JEE Redox Reactions Notes PDF

Welcome to Redox Reactions Classical Concept of Oxidation 01 Addition of Oxygen 02 Removal of Hydrogen 03 Loss of electron 04 Addition of an electronegative element 05 Removal of an electropositive element 2Mg(s) + O2(g) 2H2S(g) + O2(g) Zn(s) 2MgO(s) 2S(s) + 2H2O(l) Zn2+(aq) + 2e- Classical Concept...

Welcome to Redox Reactions Classical Concept of Oxidation 01 Addition of Oxygen 02 Removal of Hydrogen 03 Loss of electron 04 Addition of an electronegative element 05 Removal of an electropositive element 2Mg(s) + O2(g) 2H2S(g) + O2(g) Zn(s) 2MgO(s) 2S(s) + 2H2O(l) Zn2+(aq) + 2e- Classical Concept of Reduction 01 Addition of Hydrogen 02 Removal of Oxygen 03 Gain of electron 04 Addition of an electropositive element 05 Removal of an electronegative element C2H4(g) + H2(g) 2HgO(s) △ Cu2+(aq) + 2e− C2H6(g) 2Hg(l) + O2(g) Cu(s) Oxidation Number It is an imaginary or apparent charge developed over the atom of an element when it goes from its elemental free state to combined state in a molecule. It is based on the assumption that a complete transfer of an electron takes place from a less electronegative atom to a more electronegative atom. Concept of Oxidation Oxidation Process in which there is an increase in the oxidation number of the element from reactant to product in chemical reaction. Increase in oxidation number 0 H 2 + O2 +1 2H2O Increase in oxidation number 0 2Na + ½O2 +1 Na2O Concept of Reduction Reduction Process in which there is a decrease in the oxidation number of the element from reactant to product in chemical reaction. Decrease in oxidation number +2 MgH2 (s) 0 Mg (s) + H2 (g) Decrease in oxidation number 1 0 2Na (s) + O2 (g) 2 -2 Na2O (s) Rules for Finding Oxidation Number For single monatomic ions, the oxidation number is equal to the charge on the ion. O.N. of Naᐩ = Ca2+ = Negative O.N. Less E.N. = Positive O.N. +1 Examples H+ More E.N. Cl ̶ Rules for Finding Oxidation Number In general, the oxidation number of halogens (F, Cl, Br, I) is − 1 If any halogen atom is attached to a more electronegative atom, then it will have a positive oxidation number. Fluorine is the most E.N. atom (known). It always has an oxidation number equal to -1 in all its compounds. Examples Cl2O ClF3 HIO4 HCIO4 Rules for Finding Oxidation Number Generally, the oxidation number of hydrogen is +1 O.N. of H = +1 In Metal - Hydrides (As metals are electropositive) O.N. of H H2S -1 Examples Examples HCl = H2O LiH NaH CaH2 Rules for Finding Oxidation Number In neutral compounds, the sum of oxidation numbers of all the atoms of different elements is equal to zero. In complex ions or polyatomic ions, the sum of oxidation numbers of all the atoms is equal to the net charge on the ion. For elements in their native or free state, the oxidation number is zero. Example: H2, N2, Cl2, Cu, Zn etc. Rules for Finding Oxidation Number The oxidation number of oxygen is –2 in oxides in most of its compounds (generally) The oxidation number of oxygen is –1 in Peroxides (O22– ) = Examples O.N. of O –2 Examples H 2O 2 H2O CO2 SO2 Na2O2 BaO2 Rules for Finding Oxidation Number The oxidation number of oxygen is – (½) in superoxides (O2 ) Examples KO2 RbO2 CsO2 The oxidation number of oxygen is +1 or +2 in oxy-fluorides depending upon the bonding state. Rules for Finding Oxidation Number In their compounds, the oxidation number of alkali metal is +1 and the oxidation number of alkaline earth metal is +2. Examples O.N. of Na+ = +1 O.N. of Mg2+ = +2 NaCl KF CaO Rules for Finding Oxidation Number Alkali Metals (1st Group) Alkaline Earth