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ASSIGNMENTS
For each of the following, separate the skeletal (unbalanced) equation into two half reactions. For each half reaction,
balance the elements (mass balance), and then add electrons to the right or left side to make a net charge balance.
Identify which half reaction is the oxidation and which is the reduction. Then, multiply each half reaction by an
appropriate factor so that the two multiplied half reactions add together to make a balanced redox equation.
1.Hg2
+2+S2O2
–3→ Hg + S4O2
–6 Hg2
2++ S2O3
2–→ Hg+S4O6
2–
2. Al + Cr3+→Al3++Cr2+ Al+ Cr3+→Al3+ + Cr2+
3. Au3+ + I– → Au + I2
4.
5. a) Mn2+ (aq) + BiO3
- (aq) ⇒ MnO4
- (aq) + Bi3+(aq) [acidic]
b) Fe(CN)6
3- (aq) + Re(s) ⇒ Fe(CN)6
4- (aq) + ReO4
- (aq) [basic]
Balance I– + MnO4
2 –→ IO–3+MnO2I–+MnO4
2– → IO3– + MnO2 in basic aqueous solution.
©DSM](https://image.slidesharecdn.com/chem101fall2020redoxisdl2-200710124919/85/Oxidation-and-reduction-Balancing-the-redox-rections-9-320.jpg)
Oxidation and reduction, Balancing the redox rections
- 1. 1 Oxidation and reduction Sheikh Mahatabuddin, Ph. D. Associate Professor Department of NFE Daffodil International University smuddin.nfe@diu.edu.bd mahatab.chem@gmail.com
- 2. 2 Oxidation and Reduction Oxidation: (Classical Concepts) Addition of oxygen => C + O2 ⇒ CO2 Addition of electronegative elements => Na + Cl2 ⇒ NaCl Removal of electropositive elements => Zn + CuSO4 ⇒ ZnSO4 + Cu Removal of Hydrogen => Na + HCl ⇒ NaCl + H2 Electronic Concept: Removal or Donation of electrons: Na => Na+ + e- Reduction: (Classical Concepts) Removal of oxygen => Fe2O3 + CO ⇒ Fe + CO2 Removal of electronegative elements => H2S + Cl2 ⇒ HCl + S Addition of electropositive elements => Zn + CuSO4 ⇒ ZnSO4 + Cu Addition of Hydrogen => H2 + Cl2 ⇒ HCl Electronic Concept: Addition or Acceptance of electrons: Cl + e - => Cl - ©DSM
- 3. 3 Fe2O3 + CO ⇒ Fe + CO2 Reduction Oxidizing agent Oxidation Reducing agent 1. O removed from Fe2O3 => Reduction of Fe2O3 2. On the other hand, Fe2O3 oxidize CO therefore Fe2O3 is a Oxidizing agent! 1. O is added to CO=> Oxidation of CO 2. On the other hand, Fe2O3 Reduced to Fe therefore CO is a Reducing agent Red Oxi Oxi Red Reducing agent get oxidized to reduce the oxidizing agent Oxidizing agent get reduced to oxidize the reducing agent Reducing agents: All metals, NaBH4, LiAlH4, H2 Oxidizing agents: Nonmetals except H2, KMnO4, K2Cr2O7, H2O2 Oxidation Reduction Reactions Takes Place Simultaneously Na => Na+ + e- Oxidation Cl + e- => Cl- Reduction ©DSM
- 4. 4 Oxidation Number: Oxidation number, also called Oxidation State, the total number of electrons that an atom either gains (-ve) or loses (+ve) in order to form a chemical bond with another atom. Rules for calculating oxidation number: 1. The oxidation number of an atom is zero in a neutral substance that contains atoms of only one element. Thus, the atoms in O2, O3, P4, S8, and aluminum metal all have an oxidation number of 0. 2. The oxidation number of simple ions is equal to the charge on the ion. The oxidation number of sodium in the Na+ ion is +1, for example, and the oxidation number of chlorine in the Cl- ion is -1. 3. The oxidation number of hydrogen is +1 when it is combined with a nonmetal as in CH4, NH3, H2O, and HCl. 4. The oxidation number of hydrogen is -1 when it is combined with a metal as in. LiH, NaH, CaH2, and LiAlH4. 5. The metals in Group IA form compounds (such as Li3N and Na2S) in which the metal atom has an oxidation number of +1. 6. The elements in Group IIA form compounds (such as Mg3N2 and CaCO3) in which the metal atom has a +2 oxidation number. ©DSM
- 5. 5 7. Oxygen usually has an oxidation number of -2. Exceptions include molecules and polyatomic ions that contain O-O bonds, such as O2, O3, H2O2, and the O2 2- ion. 8. The elements in Group VIIA often form compounds (such as AlF3, HCl, and ZnBr2) in which the nonmetal has a -1 oxidation number. 9. The sum of the oxidation numbers in a neutral compound is zero. H2O: 2(+1) + (-2) = 0 10. The sum of the oxidation numbers in a polyatomic ion is equal to the charge on the ion. The oxidation number of the sulfur atom in the SO4 2- ion must be +6, for example, because the sum of the oxidation numbers of the atoms in this ion must equal -2. SO4 2-: (+6) + 4(-2) = -2 11. Elements toward the bottom left corner of the periodic table are more likely to have positive oxidation numbers than those toward the upper right corner of the table. Sulfur has a positive oxidation number in SO2, for example, because it is below oxygen in the periodic table. SO2: (+4) + 2(-2) = 0 FeCl2 + Cl2 ⇒ FeCl3 Fe oxidation number changes to +2 to +3 Oxidation Cl oxidation number changes to 0 to -1 Reduction Increase in +ve oxidation number Decrease in –ve oxidation number Increase in – ve oxidation number Decrease in +ve oxidation number ©DSM
