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Predicting the direction of redox reactions

Predicting the direction of redox reactions. Know that standard electrode potentials can be listed as an electrochemical series. Use E values to predict the direction of simple redox reactions and to calculate the e.m.f . of a cell. GOLDEN RULE. The more +ve electrode gains electrons

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Predicting the direction of redox reactions

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  1. Predicting the direction of redox reactions • Know that standard electrode potentials can be listed as an electrochemical series. • Use E values to predict the direction of simple redoxreactions and to calculate the e.m.f. of a cell.

  2. GOLDEN RULE The more +ve electrode gains electrons (+ charge attracts electrons)

  3. Electrodes with negative emf are better at releasing electrons (better reducing agents).

  4. – 0 + –ve electrode +ve electrode e– + 1.10 V + 0.34 V Cu2+ + 2 e- Cu – 0.76 V Zn2+ + 2 e- Zn Cu2+ + Zn → Cu + Zn2+

  5. USE OF Eo VALUES - WILL IT WORK? E° valuesCan be used to predict the feasibility of redox and cell reactions In theory ANY REDOX REACTION WITH A POSITIVE E° VALUE WILL WORK An equation with a more positive E° value reverse a less positive one

  6. USE OF Eo VALUES - WILL IT WORK? An equation with a more positive E° value reverse a less positive one What happens if an Sn(s) / Sn2+(aq) and a Cu(s) / Cu2+(aq) cell are connected? Write out the equations Cu2+(aq) + 2e¯ Cu(s) ; E° = +0.34V Sn2+(aq) + 2e¯ Sn(s) ; E° = -0.14V the half reaction with the more positive E° value is more likely to work it gets the electrons by reversing the half reaction with the lower E° value therefore Cu2+(aq) ——> Cu(s) and Sn(s) ——> Sn2+(aq) the overall reaction is Cu2+(aq) + Sn(s) ——> Sn2+(aq) + Cu(s) the cell voltage is the difference in E° values... (+0.34) - (-0.14) = + 0.48V

  7. USE OF Eo VALUES - WILL IT WORK? An equation with a more positive E° value reverse a less positive one Will this reaction be spontaneous?Sn(s) + Cu2+(aq) ——> Sn2+(aq) + Cu(s) Write out the appropriate equations Cu2+(aq) + 2e¯ Cu(s) ; E° = +0.34V as reductions with their E° values Sn2+(aq) + 2e¯ Sn(s) ; E° = - 0.14V The reaction which takes place will involve the more positive one reversing the other i.e. Cu2+(aq) ——> Cu(s) and Sn(s) ——> Sn2+(aq) The cell voltage will be the difference in E° values and will be positive...(+0.34) - (- 0.14) = + 0.48V If this is the equation you want then it will be spontaneous If it is the opposite equation (going the other way) it will not be spontaneous

  8. USE OF Eo VALUES - WILL IT WORK? An equation with a more positive E° value reverse a less positive one Will this reaction be spontaneous?Sn(s) + Cu2+(aq) ——> Sn2+(aq) + Cu(s) Split equation into two half equationsCu2+(aq) + 2e¯ ——> Cu(s) Sn(s) ——> Sn2+(aq) + 2e¯ Find the electrode potentials Cu2+(aq) + 2e¯ Cu(s) ; E° = +0.34V and the usual equationsSn2+(aq) + 2e¯ Sn(s) ; E° = - 0.14V Reverse one equation and its signSn(s) ——> Sn2+(aq) + 2e¯ ; E° = +0.14V Combine the two half equationsSn(s) + Cu2+(aq) ——> Sn2+(aq) + Cu(s) Add the two numerical values(+0.34V) + (+ 0.14V) = +0.48V If the value is positive the reaction will be spontaneous

  9. Predicting redox reactions • 5.3 exercise 2

  10. PREDICTING REDOX REACTIONS – Q1 – 0 + –ve electrode +ve electrode e– + 0.51 V –0.25 V Ni2+ + 2 e- Ni – 0.76 V Zn2+ + 2 e- Zn Ni2+ + Zn → Ni + Zn2+

  11. PREDICTING REDOX REACTIONS – Q2 0 + –ve electrode +ve electrode e– + 0.46 V + 0.80 V Ag+ + e- Ag + 0.34 V Cu2+ + 2 e- Cu 2 Ag+ + Cu → 2 Ag + Cu2+

  12. PREDICTING REDOX REACTIONS – Q3 a Mg(s)|Mg2+(aq)||V3+(aq),V2+(aq)|Pt(s) – 0 –ve electrode +ve electrode e– + 2.10 V – 0.26 V V3+ + e- V2+ – 2.36 V YES: Mg reduces V3+ to V2+ Mg2+ + 2 e- Mg