Metals (2nd Group) Lithium (Li) Beryllium (Be) Sodium (Na) Magnesium (Mg) Potassium (K) Calcium (Ca) Rubidium (Rb) Strontium (Sr) Caesium (Cs) Barium (Ba) Range of Oxidation Numbers Generally, Oxidation Number Maximum O.N. Minimum O.N. = = Electrons in valence shell Electrons in valence shell - 8 Range of Oxidation Numbers Generally, for p-block elements n - 18 ≤ Oxidation number ≤ n: Group number of element in modern periodic table n - 10 O.S. of Some Oxygen Containing Compounds H2SO5 (Peroxymonosulphuric acid) Calculated O.S. = 8 Maximum O.S. = 6 O.S. of Some Oxygen Containing Compounds H2S2O8 Calculated O.S. = 7 Maximum O.S. = 6 O.S. of Some Oxygen Containing Compounds Oxidation number of some oxygen containing compounds Case 1 Calculated O.N. > Maximum O.N. Calculated O.N. = 7 Maximum O.N. = 6 H2S2O8 Case 2 Calculated Maximum = O.N. O.N. Calculated O.N. = 6 Maximum O.N. = 6 SO3 Case 3 Calculated O.N. < Maximum O.N. Calculated O.N.= 4/3 Actual O.N. = 2,0 C3O2 Paradox of Fractional Oxidation Number In certain compounds, the oxidation number of a particular element is a fraction and not a whole number The actual oxidation numbers of these elements are whole numbers, but the average of those can be fractional. Paradox of Fractional Oxidation Number C3O2 3x - 4 = 0 Carbon suboxide ⇒ Average Oxidation number of Carbon x= = 4 3 4 3 Paradox of Fractional Oxidation Number + Average oxidation number of C atoms 4 3 + 4 3 + 4 3 The actual oxidation number of the elements can be determined on the basis of the bonds present in their molecules. Actual oxidation number of C atoms +2 +0 +2 Paradox of Fractional Oxidation Number (S4O6)2− 4x - 12 = -2 Tetrathionate ion ⇒ Average Oxidation number of Sulphur x= = 5 2 5 2 Paradox of Fractional Oxidation Number +2.5 +2.5 Average oxidation number of S atoms +2.5 +2.5 +5 0 Actual oxidation number of S atoms +5 0 Oxidising and Reducing Agent Oxidising Agent/Oxidant Reducing Agent/Reductant Substance which oxidises others and itself gets reduced Substance which reduces others and itself gets oxidised Examples Examples KMnO4, H2O2, K2Cr2O7 Na2S2O3, H2O2 Oxidising and Reducing Agent Reduction +2 Fe2+ (aq) + +7 MnO− 4 (aq) +3 Fe3+ (aq) Oxidation +2 +3 Fe2+ Fe3+ +7 +2 − MnO4 Mn2+ +2 + Mn2+(aq) Combination Decomposition Redox Reactions Displacement Disproportionation Comproportionation Combination Reactions A + B C (s) + O2 (g) CH4 (g) + 2O2 (g) C CO2 (g) CO2 (g) + 2H2O (l) Decomposition Reactions C 2 H2O (l) A + B 2H2 (g) + O2 (g) Displacement Reactions X + YZ Cu (s) + 2 AgNO3 (aq) Na (s) + 2 H2O (l) XZ + Y 2 Ag (s) + Cu(NO3)2 (aq) 2 NaOH (aq) + H2 (g) Disproportionation Reactions The redox reaction in which an element from the same oxidation state changes to two different oxidation states (one lower and other higher) Disproportionation Reactions Reduction 0 I2 (s) + NaOH (aq) −1 +5 NaI (aq) + NaIO3 (aq) Oxidation This reaction involves the change in oxidation state of I from (0) to higher (+5) and lower (−1) O.S Comproportionation Redox Reaction The redox reaction in which an element from two different oxidation states (one lower and other higher) changes to the same oxidation state It is reverse of disproportionation. Oxidation −1 I− +5 (aq) + IO − (aq) 3 Reduction 0 I2 (aq) Intramolecular Redox Reaction Oxidation +5 −2 2KClO3 (s) −1 0 2KCl (s) + 3O2 (g) Reduction In this case, oxidation and reduction takes place through a single molecule n - factor n-Factor It is defined as the change in oxidation state per molecule. It is also known as valence factor. H2SO3 H2SO4 Step 1 Find the oxidation state of the element on the reactant and the product side which changes its oxidation state in the reaction +4 H2SO3 Step 2 +6 H2SO4 Find the difference in oxidation states and take its magnitude. 