Some Basic Concepts of Chemistry Redox Reaction

Chemistry SOME BASIC CONCEPTS OF CHEMISTRY REDOX REACTION Oxidising And Reducing Agent  Oxidising Agent or Oxidant Oxidising agents are those compounds which can oxidize others and reduce itself during the chemical reaction. Those reagents in which for an element, oxidation number decreases or which undergoes gain of electrons in a redox reaction are termed as oxidants. Ex. KMnO4, K2Cr2O7, HNO3, conc. H2SO4 etc are powerful oxidising agents.  Reducing Agent or Reductant Reducing agents are those compounds which can reduce other and oxidize itself during the chemical reaction. Those reagents in which for an element, oxidation number increases or which undergoes loss of electrons in a redox reaction are termed as reductants. Ex. KI, Na2S2O3 etc are the powerful reducing agents. Note: There are some compounds also which can work both as oxidising agent and reducing agent Ex. H2O2, NO2– HOW TO IDENTIFY WHETHER A PARTICULAR SUBSTANCE IS AN OXIDISING OR A REDUCING AGENT Find the oxidation state (O.S.) of the central atom If O.S.= (Maximum O.S.) It is oxidizing agent If O.S.= (Minimum O.S.) It is a reducing agent If O.S. = (Intermediate between maximum & minimum) It can act both as reducing agent & oxidising agent It can disproportionate as well 1 Chemistry  Redox Reaction A reaction in which oxidation and reduction simultaneously take place is called a redox reaction in all redox reactions, the total increase in oxidation number must be equal to the total decrease in oxidation number. Ex. +3 +2 +2 10 FeSO4 + 2KMnO+5 4 + 8H2 SO4 ⟶ 5Fe2 (SO4 )3 + 2MnSO4 + K 2 SO4 + 8H2 O  Disproportionation Reaction A redox reaction in which same element present in a particular compound in a definite oxidation state is oxidized as well as reduced simultaneously is a disproportionation reaction. Disproportionation reactions are a special type of redox reactions. One of the reactants in a disproportionation reaction always contains an element that can exist in at least three oxidation states. The element in the form of reacting substance is in the intermediate oxidation state and both higher and lower oxidation states of that element are formed in the reaction. For example: 0 −2 2H2 O−1 2 (aq) ⟶ 2H2 O(ℓ) + O 2 (g) 2− S08 (s) + 120H− (aq) ⟶ 4 S−2 (aq) + 2 S+2 2 O3 (aq) + 6H2 O(𝜆) CI20 (g) + 2OH − (aq) ⟶ Cl+1 O− (aq) + Cl−1− (aq) + H2 O(𝜆)  Consider The Following Reactions (a) 2KClO3 2KCl + 3O2 KClO3 plays a role of oxidant and reductant both. Here, Cl present in KClO 3 is reduced and O present in KClO3 is oxidized. Since same element is not oxidized and reduced, so it is not a disproportionation reaction, although it looks like one. (b) NH4NO2 N2 + 2H2O Nitrogen in this compound has -3 and +3 oxidation number, which is not a definite value. So, it is not a disproportionation reaction. It is an example of comproportionating reaction, which is a class of redox reaction in which an element from two different oxidation state gets converted into a single oxidation state. (c) +5 −1 4KClO3 → 3 → 3KClO4 + KCl It is a case of disproportionation reaction and Cl atom is disproportional ting. 2 Chemistry List of Some Important Disproportionation Reactions 1. H2 O2 ⟶ H2 O + O2 2. X2 + OH − (dil.) ⟶ X − + XO− (X = Cl, Br, I) 3. x2 + OH − (conc.) ⟶ x + XO− 3 F2 does not undergo disproportionation as it is the most electronegative element. 1. F2 + NaOH( dil.) ⟶ F − + OF2 2. F2 + NaOH( conc.) ⟶ F − + O2 3. (CN)2 + OH − ⟶ CN − + OCN− 4. P4 + OH − ⟶ PH3 + H2 PO− 2 5. S8 + OH − ⟶ S2− + S2 O2− 3 6. MnO4 7. NH2 OH ⟶ N2 O + NH3 8. NH2 OH ⟶ N2 + NH3 9. Oxyacid’s of Phosphorus (+1, +3 oxidation number) 10. H3 PO2 ⟶ PH3 + H3 PO3 11. H3 PO3 ⟶ PH3 + H3 PO4 10. Oxyacid’s of Chlorine (Halogens)(+1, +3, +5 Oxidation number) 2− ⟶ MnO− 4 + MnO2 ClO− ⟶ Cl− + ClO− 2 − − ClO− 2 ⟶ Cl + ClO3 − − ClO− 3 ⟶ Cl + ClO4 11. HNO2 ⟶ NO + HNO3 Reverse of disproportionation is called Comproportionating. In some of the disproportionation reactions, by changing the medium (from acidic to basic or reverse), the reaction goes in backward direction and can be taken as an example of Comproportionating reaction. + I− + IO− 3 + H ⟶ I2 + H 2 O BALANCING OF REDOX REACTIONS All balanced equations must satisfy two criteria. 