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CHEMISTRY II

CHEMISTRY II. Oxidation – Reduction REDOX. Let’s Get Defined. Oxidation: Loss of electrons ( gain in positive charge ) Reduction: Gain of electrons ( gain in negative charge )

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CHEMISTRY II

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  1. CHEMISTRY II Oxidation – Reduction REDOX

  2. Let’s Get Defined • Oxidation: Loss of electrons ( gain in positive charge ) • Reduction: Gain of electrons ( gain in negative charge ) • Redox: Oxidation – Reduction chemical reaction. The number of electrons gained is equal to the number of electrons lost.

  3. Net Ionic Equations • Redox equations are written as net ionic equations. That is all substances not changed during the reaction are not written. • Example • KMnO4 ----- > K+ + MnO4-1 • HCl ----- > H+ + Cl-1 • Redox reaction • K+ + MnO4-1 + H+ + Cl-1 --- > MnO2 + Cl2 + K+ + H+ • Net Ionic Equation is: • MnO4-1 + Cl-1 ---- > MnO2 + Cl2

  4. Let’s Get Balanced • Half Reaction Method • MnO4-1 + Cl-1 ---- > MnO2 + Cl2 • Step 1 : Divide the reaction into two half reactions • MnO4-1 ---- > MnO2 • Cl-1 ---- > Cl2 • 4H+ + MnO4-1 ---- > MnO2 + 2 H2O • 2 Cl-1 ---- > Cl2 Step 2: Balance each half RXN, for oxygen add water and for hydrogen add H+

  5. Step 3: Balance the charge by adding electrons • 3e- + 4H+ + MnO4-1 ---- > MnO2 + 2 H2O • 2 Cl-1 ---- > Cl2 + 2e- • Step 4: Make electrons gained = electrons lost • 2(3e- + 4H+ + MnO4-1 ---- > MnO2 + 2 H2O) • 3(2 Cl-1 ---- > Cl2 + 2e-) • Multiply and eliminate electrons • 6e- + 8H+ + 2MnO4-1 ---- > 2MnO2 + 4 H2O • 6 Cl-1 ---- > 3 Cl2 + 6e- • ------------------------------------------ • 8H+ + Cl-1 + 2MnO4-1 ---- > 2MnO2 + 3 Cl2 + 4 H2O Step 5: Add the equations and cancel anything that appears on both sides of the equation.

  6. Let’s do that again • Example 2 • Cr + NO2-1 ----> CrO2-1 + N2O2-2 • 1. Divide • Cr ---- > CrO2-1 • NO2-1 ---- > N2O2-2 • Balance • Cr ---- > CrO2-1 • 2NO2-1 ---- > N2O2-2 2H2O + + 4 H+ 4H+ + + 2H2O

  7. + 3e- • Balance Electrons • 2 H2O + Cr ---- > CrO2-1 + 4H+ • 4H+ + 2NO2-1 ---- > N2O2-2 + 2H2O • Electrons gain = Electrons lost • 4(2 H2O + Cr ---- > CrO2-1 + 4H+ + 3e-) • 3(4e- + 4H+ + 2NO2-1 ---- > N2O2-2 + 2H2O) • Multiply 8H2O + 4Cr ---- > 4CrO2-1 + 16H+ + 12e- 12e- + 12H+ + 6NO2-1 ---- > 3N2O2-2 + 6H2O 4e- +

  8. The Grand Finale Add and eliminate 8H2O + 4Cr ---- > 4CrO2-1 + 16H+ + 12e- 12e- + 12H+ + 6NO2-1 ---- > 3N2O2-2 + 6H2O ---------------------------------------- 2H2O + 4Cr + 6NO2-1--- > 4CrO2-1 + 3N2O2-2 + 4H+ 2 4

  9. Let’s Get Back to the Basics • All redox reactions that have been demonstrated thus far have been carried out in an acid media. That is why H+ has been used to balance hydrogen. If a reaction is carried out in a basic media, OH- and H2O are used to balance oxygen and hydrogen. • Example: Let’s take the previous reaction • Cr + NO2-1 ----> CrO2-1 + N2O2-2 • Balance this equation for a basic media. Basic