6 - 4 = │2│ +4 +6 H2SO3 H2SO4 Step 3 If the number of atoms of an element is different between the reactant and the product side, balance them. Step 4 Then the balanced number should be multiplied with the difference in oxidation state to calculate the nf. 4─3 = 1 nf for H2C2O3 +3 H2C2O3 ⇒ 2x1 ⇒ 2 +4 CO2 Calculation of nf for different Species 1 For Ions 2 For Salts nf =│Charge│ ⇒ nf = 1 Na+, Cl- Al2(SO4)3 ⇒ nf = 6 ⇒ SO42- ⇒ ⇒ ⇒ Na+ NaCl Al3+, SO42- nf = 1 nf = 2 Calculation of nf for different Species 3 Oxy Acids of Phosphorus For Acids H3PO4 ⇒ nf = 3 3 −OH bond H3PO3 ⇒ nf = 2 2 −OH bond H3PO2 ⇒ nf = 1 1 −OH bond nf = Basicity Number of H+ ions an acid donates Remember!!! nf can’t be zero or negative nf can be a fraction! Balancing of Redox Reactions Balancing of Redox Reactions 1 Oxidation Number Method Atom Balance Methods 2 Charge Balance Ion Electron Method Oxidation Number Method Balancing of Redox Reactions Generally balancing is carried out in 2 media Acidic medium (H+/H2O) Basic medium (OH−/H2O) Oxidation Number Method Example MnO4− (aq) + Fe2+ (aq) Step 1 Mn2+ (aq) + Fe3+ (aq) Identifying the oxidizing/reducing agent Oxidation Number Method Oxidation Number increases by 1 unit +7 MnO4− +2 (aq) + Fe2+ (aq) Oxidation Number decreases by 5 units Oxidation +2 +3 Mn2+ (aq) + Fe3+ (aq) Reduction MnO4− is the oxidising agent and Fe2+ is the reducing agent Oxidation Number Method Step 2 +7 Finding the nf +2 MnO4− (aq) + Fe2+ (aq) +2 +3 Mn2+ (aq) + Fe3+ (aq) nf of MnO4− = │(7 − 2) × 1│ = 5 nf of Fe2+ = │(2 − 3) × 1│ = 1 Oxidation Number Method Step 3 Equalising the decrease/increase in oxidation number 1MnO4− (aq) + 5Fe2+ (aq) Mn2+ (aq) + Fe3+ (aq) Cross multiply the Oxidising and the Reducing Agents by 1 (nf of R.A.) and 5 (nf of O.A.) respectively on the reactant side. Oxidation Number Method Step 4 Balancing the elements Balance all the atoms except O and H, without changing the stoichiometric coefficients on the reactant side MnO4− (aq) + 5Fe2+ (aq) Mn2+ (aq) + 5Fe3+ (aq) Oxidation Number Method Step 5 Balancing the O atoms Balance the O atoms by adding H2O MnO4− (aq) + 5Fe2+ (aq) Mn2+ (aq) + 5Fe3+ (aq) + 4H2O (l) For an Acidic Medium As soon as we add H2O, we add twice the H+ ions on the opposite side. xH2O 2xH+ For an Acidic Medium MnO4− (aq) + 5Fe2+ (aq) + 8H+ (aq) Mn2+ (aq)+ 5Fe3+ (aq) + 4H2O (l) Charge on the reactant side = [ 1 × (−1) + 5 × (+2) + 8 × (+1) ] = +17 Charge on the product side = [1 × (+2) + 5 × (+3) + 4 × (0)] = +17 For a Basic Medium Example ClO–(aq) + CrO2– (aq) + OH–(aq) Cl–(aq) + CrO42– (aq) + H2O(l) Balance the whole reaction as in an acidic medium 3ClO− (aq) + 2CrO2─ (aq) + H2O (l) 3Cl− (aq) + 2CrO42─ (aq) + 2H+ (aq) For a Basic Medium Add to both sides the same number of OH─ as there are H+ 3ClO− + 2CrO2─ + H2O (l) ＋ 2OH─ 3Cl− (aq) + 2CrO42 ─ + 2H+ ＋ 2OH─ Combine H+ and OH─ to form H2O 3ClO−(aq) + 2CrO2─ (aq) + H2O (l) ＋ 2OH─(aq) 3Cl− (aq) + 2CrO42 ─ (aq)+ 2H2O(l) For a Basic Medium Cancel any H2O that you can 3ClO−(aq) + 2CrO2─ (aq)+ 2OH─(aq) Charge on reactant side = ─7 3Cl− (aq) + 2CrO42 ─ (aq)+ H2O (l) Charge on product side = ─7 Since the charges on both the sides are the same, this equation is overall perfectly balanced. Equivalent Mass Equivalent mass Or Gram Eq. mass = Molar Mass n - factor Earlier it was expressed by taking hydrogen, oxygen, chlorine and silver as the reference elements. Number of