Redox Titrations Lecture 2 - PDF

 Pharmaceutical Analytical Chemistry-2 Level I Students Redox Titrations [ Lecture 2 ]  Part# 1 Rules for assigning oxidation numbers Examples: ◼ In H O :...

 Pharmaceutical Analytical Chemistry-2 Level I Students Redox Titrations [ Lecture 2 ]  Part# 1 Rules for assigning oxidation numbers Examples: ◼ In H O : 2 The sum of the oxidation numbers of each atom in the water molecule must equal the net charge of the molecule: Two hydrogens = +2, One oxygen = -2, Net charge = 0 ◼ In SO 2- : 4 we can calculate the oxidation number of sulfur (S) as follows: Four oxygens = -8, Net charge = -2 If the oxidation number of S is X : X + (-8) = -2 so the oxidation number of S in SO42- equals +6 ◼ In S O 2- (tetrathionate): 4 6 we can calculate the oxidation number of sulfur (S) as follows: Six oxygens = -12, Net charge = -2 If the oxidation number of S is X : 4X + (-12) = -2 + 10 4X = +10 so the oxidation number of S in S O 2- equals = +2.5 4 6 4 2  Part# 1 BALANCING REDOX EQUATIONS WHY ???? Balancing redox equation is necessary because the net charge must be equal on both sides of the redox equation. Number of electrons lost = Number of electrons gained. There is no electrons entering or leaving the system during redox reactions. 3  Part# 1 2+ - Shortened way to balance a redox equation (e.g. Fe & MnO4 reaction) 1. Write the unbalanced equation for the reaction in ionic form to determine which substance will be oxidized and which will be reduced (i.e. determine the reactants and products): Fe2+ + MnO4- ⇌ Fe3+ + Mn2+ 2. Balance [if needed] the number of atoms that are oxidized or reduced (usually atoms other than H & O): Fe2+ + MnO4- ⇌ Fe3+ + Mn2+ 3. Determine the number of electrons lost or gained for each oxidized or reduced atom by noting the change in the oxidation number per atom and then multiply this by the total number of atoms. Then adjust the number of electrons lost or gained to be equal in both half-reactions by multiplying by appropriate coefficients as illustrated: {in this example, Fe is converted from (+2) to (+3) so it loses 1 electron while Mn is converted from (+7) to (+2) so it gains 5 electron} +7 +2 Fe2+ + MnO4- ⇌ Fe3+ + Mn2+ Modify in the left side 1 e- 5 e- 5 Fe2+ + MnO4- ⇌ Fe3+ + Mn2+ 4. Re-balance [if needed] the number of atoms reduced or oxidized (because the first balancing in step 2 is altered). 5 Fe2+ + MnO4- ⇌ 5 Fe3+ + Mn2+ Modify in the right side + - 5. Balance oxygens and hydrogens [if needed] by adding H or OH and then H O to the appropriate side of the 2 + reaction (for reactions in acidic medium, add H to the left side double the number of excess O in this side to produce equivalent amount of H O in the right side). 2 2+ - + 3+ 2+ 5 Fe + MnO4 + 8 H ⇌ 5 Fe + Mn + 4 H O2 +17 6. Verify that both sides contains the same types and numbers of atoms and have the same net charge. 4  Part# 1 2- - Shortened way to balance a redox equation (e.g. C O & MnO reaction) 2 4 4 1. Write the unbalanced equation for the reaction in ionic form to determine which substance will be oxidized and which will be reduced (i.e. determine the reactants and products): C O 2- + MnO - ⇌ CO + Mn2+ 2 4 4 2 2. Balance [if needed] the number of atoms that are oxidized or reduced (usually atoms other than H & O): C O 2- + MnO - ⇌ 2 CO + Mn2+ 2 4 4 2 3. Determine the number of electrons lost or gained for each oxidized or reduced atom by noting the change in the oxidation number per atom and then multiply this by the total number of atoms. Then adjust the number of electrons lost or gained to be equal in both half-reactions by multiplying by appropriate coefficients as illustrated: {in this example, each carbon is converted from (+3) to (+4) so each carbon loses 1 electron so the two carbons loses 2 electrons} +7 +2 +3 +4 C O 2- + MnO - ⇌ 2 C O + Mn2+ 2 4 4 2 Modify in the left side 2 e- 5 e- 5 C O 2- + 2 MnO - ⇌ 2 CO + Mn2+ 2 4 4 2 4. Re-balance [if needed] the number of atoms reduced or oxidized (because the first balancing in step 2 is altered). 