- 6. 6 Find out the oxidation and reducing agents from following reactions:- 1. Mg + PbO MgO + Pb 2. CuO + C Cu + CO2 3. Fe2O3 + C Fe + CO 4. Mg + CuO MgO + Cu 5. Mg + O2 MgO 6. Al + Fe2O3 Fe + Al2O3 7. CuO + H2 Cu + H2O 8. Cu + AgNO3 CuNO3 + Ag 9. Ce4+ + Fe2+ Ce3+ + Fe3+ 10. Sn4+ + Cr2+ Sn2+ + Cr3+ ©DSM
- 7. 7 The method used to balance redox reactions is called the Half Equation Method. The equation is separated into two half-equations; one for oxidation and one for reduction. Each equation is balanced by adjusting coefficients and adding H2O, H+, and e- in this order: 1. Balance elements in the equation other than O and H. 2. Balance the oxygen atoms by adding the appropriate number of water (H2O) molecules to the opposite side of the equation. 3. Balance the hydrogen atoms (including those added in step 2 to balance the oxygen atom) by adding H+ ions to the opposite side of the equation. 4. Add up the charges on each side. Make them equal by adding enough electrons (e-) to the more positive side. (Rule of thumb: e- and H+ are almost always on the same side.) 5. The e- on each side must be made equal; if they are not equal, they must be multiplied by appropriate integers (the lowest common multiple) to be made the same. 6. The half-equations are added together, canceling out the electrons to form one balanced equation. Common terms should also be canceled out. • (If the equation is being balanced in a basic solution, through the addition of one more step, the appropriate number of OH- must be added to turn the remaining H+ into water molecules.) • The equation can now be checked to make sure that it is balanced. Balancing Oxidation-Reduction reactions: ©DSM
- 8. 8 Balance the following redox reaction in acidic conditions. FeSO4 (aq) + Cr2O7 2- (aq) → Cr3+ (aq) + Fe3+ (aq) Here the change of oxidation state: Fe2+ Fe3+ increase in positive oxidation number, therefore oxidation Cr6+ Cr3+ Decrease in positive oxidation number, therefore reduction Reduction half reaction Cr2O7 2- + 6e- => 2Cr3+ + 7H2O – (i) Oxidation half reaction Fe2+ => Fe3+ + e- as we need 6 e- while the oxidation process provide 1e- therefore multiplying the oxidation half reaction by 6 => 6 Fe2+ => 6 Fe3+ + 6 e- ------ (ii) Addition of equation i with ii Cr2O7 2- + 6e- => 2Cr3+ + 7H2O 6 Fe2+ => 6 Fe3+ + 6 e- (Addition) Cr2O7 2- + 6 Fe2+ = 2 Cr3+ + 6 Fe3+ + 7H2O---(iii) As it was a acidic solution and FeSO4 used as oxidizing agent, therefore to provide opposite ion to the equation iii K2Cr2O7 + 6 FeSO4 + 7 H2SO4 = K2SO4 + Cr2(SO4)3 + 3 Fe2(SO4)3 + 7H2O After providing opposite ions the balanced equation will be-- ©DSM
- 9. 9 ASSIGNMENTS For each of the following, separate the skeletal (unbalanced) equation into two half reactions. For each half reaction, balance the elements (mass balance), and then add electrons to the right or left side to make a net charge balance. Identify which half reaction is the oxidation and which is the reduction. Then, multiply each half reaction by an appropriate factor so that the two multiplied half reactions add together to make a balanced redox equation. 1.Hg2 +2+S2O2 –3→ Hg + S4O2 –6 Hg2 2++ S2O3 2–→ Hg+S4O6 2– 2. Al + Cr3+→Al3++Cr2+ Al+ Cr3+→Al3+ + Cr2+ 3. Au3+ + I– → Au + I2 4. 5. a) Mn2+ (aq) + BiO3 - (aq) ⇒ MnO4 - (aq) + Bi3+(aq) [acidic] b) Fe(CN)6 3- (aq) + Re(s) ⇒ Fe(CN)6 4- (aq) + ReO4 - (aq) [basic] Balance I– + MnO4 2 –→ IO–3+MnO2I–+MnO4 2– → IO3– + MnO2 in basic aqueous solution. ©DSM
- 10. 10 Dependence of the electropositivity and oxidation potential for the displacement of any metal ions in a solution The electrochemical series Equilibrium Eo (Volts) Li (aq) + e- ⇌ Li (s) - 3.03 K+ (aq) + e- ⇌ K (s) - 2.92 Ca2+ (aq) + 2e- ⇌ Ca (s) - 2.87 Na+ (aq) + e- ⇌ Na (s) - 2.71 Mg2+ (aq) + 2e- ⇌ Mg (s) - 2.37 Al3+ (aq) + 3e- ⇌ Al (s) - 1.66 Zn2+ (aq) + 2e- ⇌ Zn (s) - 0.76 Fe2+ (aq) + 2e- ⇌ Fe (s) - 0.44 Pb2+ (aq) + 2e- ⇌ Pb (s) - 0.13 2 H+ (aq) + 2e- ⇌ Ca (s) 0 Cu2+ (aq) + 2e- ⇌ Cu (s) + 0.34 Ag+ (aq) + e- ⇌ Ag (s) + 0.80 Au3+ (aq) + 3e- ⇌ Au (s) + 1.50 A series of electrodes or half cells arranged in order of their increasing standard oxidation potentials or in the decreasing order of their standard reduction potentials is called an electromotive force series or electrochemical series is also known as electromotive force series e. m. f. series. ©DSM https://thefactfactor.com/facts/pure_science/chemistry/physical-chemistry/electrochemical-series/5877/
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