  10. Balance • Cr + NO2-1 ----> CrO2-1 + N2O2-2 • Divide • Cr ---- >CrO2-1 • NO2-1 ---->N2O2-2 • Let me give you a secret: Balance the equation as if it were an acid and then……………. Basic

  11. Same oh, Same oh • Cr ---- >CrO2-1 • 2NO2-1 ---->N2O2-2 • Now flip the water and hydrogen ion from one side to the other and change the H+ to OH-. Balance Charge • Cr ---- >CrO2-1 • 2NO2-1 ---->N2O2-2 2H2O + + 4H+ 4H+ + + 2H2O 4OH- + + 2H2O + 3e- 4e- + 2H2O + + 4OH-

  12. Electrons Gain = Electrons Lost • 4(4OH- + Cr ---- > CrO2-1 + 2H2O + 3e-) • 3(4e- + 2H2O + 2NO2-1 ---- > N2O2-2 + 4OH-) • -------------------------------------------- 12e-+16OH- +6H2O +6NO2-1+4Cr --- > 4CrO2-1+3N2O2-2+8H2O+12OH-+ 12e- • 4OH- + 6NO2-1 + 4Cr --- > 4CrO2-1 + 3N2O2-2 + 2H2O 4 2 Basic

  13. Oxidation Number Method • This is a method for balancing Redox equations that are not net ionic equations. • Example: • KMnO4 + HCl ---- > KCl + MnO2 + Cl2 + H2O • Step 1: Assign oxidation numbers to each element in the equation. • K+1Mn+7O4-2 + H+1Cl-1 ---- > K+1Cl-1 + Mn+4O2-2 + Cl20 + H2+1O-2

  14. Step 2: Identify the elements that have a change in charge. A gain in positive charge means a loss of electrons, thus Oxidation. A gain in negative charge means a gain of electrons, thus reduction. • K+1Mn+7O4-2 + H+1Cl-1 ---- > K+1Cl-1 + Mn+4O2-2 + Cl20 + H2+1O-2 • Step 3: Divide the elements that change into two half reactions and balance. • 3e- + Mn+7 ----- > Mn+4 • 2Cl- ---- > Cl2 + 2e- Gain 3 e- Reduction Lose 1 e- Oxidation

  15. 2 • Balance the electrons gained with the electrons lost and add the equations. • (3e- + Mn+7 ----- > Mn+4) • (2Cl- ---- > Cl2 + 2e-) 6e- + 2Mn+7 + 6Cl- ----- > 2 Mn+4 + 3Cl2 + 6e- Use the same coefficients from the half reactions in the full equation. 2KMnO4 + 6HCl ---- > KCl + 2MnO2 + 3Cl2 + H2O Balance the remaining by inspection. 2KMnO4 + 8HCl ---- > 2KCl + 2MnO2 + 3Cl2 + 4H2O 3

  16. Electrochemistry

  17. Voltaic Cells and Standard Reduction Potentials Mg+2 + 2e- --- > Mg (reduced) ,E0 = -2.37 v Cu --- > Cu+2 + 2e-(oxidized) ,E0 = -0.34 v Cu + Mg+2 --- > Mg + Cu+2,E0 = -2.71 v Salt Bridge Anode Cathode Cu Mg

  18. An Equation for Nerds E0 = Standard voltage from table. n = the number of electrons transferred • The Nerst Equation: • E = E0 -0.0591 x [Products] • n [Reactants] Cu + Mg+2 --- > Mg + Cu+2,E0 = -2.71 v If the Concentrations of [Cu+2] = 0.002 M and [Mg+2] = 0.001 M, what is the maximum voltaic energy that can be produced?

  19. Continue • Solve: Cu + Mg+2 --- > Mg + Cu+2,E0 = -2.71 v If the Concentrations of [Cu+2] = 0.002 M and [Mg+2] = 0.001 M, what is the maximum voltaic energy that can be produced? E = E0 -0.0591/n [Cu+2]/[Mg+2] = 2.71v -0.0591/2 [0.002]/[0.001] = 2.65 v

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