5 C O 2- + 2 MnO - ⇌ 10 CO + 2 Mn2+ Modify in the right side 2 4 4 2 + - 5. Balance oxygens and hydrogens [if needed] by adding H or OH and then H O to the appropriate side of the 2 + reaction (for reactions in acidic medium, add H to the left side double the number of excess O in this side to produce equivalent amount of H O in the right side). 2 2- - + 2+ 5 C O2 4 + 2 MnO4 + 16 H ⇌ 10 CO + 2 Mn2 + 8 H O2 +4 6. Verify that both sides contains the same types and numbers of atoms and have the same net charge. 5  Part# 1 2- Shortened way to balance a redox equation (e.g. I & S O reaction) 2 2 3 1. Write the unbalanced equation for the reaction in ionic form to determine which substance will be oxidized and which will be reduced (i.e. determine the reactants and products): 2- - 2- I + S O ⇌ I + S O 2 2 3 4 6 2. Balance [if needed] the number of atoms that are oxidized or reduced (usually atoms other than H & O): 2- - 2- I + 2 S O ⇌ 2 I + S O 2 2 3 4 6 3. Determine the number of electrons lost or gained for each oxidized or reduced atom by noting the change in the oxidation number per atom and then multiply this by the total number of atoms. Then adjust the number of electrons lost or gained to be equal in both half-reactions by multiplying by appropriate coefficients as illustrated: {in this example, each I is converted from (0) to (-1) so each I gains 1 electron so the two I gains 2 electrons & each S is converted from (+2) to (+2.5) so each S loses 0.5 electron so the four S atoms (2 S ) loses 2 electrons} 2 0 +2 -1 +2.5 2- - 2- I + 2 S O ⇌ 2 I + S O 2 2 3 4 6 2 e- 2 e- 4. Re-balance [if needed] the number of atoms reduced or oxidized (because the first balancing in step 2 is altered). 2- - 2- I + 2 S O ⇌ 2 I + S O 2 2 3 4 6 + - 5. Balance oxygens and hydrogens [if needed] by adding H or OH and then H O to the appropriate side of the 2 + reaction (for reactions in acidic medium, add H to the left side double the number of excess O in this side to produce equivalent amount of H O in the right side). 2 2- - 2- I + 2 S O ⇌ 2 I + S O 2 2 3 4 6 -4 6. Verify that both sides contains the same types and numbers of atoms and have the same net charge. 6  Part# 1 Summary of common oxidation states of some important elements Element Studied oxidation states Fe 2+ Ferrous 3+ Ferric Fe Fe (Iron) or Iron (II) or Iron (III) Ce 3+ Cerous 4+ Ceric Ce Ce (Cerium) or Cerium (III) or Cerium (IV) Sn 2+ Stannous 4+ Stannic Sn Sn (Tin) or Tin (II) or Tin (IV) Hg Hg+ or Hg22+ Mercurous Hg2+ Mercuric (Mercury) or Mercury (I) or Mercury (II) Cu Cu+ or Cu 2+ Cuprous Cu2+ Cupric (Copper) 2 or Copper (I) or Copper (II) S 2- 2- S O Thiosulfate S O Tetrathionate (Sulfur) 2 3 4 6 7  Part# 1 Summary of common oxidation states of some important elements Element Studied oxidation states 2+ Manganous 3+ Manganic Mn Mn or Manganese (II) or Manganese (III) Mn - 2- MnO4 Permanganate MnO4 Manganate (Manganese) Manganese MnO 2 dioxide 2- 2- Cr O Dichromate CrO Chromate 2 7 4 Cr (Chromium) 3+ Chromic Cr or Chromium (III) 8 Thank You