parts by mass of an element which reacts or displaces from a compound 1.008 parts by mass of hydrogen, 8 parts by mass of oxygen and 35.5 parts by mass of chlorine, is known as the equivalent mass of that element. Number of Gram Equivalents No. of Gram Eq. = = = = Given Mass Gram Equivalent Mass Given Mass Molar Mass/nf Given Mass Molar Mass Moles × nf × nf Normality Normality Number of gram equivalent of solute present in 1 litre of the solution = Number of gram equivalents of solute Volume of the solution (L) = Number of moles of solute = Molarity × nf Volume of the solution (L) × nf Law of Equivalence The law states that one equivalent of an element always combines with one equivalent of the other. In a chemical reaction, equivalents or milliequivalents of the reactants react in equal amount to give the same number of equivalents or milliequivalents of the products separately. n1 A + n2 B n 3 C + n4 D Eq. of A = Eq. of B = Eq. of C = Eq. of D Recall !!! Equivalence point is the point where: No. of equivalent of the analyte = No. of equivalent of the titrant Analyte: Solution with unknown concentration Titrant: Solution with known concentration In this titration, one oxidizing agent (O.A.) reacts with one reducing agent (R.A.) O.A. + R.A. Products Redox Titration At the equivalence point Equivalence of O.A. = Equivalence of R.A N1V1 = N2V2 N1 and V1= Normality and Volume of the Analyte Respectively N2 and V2= Normality and Volume of the titrant Respectively Titration using Potassium Permanganate Strong Oxidizing Agent For acidification, H2SO4 is generally used. Acts in both acidic as well as basic medium Self Indicator Use of HCl is avoided as KMnO4 oxidises HCl to Cl2 gas. Titration using Potassium Permanganate Purple Pale pink MnO4− (aq) + 5Fe2+ (aq) + 8H+ (aq) Mn2+ (aq) + 5Fe3+ (aq) + 4H2O (l) Beyond the equivalence point, pink color re-appears Titration Using Potassium Dichromate Oxidizing agent only in acidic medium External redox indicators are used in titrations. Example : Diphenylamine, Potassium Ferricyanide For acidification, H2SO4 and HCl can be used as K2Cr2O7 does not oxidize HCl at room temperature. K2Cr2O7 is less powerful oxidizing agent than KMnO4. Titration Using Potassium Dichromate Cr2O72− (aq) + 14H+ (aq) + 6e− 2Cr3+ (aq) + 7H2O (l) Orange K2Cr2O7 oxidises the diphenylamine indicator just after the equivalence point giving intense blue colour indicating the end point. Back Titration Reaction 1 Z + X (Excess) Product 1 Reaction 2 X (Remaining) + Y Product 2 Back Titration The substance or solution of unknown concentration (1) is made to react with known volume and concentration of intermediate reactant solution (2). 4 gm of contaminated chalk, CaCO3 (1) + 200 ml, 0.5N HCl (2) Back Titration The reaction goes past the equivalence point. The amount of intermediate reactant is in excess of that required for completing reaction with analyte. After completing the reaction with analyte, the resulting solution containing excess of intermediate reactant is titrated with known volume and concentration of titrant Let us assume 50 ml of 0.5N NaOH Back Titration N1V1 = N2V2 0.5 × V1 = 0.5 × 50 N1 and V1 are normality and volume of intermediate reactant N2 and V2 are normality and volume of titrant Volume of excess HCl, V1 m.eq. of excess HCl = 50 mL = m.eq. of titrant (NaOH) = 0.5 × 50 = 25 m.eq. Back Titration Determination of excess volume or excess m.eq. of intermediate reactant allows us to determine the volume or m.eq. of intermediate reactant which reacted with analyte m.eq. of Chalk = Total m.eq. of HCl m.eq. of