slide1 n.
Download
Skip this Video
Loading SlideShow in 5 Seconds..
Predicting the direction of redox reactions PowerPoint Presentation
Download Presentation
Predicting the direction of redox reactions

Loading in 2 Seconds...

play fullscreen
1 / 25

Predicting the direction of redox reactions - PowerPoint PPT Presentation


  • 181 Views
  • Uploaded on

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

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about 'Predicting the direction of redox reactions' - tamber


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.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
slide1

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.
golden rule
GOLDEN RULE

The more +ve electrode gains electrons

(+ charge attracts electrons)

slide4

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

slide5

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+

slide6

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

slide7

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

slide8

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

slide9

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

slide11

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+

slide12

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+

slide13

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

slide14

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+

slide15

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

slide16

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

slide17

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

slide18

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

slide19

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

slide20

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

slide21

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

slide22

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+

slide23

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+

slide24

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-

slide25

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+