1 / 24

Balancing Equations for Redox Reactions

Balancing Equations for Redox Reactions. Some redox reactions have equations that must be balanced by special techniques. MnO 4 - + 5 Fe 2+ + 8 H + ---> Mn 2+ + 5 Fe 3+ + 4 H 2 O. Mn = +7. Fe = +2. Mn = +2. Fe = +3. Reduction of VO 2 + with Zn. Balancing Equations.

blade
Télécharger la présentation

Balancing Equations for Redox Reactions

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Balancing Equations for Redox Reactions Some redox reactions have equations that must be balanced by special techniques. MnO4- + 5 Fe2+ + 8 H+---> Mn2+ + 5 Fe3+ + 4 H2O Mn = +7 Fe = +2 Mn = +2 Fe = +3

  2. Reduction of VO2+ with Zn

  3. Balancing Equations Balance the following in acid solution— VO2+ + Zn ---> VO2+ + Zn2+ Step 1: Write the half-reactions Ox Zn ---> Zn2+ Red VO2+ ---> VO2+ Step 2: Balance each half-reaction for mass. Ox Zn ---> Zn2+ Red 2 H++ VO2+ ---> VO2+ + H2O Add H2O on O-deficient side and add H+ on other side for H-balance.

  4. Balancing Equations Step 3: Balance half-reactions for charge. Ox Zn ---> Zn2+ + 2e- Red e- + 2 H+ + VO2+ ---> VO2+ + H2O Step 4: Multiply by an appropriate factor. Ox Zn ---> Zn2+ +2e- Red 2e-+ 4 H+ + 2 VO2+ ---> 2 VO2+ + 2 H2O Step 5: Add balanced half-reactions Zn + 4 H+ + 2 VO2+ ---> Zn2+ + 2 VO2+ + 2 H2O

  5. Tips on Balancing Equations • Never add O2, O atoms, or O2- to balance oxygen ONLY add H2O or OH-. • Never add H2 or H atoms to balance hydrogen ONLY add H+ or H2O. • Be sure to write the correct charges on all the ions. • Check your work at the end to make sure mass and charge are balanced. • PRACTICE!

  6. Using Standard Potentials, Eo • In which direction do the following reactions go? • Cu(s) + 2 Ag+(aq) ---> Cu2+(aq) + 2 Ag(s) • Goes right as written • 2 Fe2+(aq) + Sn2+(aq) ---> 2 Fe3+(aq) + Sn(s) • Goes LEFT opposite to direction written • What is Eonet for the overall reaction?

  7. Eo for a Voltaic Cell Cd --> Cd2+ + 2e- or Cd2+ + 2e- --> Cd Fe --> Fe2+ + 2e- or Fe2+ + 2e- --> Fe All ingredients are present. Which way does reaction proceed? Calculate Eo for this cell.

  8. E at Nonstandard Conditions • The NERNST EQUATION • E = potential under nonstandard conditions • n = no. of electrons exchanged • F = Faraday’s constant • R = gas constant • T = temp in Kelvins • ln = “natural log” • Q = reaction quotient

  9. Eo and Thermodynamics • Eo is related to ∆Go, the free energy change for the reaction. • ∆G˚ is proportional to –nE˚ ∆Go = -nFEo where F = Faraday constant = 9.6485 x 104 J/V•mol of e- (or 9.6485 x 104 coulombs/mol) and n is the number of moles of electrons transferred

  10. Eo and ∆Go ∆Go = - n F Eo For a product-favored reaction Reactants ----> Products ∆Go < 0 and so Eo > 0 Eo is positive For a reactant-favored reaction Reactants <---- Products ∆Go > 0 and so Eo < 0 Eo is negative

  11. Eo and Equilibrium Constant DGo = -RT ln K DGo = -nFEo

  12. Electrolysis Using electrical energy to produce chemical change. Sn2+(aq) + 2 Cl-(aq) ---> Sn(s) + Cl2(g)

  13. ElectrolysisElectric Energy ---> Chemical Change •  Electrolysis of molten NaCl. •  Here a battery “pumps” electrons from Cl- to Na+. •  NOTE:Polarity of electrodes is reversed from batteries.

  14. Electrolysis of Molten NaCl Figure 20.18

  15. Electrolysis of Molten NaCl Anode (+) 2 Cl- ---> Cl2(g) + 2e- Cathode (-) Na+ + e- ---> Na Eo for cell (in water) = - 4.07 V (in water) External energy needed because Eo is (-).

  16. Electrolysis of Aqueous NaI Anode (+): 2 I- ---> I2(g) + 2e- Cathode (-): 2 H2O + 2e- ---> H2 + 2 OH- Eo for cell = -1.36 V

  17. Electrolysis of Aqueous CuCl2 Anode (+) 2 Cl- ---> Cl2(g) + 2e- Cathode (-) Cu2+ + 2e- ---> Cu Eo for cell = -1.02 V Note that Cu2+ is more easily reduced than either H2O or Na+.

  18. Michael Faraday1791-1867 Originated the terms anode, cathode, anion, cation, electrode. Discoverer of • electrolysis • magnetic props. of matter • electromagnetic induction • benzene and other organic chemicals Was a popular lecturer.

  19. Quantitative Aspects of Electrochemistry Consider electrolysis of aqueous silver ion. Ag+ (aq) + e- ---> Ag(s) 1 mol e- ---> 1 mol Ag If we could measure the moles of e-, we could know the quantity of Ag formed. But how to measure moles of e-?

  20. Michael Faraday1791-1867 Quantitative Aspects of Electrochemistry But how is charge related to moles of electrons? = 96,500 C/mol e- = 1 Faraday

  21. Quantitative Aspects of Electrochemistry 1.50 amps flow through a Ag+(aq) solution for 15.0 min. What mass of Ag metal is deposited? Solution (a) Calc. charge Charge (C) = current (A) x time (t) = (1.5 amps)(15.0 min)(60 s/min) = 1350 C

  22. Quantitative Aspects of Electrochemistry 1.50 amps flow through a Ag+(aq) solution for 15.0 min. What mass of Ag metal is deposited? Solution (a) Charge = 1350 C (b) Calculate moles of e- used (c) Calc. quantity of Ag

  23. Quantitative Aspects of Electrochemistry The anode reaction in a lead storage battery is Pb(s) + HSO4-(aq) ---> PbSO4(s) + H+(aq) + 2e- If a battery delivers 1.50 amp, and you have 454 g of Pb, how long will the battery last? Solution a) 454 g Pb = 2.19 mol Pb b) Calculate moles of e- c) Calculate charge 4.38 mol e- • 96,500 C/mol e- = 423,000 C

  24. Quantitative Aspects of Electrochemistry The anode reaction in a lead storage battery is Pb(s) + HSO4-(aq) ---> PbSO4(s) + H+(aq) + 2e- If a battery delivers 1.50 amp, and you have 454 g of Pb, how long will the battery last? Solution a) 454 g Pb = 2.19 mol Pb b) Mol of e- = 4.38 mol c) Charge = 423,000 C d) Calculate time About 78 hours

More Related