Chalk = − m.eq. of excess HCl (0.5 × 200) − 25 Back Titration m.eq. of Chalk Mass of chalk E × 1000 = = m.eq. of HCl used for chalk = Mass of chalk × 1000 Mass of chalk = 3.75 g 50 75 = 75 This means chalk contained 0.25 gm of impurities in it. Double Indicator Titration NaOH Na2CO3 NaHCO3 Reaction 1 NaOH + HCl NaCl + H2O Reaction 2 Na2CO3 + HCl NaHCO3 + H2O Reaction 3 NaHCO3 + HCl H2CO3 + H2O Double Indicator Titration Suitable indicator is employed to identify completion of individual reaction. The indicator is selected based on the range of pH change at equivalence volume of the titrant. Indicator Color Transition Range Phenolphthalein (HPh) 8.3 − 10 Methyl Orange (MeOH) 3.2 − 4.4 Double Indicator Titration The reaction is completed in two stages. The first stage reaction is : Na2CO3 (aq) + HCl (aq) NaCl (aq) + NaHCO3 (aq) Titrant It means that pH value remains above 7. Intermediate Product (Basic Salt) Double Indicator Titration The second stage reaction is : NaHCO3 (aq) + HCl (aq) NaCl (aq) + CO2 (g) + H2O (l) Acidic pH of the solution further goes down. Remeber !!! Double Indicator Titration m.eq. of HCl = m.eq. of NaOH + m.eq. of Na2CO3 + meq. of NaHCO3 Indicator - Methyl Orange m.eq. of HCl = m.eq. of NaOH + m.eq. of Na2CO3 Indicator - Phenolphthalein Iodometric Titration Iodometric titration is carried out in two steps This titration is used to determine the strength or the concentration of an oxidising agent. nf = 2 Step 1 −1 I − (aq) + O.A 0 I2 (g) + Product unknown Oxidising agent oxidises I − to I2 Eq. of oxidising agent = Eq. of I2 The iodine liberated is then titrated with standard hypo solution (Na2S2O3) Step 2 nf = 2 0 2− I2 (g) + S2O3 (aq) −1 I − (aq) + S4O6 (aq) 2− Eq. of I2 = Eq. of S2O2− 3 In both the cases, the n-factor of I2 is the same Eq. of O.A. = Eq. of I2 = Eq. of S2O2− 3 Iodimetric Titration Iodimetric titration is carried out in a single step It is used to determine the strength or the concentration of a reducing agent, which is directly titrated with I2 Iodimetric Titration I2 (g) + R.A unknown I– (aq) + Product Both iodometric and iodimetric titrations are carried out in an acidic medium only, as I2 will undergo disproportion reaction in basic medium. I – (aq) + OH – (aq) I2 (g) + IO3– (aq) H2O2 H2O2 as reducing agent. Acidic Medium H2O2 O2 + 2H+ + 2e- Basic Medium H2O2 + 2OH- O2 + 2H2O + 2e- H2O2 as oxidising agent. Acidic Medium H2O2 + 2H+ + 2e- 2H2O Basic Medium H2O2 + 2e- 2OH- Volume Strength of H2O2 Decompostion of H2O2 : 2H2O2 (aq) Volume Strength of H2O2 2H2O (l) + O2 (g) The concentration of H2O2 is usually represented in term of volume (vol). If a sample of H2O2 is labelled as ‘x V’, it means that 1 vol of H2O2 solution gives x vol of O2 gas at STP on complete decomposition = N × 5.6 = M × 11.2 Redox Reactions in Electrochemistry Electrochemistry The area of chemistry concerned with the interconversion of chemical energy and electrical energy. Let's define some basic terms in Electrochemistry like Electrode , Electrode potential etc. Electrodes In order to pass the current through an electrolytic solution two rods or plates (metallic conductors) are always needed which are connected with the terminals of a battery. These rods are known as electrodes. Types of Electrodes Anode The electrode through which the electron leaves the electrolytic solution (positive electrode). Cathode The electrode through which the electron enter the electrolytic solution (negative electrode). Based on the tendency to get oxidised or reduced Electrolyte The substances whose aqueous solution allow the