  13. PREDICTING REDOX REACTIONS – Q3 b + 0 –ve electrode +ve electrode e– + 0.59 V + 1.36 V Cl2 + 2 e- 2 Cl- + 0.77 V NO: Cl- won’t reduce Fe3+ to Fe2+ Fe3+ + e- Fe2+

  14. PREDICTING REDOX REACTIONS – Q3 c Pt(s)|Br-(aq),Br2(aq)||Cl2(g)|Cl-(aq)|Pt(s) + 0 –ve electrode +ve electrode e– + 0.27 V + 1.36 V Cl2 + 2 e- 2 Cl- + 1.09 V Br2 + 2 e- 2 Br- YES: Cl2 oxidises Br- to Br2

  15. PREDICTING REDOX REACTIONS – Q3 d Sn(s)|Sn2+(aq)||Fe3+(aq),Fe2+(aq)|Pt(s) – + 0 –ve electrode +ve electrode e– + 0.91 V + 0.77 V Fe3+ + e- Fe2+ – 0.14 V YES: Sn reduces Fe3+ to Fe2+ Sn2+ + 2 e- Sn

  16. PREDICTING REDOX REACTIONS – Q3 e + 0 –ve electrode +ve electrode e– + 0.03 V + 1.36 V + 1.33 V Cl2 + 2 e- 2 Cl- Cr2O72- + 14 H+ + 6 e- 2 Cr3+ + 7 H2O NO: H+/Cr2O72- won’t oxidise Cl- to Cl2

  17. PREDICTING REDOX REACTIONS – Q3 f Pt(s)|Cl-(aq)|Cl2(g)||MnO4-(aq),H+(aq),Mn2+(aq)|Pt(s) + 0 –ve electrode +ve electrode e– + 0.03 V + 1.51 V + 1.36 V MnO4- + 8 H+ + 5 e- Mn2+ + 4 H2O Cl2 + 2 e- 2 Cl- YES: H+/MnO4- oxidises Cl- to Cl2

  18. PREDICTING REDOX REACTIONS – Q3 g Fe(s)|Fe2+(aq)||H+(aq)|H2(g)|Pt(s) – 0 –ve electrode +ve electrode e– + 0.44 V 0.00 V 2 H+ + 2 e- H2 – 0.44 V YES: H+ oxidises Fe to Fe2+ Fe2+ + 2 e- Fe

  19. PREDICTING REDOX REACTIONS – Q3 h 0 + –ve electrode +ve electrode e– + 0.34 V + 0.34 V Cu2+ + 2 e- Cu 0.00 V NO: H+ won’t oxidise Cu to Cu2+ 2 H+ + 2 e- H2

  20. PREDICTING REDOX REACTIONS – Q4 + 0 + 1.36 V Cl2 + 2 e- 2 Cl- + 0.77 V + 1.51 V Fe3+ + e- Fe2+ MnO4- + 8 H+ + 5 e- Mn2+ + 4 H2O NO + 1.33 V YES Cr2O72- + 14 H+ + 6 e- 2 Cr3+ + 7 H2O NO

  21. PREDICTING REDOX REACTIONS – Q5a 0 + –ve electrode 2.19 = 0.34 - Eleft +ve electrode Eleft = 0.34 – 2.19 = – 1.85 V e– + 2.19 V + 0.34 V Cu2+ + 2 e- Cu ? V Be2+ + 2 e- Be Be2+ + Cu → Be + Cu2+

  22. PREDICTING REDOX REACTIONS – Q5b – 0 –ve electrode When using SHE +ve electrode E= cell emf = – 1.90 V e– 1.90 V + 0.00 V 2 H+ + 2 e- H2 ? V Th4+ + 4 e- Th 4 H+ + Th → 2 H2 + Th4+

  23. PREDICTING REDOX REACTIONS – Q6a Pt(s)|H2(g)|H+(aq)||Br2(aq),Br-(aq)|Pt(s) + 0 –ve electrode +ve electrode e– + 1.09 V + 1.09 V Br2 + 2 e- 2 Br- 0.00 V 2 H+ + 2 e- H2 H2 + Br2→ 2 H+ + 2 Br-

  24. PREDICTING REDOX REACTIONS – Q6b Cu(s)|Cu2+(aq)||Fe3+(aq),Fe2+(aq)|Pt(s) + 0 –ve electrode +ve electrode e– + 0.43 V + 0.77 V Fe3+ + e- Fe2+ + 0.34 V Cu2+ + 2 e- Cu 2 Fe3+ + Cu → 2 Fe2+ + Cu2+

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