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.

Standard electrode potentials E/V

F2(g) + 2 e- 2 F-(aq) + 2.87

MnO42-(aq) + 4 H+(aq) + 2 e- MnO2(s) + 2 H2O(l) + 1.55

MnO4-(aq) + 8 H+(aq) + 5 e- Mn2+(aq) + 4 H2O(l) + 1.51

Cl2(g) + 2 e- 2 Cl-(aq) + 1.36

Cr2O72-(aq) + 14 H+(aq) + 6 e- 2 Cr3+(aq) + 7 H2O(l) + 1.33

Br2(g) + 2 e- 2 Br-(aq) + 1.09

Ag+(aq) + e- Ag(s) + 0.80

Fe3+(aq) + e- Fe2+(aq) + 0.77

MnO4-(aq) + e- MnO4

2-(aq) + 0.56

I2(g) + 2 e- 2 I-(aq) + 0.54

Cu2+(aq) + 2 e- Cu(s) + 0.34

Hg2Cl2(aq) + 2 e- 2 Hg(l) + 2 CI-(aq) + 0.27

AgCl(s) + e- Ag(s) + Cl-(aq) + 0.22

2 H+(aq) + 2 e- H2(g) 0.00

Pb2+(aq) + 2 e- Pb(s) - 0.13

Sn2+(aq) + 2 e- Sn(s) - 0.14

V3+(aq) + e- V2+(aq) - 0.26

Ni2+(aq) + 2 e- Ni(s) - 0.25

Fe2+(aq) + 2 e- Fe(s) - 0.44

Zn2+(aq) + 2 e- Zn(s) - 0.76

Al3+(aq) + 3 e- Al(s) - 1.66

Mg2+(aq) + 2 e- Mg(s) - 2.36

Na+(aq) + e- Na(s) - 2.71

Ca2+(aq) + 2 e- Ca(s) - 2.87

K+(aq) + e- K(s) - 2.93

Increasingreducing

power

Increasingoxidising

power

GOLDEN RULE

The more +ve electrode gains electrons

(+ charge attracts electrons)

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

– + 0

– 0.76 V

–ve electrode

Zn2+ + 2 e- Zn

+ 0.34 V

+ve electrode

Cu2+ + 2 e- Cu

+ 1.10 V

e–

Cu2+ + Zn → Cu + Zn2+

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

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 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 workit 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

An equation with a more positive E° value reverse a less positive one

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 differencein E° values and will be positive... (+0.34) - (- 0.14) = + 0.48V

If this is the equation you want then it will be spontaneousIf it is the opposite equation (going the other way) it will not be spontaneous

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 equations Cu2+(aq) + 2e¯ ——> Cu(s)

Sn(s) ——> Sn2+(aq) + 2e¯

Find the electrode potentials Cu2+(aq) + 2e¯ Cu(s) ; E° = +0.34V

and the usual equations Sn2+(aq) + 2e¯ Sn(s) ; E° = - 0.14V

Reverse one equation and its sign Sn(s) ——> Sn2+(aq) + 2e¯ ; E° = +0.14V

Combine the two half equations Sn(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

• Predicting redox reactions

• 5.3 exercise 2

– + 0

– 0.76 V

–ve electrode

Zn2+ + 2 e- Zn

– 0.25 V

+ve electrode

Ni2+ + 2 e- Ni

+ 0.51 V

e–

Ni2+ + Zn → Ni + Zn2+

PREDICTING REDOX REACTIONS – Q1

+ 0

+ 0.34 V

–ve electrode

Cu2+ + 2 e- Cu

+ 0.80 V

+ve electrode

Ag+ + e- Ag

+ 0.46 V

e–

2 Ag+ + Cu → 2 Ag + Cu2+

PREDICTING REDOX REACTIONS – Q2

0

– 2.36 V

–ve electrode

Mg2+ + 2 e- Mg

– 0.26 V

+ve electrode

V3+ + e- V2+

+ 2.10 V

e–

Mg(s)|Mg2+(aq)||V3+(aq),V2+(aq)|Pt(s)