passage of electric current and is chemically decomposed into ions, are termed electrolytes. Zn (s) Zn2+ (aq) + 2e- Electrode Potential The potential difference developed between an electrode and an electrolyte in a solution. Unit: Volt Standard Electrode Potential Potential difference developed between metal electrodes and the solution of its ions at 1 M concentration at 1 bar pressure and at a particular temperature. By convention, Standard electrode potential of hydrogen electrode = 0.00 volts Types of Electrode Potential Electrode Potential Oxidation Potential (OP) Reduction Potential (RP) Oxidation Potential Electrode potential for oxidation half reaction M (s) Mn+ (aq) + ne− Indicates standard state Under standard conditions E 0M(s) / Mn+(aq) Termed as negative electrode Shorthand notation E M(s) / Mn+(aq) Standard Oxidation Potential (SOP) Example of Oxidation Potential H2O 2 ⇌ O2 + 2H+ + 2e− Cu (s) ⇌ Cu2+ (aq) + 2e− Oxidation half Oxidation half Representation Representation E o H2O2 / O2 o E Cu(s) / Cu2+(aq) = −0.34 V Example of Oxidation Potential Zn (s) ⇌ Zn2+ (aq) + 2e− Oxidation half Representation o E Zn(s) / Zn2+(aq) = +0.76 V Reduction Potential Electrode potential for reduction half reaction Mn+ (aq) + ne− M (s) Indicates standard state Under standard conditions o E Mn+(aq) / M(s) Termed as positive electrode Shorthand notation E Mn+(aq) / M(s) Standard Reduction Potential (SRP) Example of Reduction Potential Cu2+ (aq) + 2e− ⇌ Cu (s) Zn2+ (aq) + 2e− ⇌ Reduction half Reduction half Representation Representation o E Cu 2+(aq) / Cu(s) = 0.34 V Zn (s) o E Zn 2+(aq) / Zn(s) = −0.76 V Example of Reduction Potential H2O2 + 2H+ + 2e− ⇌ Reduction half Representation E o H2O2 / H2O 2H2O Electrochemical Cell It is a device for the interconversion of chemical energy & electrical energy. Electrochemical cells Galvanic cell Electrolytic cell Galvanic Cell A galvanic cell is an electrochemical cell that produces electricity as a result of the spontaneous reaction occurring inside it. Δ rG < 0 Example Daniel cell Chemical energy is converted into electrical energy. Electrolytic Cell Electrical energy is being used to carry out chemical reactions (chemical energy). An electric current drives a non-spontaneous reaction. Salt Bridge U-shaped tube containing a solution of KCl or NH4NO3 usually solidified by boiling with agar agar and later cooled to form a jelly like substance. Functions of Salt Bridge 01 02 03 Completes the circuit. Maintains electrical neutrality in the solution of both half cells. Provides anions to the oxidation half cell and cations to the reduction half cell. Redox Couple Redox couple is the combination of oxidized and reduced forms of the same substance taking part in oxidation or reduction half reaction. Represented as Ma+/M Eg: Zn2+/Zn and Cu2+/Cu Electrochemical Series A list of oxidizing agents arranged in the decreasing order of their standard reduction potential Reduced form) Eo / V F2 (g) + 2e- → 2F- 2.87 O2 (g) + 4H+ + 4e- → 2H2O 1.23 Cu2+ + 2e- → Cu (s) 2H+ + 2e- → H2 (g) Fe2+ + 2e- → Fe (s) Zn2+ + 2e- → Zn (s) Li+ + e- → Li (s) Increasing strength of reducing agent Increasing strength of oxidizing agent Reaction (Oxidized form + ne- 0.34 0.00 -0.44 -0.76 -3.05 Zn Rod in CuSO4 Solution Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) This reaction is spontaneous as Reduction potential of Cu2+ is more than Zn2+ Applications of Electrochemical Series 01 Predicting the feasibility of the redox reaction. Calculation of EMF of the cells. 03 Comparison of strength of oxidising and reducing agent. 02