1. Atom Balance (Mass Balance) There should be the same number of atoms of each kind on reactant and product side. 3 Chemistry 2. Charge Balance The sum of actual charges on both sides of the equation must be equal. There are two methods for balancing the redox equations: 1. Oxidation - number change method 2. Ion electron method or half-cell method Since First method is not very much fruitful for the balancing of redox reactions, students are advised to use second method (Ion electron method) to balance the redox reactions Ion electron method: By this method redox equations are balanced in two different medium. (a) Acidic medium (b) Basic medium Balancing in Acidic Medium Students are advised to follow the following steps to balance the redox reactions by Ion electron method in acidic medium Ex. Balance the following redox reaction: FeSO4 + KMnO4 + H2 SO4 ⟶ Fe2 (SO4 )3 + MnSO4 + H2 O + K 2 SO4 Sol. Step–I Assign the oxidation number to each element present in the reaction. ++ + + Step II: Now convert the reaction in Ionic form by eliminating the elements or species, which are not undergoing either oxidation or reduction. Fe2+ +⟶ Fe3+ + Mn2+ Step III: Now identify the oxidation / reduction occuring in the reaction undergoes reduction . Fe2+ + MnO− 4 Fe3+ + Mn2+ undergoes oxidation. Step IV: Spilt the Ionic reaction in two half, one for oxidation and other for reduction. Fe2+ Step V: oxidation ⟶ Fe3+ Reduction MnO4-   Mn 2+ Balance the atom other than oxygen and hydrogen atom in both half reactions Fe2+ ⟶ Fe3+ 2+ MnO− 4 ⟶ Mn 4 Chemistry Fe & Mn atoms are balanced on both side. Step VI: Now balance O & H atom by H2O & H+ respectively by the following way: For one excess oxygen atom, add one H2O on the other side and two H+ on the same side. Step VII: Fe2+ ⟶ Fe3+ (no oxygen atom) .................(i) 2+ 8H + + MnO− + 4H2 O 4 ⟶ Mn ................(ii) Equation (i) & (ii) are balanced atomwise. Now balance both equations chargewise. To balance the charge, add electrons to the electrically positive side. Fe2+ oxidation ⟶ Fe3+ + e− 5e− + 8H + + MnO− 4 Step VIII: ................(1) Reduction ⟶ Mn2+ + 4H2 O ................(2) The number of electrons gained and lost in each half -reaction are equalized by multiplying both the half reactions with a suitable factor and finally the half reactions are added to give the overall balanced reaction. Here, we multiply equation (1) by 5 and (2) by 1 and add them: Fe2+ ⟶ Fe3+ + e− .........(1) × 5 2+ 5e− +8H+ +MnO− 4 ⟶Mn +4H2 O ........(2) ×1 3+ 2− 5Fe2+ +8H+ +MnO− 4 ⟶5Fe +Mn +4H2 O (Here, at his stage, you will get balanced redox reaction in Ionic form) Step IX: Now convert the Ionic reaction into molecular form by adding the elements or species, which are removed in step (2). Now, by some manipulation, you will get: 5 1 5FeSO4 + KMnO4 + 4H2 SO4 ⟶ 2 Fe2 (SO4 )3 + MnSO4 + 4H2 O + 2 K 2 SO4 or 10FeSO4 + 2KMnO4 + 8H2 SO4 ⟶ 5Fe2 (SO4 )3 + 2MnSO4 + 8H2 O + K 2 SO4 5 Chemistry Balancing in Basic Medium In this case, except step VI, all the steps are same. We can understand it by the following example: Ex. Balance the following redox reaction in basic medium: 2− − − ClO− + CrO− 2 + OH ⟶ Cl + CrO4 + H2 O Sol. By using up to step V, we will get: +1 ClO− Reduction ⟶ Cl− ⬚+3 CrO−3 2 0xidation ⟶ ⬚+6 CrO2− 4 Now, students are advised to follow step VI to balance ‘O’ and ‘H’ atom. + 2H + + ClO− ⟶ Cl− + H2 O | 2H2 O + CrO2 − ⟶ CrO2− 4 + 4H Now, since we are balancing in basic medium, therefore add as many as OH – on both side of equation as there are H+ ions in the equation. 2OH − + 2H + + ClO− ⟶ Cl− + H2 O + 2OH − 2− + − 4OH − + 2H2 O + CrO− 2 ⟶ CrO4 + 4H + 4OH Finally, you will get Finally you will get H2 O + ClO− ⟶ Cl− + 2OH − ...........(i) 4OH– + CrO2– CrO42– + 2H2O ........... (ii) Now see equation (i) and (ii) in which O and H atoms are balanced by OH – and H2O Now from step VIII Adding: 2e− + H2 O + ClO− ⟶ Cl− + 2OH − ............. (i) ×3 4OH − + CrO− 2 ............. (ii) ×2 − ⟶ CrO2− 4 + 2H2 O + 3e ––––––––––––––––––––––––––––––––––––––– 2− − − 3ClO− + 2CrO− 2 + 2OH ⟶ 3Cl + 2CrO4 + H2 O 6