Oxidation Number

Oxidation Number

In chemistry, the oxidation number is used to keep track of the positive or negative character of atoms or ions. When electrons are removed or shifted away from an atom during a chemical reaction, the atom is given a more positive oxidation number.

When electrons are gained or shifted toward an atom during a reaction, the atom is given a more negative oxidation number. The actual numerical value of the oxidation number depends on the number of electrons shifted or transferred.

For simple ions the oxidation number is equal to the charge on the ion. In NaF, the sodium ion has an oxidation number of +1 because it lost one electron. The fluoride ion has an oxidation number of -1 because it gained one electron. The elements of Group I of the periodic table have oxidation numbers of +1 since they have one valence electron which they can lose. In compounds involving only two elements (binary compounds) the elements of Group VII of the periodic table usually have oxidation number of -1 since they have seven valence electrons and can gain one more electron to complete a stable octet.

In compounds the Group II elements have oxidation numbers of +2 and the elements of Group III often have oxidation numbers of +3. In binary compounds the elements of Group VI usually have oxidation number of -2 and the elements of Group V can have oxidation number of -3

In covalent substances, the more electronegative element is assigned a negative oxidation number, and the less electronegative element is assigned a positive oxidation number. In HCl, the hydrogen atom has an oxidation number of +1 due to a shift of the valence electron of the hydrogen atom towards the more electronegative chlorine atom. The chlorine atom has an oxidation number of -1.

In H2O, the oxygen atom is more electronegative than the hydrogen atoms. Each hydrogen atom has an oxidation number of +1 and the oxygen is -2. In NH3 the oxidation number of each hydrogen is +1 and the oxidation number is -3.

Many elements have more than one oxidation number depending on their various compounds. Iron has an oxidation number of +2 in FeCl2 and +3 in FeCl3.

Tin has an oxidation number of +2 and +4 in SnCl2 and SnCl4, respectively.

 

USUAL OXIDATION NUMBERS OF SOME COMMON ELEMENTS

ELEMENTS PERIODIC TABLE PLACEMENTS OXIDATION NUMBER
H   +1 (except -1 with metals)
Li Group I +1
Na Group I +1
K Group I +1
Ag Transition Metal +1
Mg Group II +2
Ba Group II +2
Ca Group II +2
Ni Transition Metal +2
Zn Transition Metal +2
Pb Group IV +2
Al Group III +2
Cl Group VII -1 (with metals and many nonmetals)
Br Group VII -1(with metals and many nonmetals)
I Group VII -1 (with metals and many nonmetals)
F Group VII -1
O Group VI -2 (except -1 in peroxides)
S Group VI -2 (also +4 and +6)
Hg Transition Metal +1 and +2
Cu Transition Metal +1 and +2
Fe Transition Metal +2 and +3
Co Transition Metal +2 and +3
Sn Group VI +2 and +4

 

Types of Solutions

Solutions

When a lump of sugar is placed in a cup of tea, the sugar disappears but it remains unchanged. We know that sugar is unchanged when it mixes with the water of the tea because we can still taste it. This is an example of a solution. The sugar has dissolved in the hot water of the tea.

A solution is a homogenous (same through-out) mixture of two or more substances. The composition of a solution can vary within certain limits. Two words that we should consider are solvent and solute. Generally we refer to the substance that is present in the larger quantity as the solvent. The solute is present in smaller amounts. In our example the sugar is the solute and the water is the solvent. The sugar dissolves in the water. In solutions where one substance is a liquid the other substance is a gas or solid, the liquid is usually the solvent.

Types of Solutions

Liquid solutions are made by dissolving solids, liquids or gases in liquids. Sugar in water is an example of a solution of a solid dissolved in a liquid. Salt water is another example of a common type of solution.

Gaseous solutions are made by dissolving a gas in another gas. All gases mix in all proportions to produce a homogenous solution. Air is an example of a gaseous solution, which is a solution of oxygen and many other gases dissolved in nitrogen.

A solid solution is formed when one solid substance is mixed with another solid substance to produce a homogenous mixture. Many alloys are solid solutions. For example brass is a solid solution in which zinc atoms have been mixed into solid crystal of copper atoms.

The enthalpy change for a reaction between a strong acid and a strong alkali

Aim / Objective: 

To determine the enthalpy change for a reaction between a strong acid and a strong alkali.

Introduction:  Enthalpy is defined as the total energy in a system.  The change in energy ∆H can be positive in heat absorbing (endothermic reactions) or negative in heat releasing (exothermic reactions). This experiment focuses on one form of enthalpy change which is enthalpy of neutralization (∆Hn).  Enthalpy of Neutralization is the enthalpy change observed when one mole of water is formed when a base reacts with an acid in a thermodynamic system.

The literature standard enthalpy for a strong acid-base reaction is -57.1kJ/mol. For weak acids and bases the heat of neutralization is different as they are not fully dissociated and hence some heat will be absorbed.

Materials/ Apparatus:

1.0M HCl solution, 1.0M NaOH solution, 2 measuring cylinders, 2 Styrofoam cups, 2 beakers, 1 thermometer, 1 glass stirring rod.

Method / Procedure:

  1. Use a measuring cylinder to measure 50cm3 of sodium hydroxide (NaOH) and pour into a Styrofoam cup.
  2. Use a thermometer to measure the temperature of the NaOH. Record the reading.
  3. Use a measuring cylinder to measure 50cm3 of hydrochloric acid (HCl) and pour into a Styrofoam cup.
  4. Use a thermometer to measure the temperature of the HCl. Record the reading.
  5. Mix the contents of both cups in a beaker and stir the contents using the glass stirring rod.
  6. Take the temperature after stirring the mixture then record the reading.
  7. Repeat steps 1 to 6 TWO times and then tabulate the results.

Suggested Results:

Experiment number Initial temperature / °C Final temperature / °C
NaOH HCl NaOH+ HCl
1 32 33.5 46
2 30 30.5 46
3 30 30.5 46

Discussion / Calculations:

Heat released = mcΔT

  1. Calculate mass

1000g = 1kg

1000g ÷1000

x = 1kg

50cm3 of acid and 50cm3 alkali = 100cm3 = 1kg

  1. Average temperature = (32+30+30+33.5+30.5+30.5) / 6

= 186.5/6

= 31.08°C

  1. Change in temperature (ΔT) = final temperature – initial temperature

= (46 – 31.08) °C

= 14.92°C

Change °C to K = 14.92 + 273 =287.92

  1. Specific heat capacity = 4.187JK-1kg-1
  1. Heat energy released = mcΔT

= 1kg x 4.187JK-1kg-1 x 287.92K

= 1205.52104J

  1. Enthalpy change = mcΔT / number of moles

50cm3 of acid & base contains (2/1000) x 50

=0.1 moles

0.1 moles H20 = 727.92J

1.0 moles H2O = 727.92J / 0.1 moles

= 7279.2J

Change from J to kJ

7279.2J / 1000 = 7.279kJ

Write the molecular and ionic equation for the reaction

NaOH(aq)   +  HCL(aq)                NaCl(aq) +  H2O(l)

H+(aq)   +    OH(aq)                 H2O(l)

Explain whether the reaction was exothermic or endothermic?

Heat was lost from the mixture to the environment can be said to be exothermic.

What was the use of the Styrofoam cups?

They were used to minimize heat loss of the mixture to the environment.