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Predicting the direction of redox reactions  Know that standard electrode potentials can be listed as an electrochemical series.  Use E values to predict.

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Presentation on theme: "Predicting the direction of redox reactions  Know that standard electrode potentials can be listed as an electrochemical series.  Use E values to predict."— Presentation transcript:

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 redox reactions and to calculate the e.m.f. of a cell.

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

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

5 –+0 – 0.76 V –ve electrode Zn e -  Zn V +ve electrode Cu e -  Cu V e–e– Cu 2+ + Zn → Cu + Zn 2+

6 USE OF Eo VALUES - WILL IT WORK? E° values Can 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

7 USE OF Eo VALUES - WILL IT WORK? What happens if an Sn(s) / Sn2+(aq) and a Cu(s) / Cu2+(aq) cell are connected? Write out the equationsCu2+(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 isCu2+(aq) + Sn(s) ——> Sn2+(aq) + Cu(s) the cell voltage is the difference in E° values... (+0.34) - (-0.14) = V An equation with a more positive E° value reverse a less positive one

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) 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° = V 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) = V 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

9 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 potentialsCu2+(aq) + 2e¯ Cu(s); E° = +0.34V and the usual equationsSn2+(aq) + 2e¯ Sn(s); E° = V 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

10 Predicting redox reactions 5.3 exercise 2

11 –+0 – 0.76 V –ve electrode Zn e -  Zn – 0.25 V +ve electrode Ni e -  Ni V e–e– Ni 2+ + Zn → Ni + Zn 2+ PREDICTING REDOX REACTIONS – Q1

12 V –ve electrode Cu e -  Cu V +ve electrode Ag + + e -  Ag V e–e– 2 Ag + + Cu → 2 Ag + Cu 2+ PREDICTING REDOX REACTIONS – Q2

13 0 – 2.36 V –ve electrode Mg e -  Mg – 0.26 V +ve electrode V 3+ + e -  V V e–e– Mg(s)|Mg 2+ (aq)||V 3+ (aq),V 2+ (aq)|Pt(s) PREDICTING REDOX REACTIONS – Q3 a – YES: Mg reduces V 3+ to V 2+

14 V –ve electrode Fe 3+ + e -  Fe V +ve electrode Cl e -  2 Cl V e–e– PREDICTING REDOX REACTIONS – Q3 b + NO: Cl - won’t reduce Fe 3+ to Fe 2+

15 V –ve electrode V +ve electrode Cl e -  2 Cl V e–e– PREDICTING REDOX REACTIONS – Q3 c + YES: Cl 2 oxidises Br - to Br 2 Br e -  2 Br - Pt(s)|Br - (aq),Br 2 (aq)||Cl 2 (g)|Cl - (aq)|Pt(s)

16 0 – 0.14 V –ve electrode Sn e -  Sn V +ve electrode Fe 3+ + e -  Fe V e–e– PREDICTING REDOX REACTIONS – Q3 d – YES: Sn reduces Fe 3+ to Fe 2+ + Sn(s)|Sn 2+ (aq)||Fe 3+ (aq),Fe 2+ (aq)|Pt(s)

17 V –ve electrode Cr 2 O H e -  2 Cr H 2 O V +ve electrode Cl e -  2 Cl V e–e– PREDICTING REDOX REACTIONS – Q3 e + NO: H + /Cr 2 O 7 2- won’t oxidise Cl - to Cl 2

18 V –ve electrode MnO H e -  Mn H 2 O V +ve electrode Cl e -  2 Cl V e–e– PREDICTING REDOX REACTIONS – Q3 f + YES: H + /MnO 4 - oxidises Cl - to Cl 2 Pt(s)|Cl - (aq)|Cl 2 (g)||MnO 4 - (aq),H + (aq),Mn 2+ (aq)|Pt(s)

19 0 – 0.44 V –ve electrode Fe e -  Fe 0.00 V +ve electrode 2 H e -  H V e–e– PREDICTING REDOX REACTIONS – Q3 g – YES: H + oxidises Fe to Fe 2+ Fe(s)|Fe 2+ (aq)||H + (aq)|H 2 (g)|Pt(s)

20 V –ve electrode Cu e -  Cu V +ve electrode 2 H e -  H V e–e– PREDICTING REDOX REACTIONS – Q3 h + NO: H + won’t oxidise Cu to Cu 2+

21 V MnO H e -  Mn H 2 O V Cl e -  2 Cl - PREDICTING REDOX REACTIONS – Q V Cr 2 O H e -  2 Cr H 2 O V Fe 3+ + e -  Fe 2+ YES NO

22 + 0 ? V –ve electrode Be e -  Be V +ve electrode Cu e -  Cu V e–e– Be 2+ + Cu → Be + Cu 2+ PREDICTING REDOX REACTIONS – Q5a 2.19 = E  left E  left = 0.34 – 2.19 = – 1.85 V

23 – 0 ? V –ve electrode Th e -  Th V +ve electrode 1.90 V e–e– 4 H + + Th → 2 H 2 + Th 4+ PREDICTING REDOX REACTIONS – Q5b When using SHE E  = cell emf = – 1.90 V 2 H e -  H 2

24 V –ve electrode V +ve electrode V e–e– PREDICTING REDOX REACTIONS – Q6a + Br e -  2 Br - Pt(s)|H 2 (g)|H + (aq)||Br 2 (aq),Br - (aq)|Pt(s) 2 H e -  H 2 H 2 + Br 2 → 2 H Br -

25 V –ve electrode V +ve electrode V e–e– PREDICTING REDOX REACTIONS – Q6b + Fe 3+ + e -  Fe 2+ Cu(s)|Cu 2+ (aq)||Fe 3+ (aq),Fe 2+ (aq)|Pt(s) Cu e -  Cu 2 Fe 3+ + Cu → 2 Fe 2+ + Cu 2+


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