experimentally the constant of water had been found and is 1.0 x 10^-14 mol²dm^-6 at 25^oC.
Thus K_w = [H⁺][OH^-] = 1.0 x 10^-14 mol²dm^-6 at 25^oC

Since water is neautral (i.e. amount of acid = amount of base)

[H⁺] = [OH^-]
[H⁺][OH^-] = 1.0x10^-14
X² = 1.0x10^-14
X = 1.0x10^-7
Thus [H⁺] = 1.0 x 10^-7
So the concentration of acid in water is 1.0 x 10^-7

pH

As we have calculated earlier, the concentration of acid in water is 1.0 x 10^-7 so if we put this into the PH equation
pH = -log [1.0 x 10^-7]
pH = 7

The table below shows how much H⁺ is present in each pH.

Acid
pH	0	1	2	3	4	5	6
Concentration of H+	1 x 100	1 x 10-1	1 x 10-2	1 x 10-3	1 x 10-4	1 x 10-5	1 x 10-6

Note: when calculating pH of an Alkaline, simply substitute the concentration of OH^- with H⁺ (i.e. pH = -log[OH^-])and subscribe by 14.
For instance 0.1 mole of N_aOH will be
pH = -log[0.1]
pH = 1.0
14-1=13
Thus, pH for 0.1 mole of N_aOH is 13.

Here is the table for Alkaline

Base
pH	14	13	12	11	10	9	8
Concentration of OH-	1 x 100	1 x 10-1	1 x 10-2	1 x 10-3	1 x 10-4	1 x 10-5	1 x 10-6

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