PREDICTING REDOX REACTIONS – Q3 a

–

YES: Mg reduces V3+ to V2+

0

+ 0.77 V

–ve electrode

Fe3+ + e- Fe2+

+ 1.36 V

+ve electrode

Cl2 + 2 e- 2 Cl-

+ 0.59 V

e–

PREDICTING REDOX REACTIONS – Q3 b

+

NO: Cl- won’t reduce Fe3+ to Fe2+

0

+ 1.09 V

–ve electrode

+ 1.36 V

+ve electrode

Cl2 + 2 e- 2 Cl-

+ 0.27 V

e–

PREDICTING REDOX REACTIONS – Q3 c

+

YES: Cl2 oxidises Br- to Br2 Br2 + 2 e- 2 Br-

Pt(s)|Br-(aq),Br2(aq)||Cl2(g)|Cl-(aq)|Pt(s)

0

– 0.14 V

–ve electrode

Sn2+ + 2 e- Sn

+ 0.77 V

+ve electrode

Fe3+ + e- Fe2+

+ 0.91 V

e–

PREDICTING REDOX REACTIONS – Q3 d

–

YES: Sn reduces Fe3+ to Fe2+

+ Sn(s)|Sn2+(aq)||Fe3+(aq),Fe2+(aq)|Pt(s)

0

+ 1.33 V

–ve electrode

Cr2O72- + 14 H+ + 6 e- 2 Cr3+ + 7 H2O

+ 1.36 V

+ve electrode

Cl2 + 2 e- 2 Cl-

+ 0.03 V

e–

PREDICTING REDOX REACTIONS – Q3 e

+

NO: H+/Cr2O72- won’t oxidise Cl- to

Cl2

0

+ 1.36 V

–ve electrode

MnO4- + 8 H+ + 5 e- Mn2+ + 4 H2O

+ 1.51 V

+ve electrode

Cl2 + 2 e- 2 Cl-

+ 0.03 V

e–

PREDICTING REDOX REACTIONS – Q3 f

+

YES: H+/MnO4- oxidises Cl- to Cl2

Pt(s)|Cl-(aq)|Cl2(g)||MnO4- (aq),H+(aq),Mn2+(aq)|

Pt(s)

0

– 0.44 V

–ve electrode

Fe2+ + 2 e- Fe

0.00 V

+ve electrode

2 H+ + 2 e- H2

+ 0.44 V

e–

PREDICTING REDOX REACTIONS – Q3 g

–

YES: H+ oxidises Fe to Fe2+

Fe(s)|Fe2+(aq)||H+(aq)|H2(g)|Pt(s)

0

0.00 V

–ve electrode

Cu2+ + 2 e- Cu+ 0.34 V

+ve electrode

2 H+ + 2 e- H2

+ 0.34 V

e–

PREDICTING REDOX REACTIONS – Q3 h

+

NO: H+ won’t oxidise Cu to Cu2+

0

+ 1.36 V

MnO4- + 8 H+ + 5 e- Mn2+ + 4 H2O

+ 1.51 V

Cl2 + 2 e- 2 Cl-

PREDICTING REDOX REACTIONS – Q4

+

+ 1.33 V Cr2O7

2- + 14 H+ + 6 e- 2 Cr3+ + 7 H2O

+ 0.77 V Fe3+ + e- Fe2+

YES

NO

NO

+ 0

? V

–ve electrode

Be2+ + 2 e- Be

+ 0.34 V

+ve electrode

Cu2+ + 2 e- Cu

+ 2.19 V

e–

Be2+ + Cu → Be + Cu2+

PREDICTING REDOX REACTIONS – Q5a

2.19 = 0.34 - Eleft

Eleft = 0.34 – 2.19 = – 1.85 V

– 0

? V

–ve electrode

Th4+ + 4 e- Th

+ 0.00 V

+ve electrode

1.90 V

e–

4 H+ + Th → 2 H2 + Th4+

PREDICTING REDOX REACTIONS – Q5b

When using SHEE = cell emf = – 1.90 V

2 H+ + 2 e- H2

0

0.00 V

–ve electrode

+ 1.09 V

+ve electrode

+ 1.09 V

e–

PREDICTING REDOX REACTIONS – Q6a

+

Br2 + 2 e- 2 Br-

Pt(s)|H2(g)|H+(aq)||Br2(aq),Br-(aq)|Pt(s)

2 H+ + 2 e- H2 H2 + Br2 → 2 H+ + 2 Br-

0

+ 0.34 V

–ve electrode

+ 0.77 V

+ve electrode

+ 0.43 V

e–

PREDICTING REDOX REACTIONS – Q6b

+

Fe3+ + e- Fe2+

Cu(s)|Cu2+(aq)||Fe3+(aq),Fe2+(aq)|Pt(s)

Cu2+ + 2 e- Cu 2 Fe3+ + Cu → 2 Fe2+ + Cu2+