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ELECTROCHEMISTRY By Mr. Abdulwakil Daudi (Msc. BSc.Ed) Oxidation Numbers or Oxidation State.

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Presentation on theme: "ELECTROCHEMISTRY By Mr. Abdulwakil Daudi (Msc. BSc.Ed) Oxidation Numbers or Oxidation State."— Presentation transcript:

1 ELECTROCHEMISTRY By Mr. Abdulwakil Daudi (Msc. BSc.Ed) Oxidation Numbers or Oxidation State

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3 Oxidation Number Oxidation number or Oxidation state is an apparent charge on an element will have when in a compound. Oxidation number is assigned to an atom to indicate its degree of oxidation or reduction. Example:- Fe 2+ = +2; Al 3+ = +3; NH 4 + = +1 Cl - =-1; NO 3 - = -1; SO 4 2- = -2

4 Rules for Assigning Oxidation Numbers to Elements Rule 1: The oxidation number of an element in its uncombined state is zero. For examples: Simple elements:- Al(s), Zn(s), Ar (g). Ox. No., X=0 Simple Molecules:- Diatomic elements. Like H 2 (g), O 2 (g), N 2 (g), F 2 (g), Br 2 (l), I 2 (s). Ox. No. of X 2 = 0 and other molecules, S 8 =0, P 4 = 0 Rule 2: The oxidation number of a simple charged element is the same is the same as the charge on the ion, for example: Cations: Na + = +1; H + = +1; Ca 2+ = +2; Mg 2+ = +2; Al 3+ = +3; Cr 3+ = +3; Mn 7+ = +7 etc Anions: Cl - = -1; Br - = -1; O 2- = -2; S 2- = -2; N 3- = -3 etc

5 Rules for Assigning Oxidation Numbers to Elements Rule 3: The oxidation number of hydrogen in all compound is always +1 except in a binary metal hydride where the oxidation number of hydrogen is –1. Examples (H= +1): In H 2 O, H 2 S, HCl, NH 3, H 2 SO 4, NH 3 Note- Non-metal with Hydrogen Examples (H=-1): In hydrides like NaH, CaH 2, AlH 3, Note- Metallic hydrides. Rule 4: The oxidation number of oxygen in a compound always –2. Except in peroxides where its –1 and OF 2 where it is +2. Examples: Oxides (-2) NaOH, H 2 O, CO 2, H 2 SO 4, MgO etc Exceptions: Peroxides (-1) Na 2 O 2, H 2 O 2, BaO 2 and (+2) OF 2

6 Rules for Assigning Oxidation Numbers to Elements Rule 5: The sum of all oxidation numbers of all elements in a neutral compound is zero. E.g. NaOH= Na+O+H=(+1)+(-2)+(+1)=0 Rule 6: The sum of all oxidation numbers in a polyatomic (Radical) ion is equal to the overall charge on the ion. E.g. SO 4 2- = S+4O=(+6)+4(-2)=+6-8= -2

7 Rules for Oxidation Number or Oxidation State ElementOxidation Number ExceptionExample Group 1 metals Always +1-Na +, K + Group 2 metals Always +2Ca 2+, Mg 2+ HalogensAlways -1F -, Cl -, Br - OxygenUsually -2Peroxide (-1), OF 2 (+2) Na 2 O (-2); Na 2 O 2 (-1) HydrogenUsually +1Metal hydrides (-1) H 2 O, NH 3 (+1); NaH (-1) CompoundsAlways sum = 0CO, Al 2 (SO 4 ) 3, Polyatomic ion Sum = chargePO 4 3- (-3); Al 3+ (+3)

8 Functions of Oxidation Numbers Naming inorganic compounds Tracking electron movement in redox reaction Calculation of oxidation number of an element in a compound or polyatomic ions.

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10 Calculation of oxidation number of elements in Neutral Compounds

11 Calculation of oxidation number of elements Polyatomic Ions

12 Exercises on calculations of oxidation numbers (i)NH 3 (ii)NaNO 2 (iii)Ca(NO 3 ) 2 (iv)MnO 2 (v) CO 2 (vi)CO (vii) NO (viii)NO 2 (ix) N 2 O (x) MgCO 3 (xi) K 2 Cr 2 O 7 (i)PH 4 + (ii)SO 4 2- (iii)Cr 2 O 7 2- (iv)ClO - (v)MnO 4 2- (vi)PO 4 3- (vii)NO 2 - (viii)ClO 3 - (ix)ClO 3 - (x) HCO 3

13 Tracking electron movement in redox reaction

14 Tracking electron movement in redox reaction

15 FURTHER REDOX REACTIONS Prepare a fresh solution of iron(II)sulphate and divided into two test tubes. In the 1 st tt, NaOH confirms presence of Fe 2+ ions. In the 2 nd tt, H 2 O 2 oxidizes Fe 2+ to Fe 3+ ions. In the oxidized 2 nd tt, NaOH confirms presence of Fe 3+ ions. The observations made and inference were recorded in the table below:

16 Results and Discussions

17 Results from Oxidation Experiment Observations and Ionic equation for 1 st test tube: Fe 2+ (aq) + 2OH - ➙ Fe(OH) 2 (s), green precipitate. Half equations on effect of acidified H 2 O 2 in 2 nd test tube. Oxidation- increase in oxidation number 2Fe 2+ (aq) ➙ 2Fe 3+ (aq) + 2e - (l) Reduction – decrease in oxidation number H 2 O 2(aq) + 2H + (aq) + 2e - ➙ 2H 2 O (l) Overall equation from the two halve equations in the above 2Fe 2+ (aq) + H 2 O 2(aq) + 2H + (aq) ➙ 2Fe 3+ (aq) + 2H 2 O (l) light-green Yellow Effect of NaOH on Fe 3+ in excess Fe 3+ (aq) + 3OH - ➙ Fe(OH) 3 (s), brown precipitate.

18 Determining Oxidation (reducing agent) or reduction (oxidizing agent) of substances

19 Redox in Terms of Oxidation Number

20 Exercise for Redox and Oxidation Number


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