© a2 046 20-jul-12. zn 2+ (aq) + 2 e – zn(s)

© www.chemsheets.co.uk A2 046 20-Jul-12

Zn2+(aq) + 2 e– Zn(s)

Zn

Zn Zn2+ + 2 e-

oxidation

Cu2+ + 2 e- Cureduction

- electrode

anodeoxidation

+ electrodecathode

reductionelectron flow

At this electrode the metal loses

electrons and so is oxidised to metal

ions.

These electrons make the electrode

negative.

At this electrode the metal ions gain

electrons and so is reduced to metal

atoms.

As electrons are used up, this makes the electrode positive.

Cu

Standard Conditions

Concentration 1.0 mol dm-3 (ions involved in ½ equation)

Temperature 298 K

Pressure 100 kPa (if gases involved in ½ equation)

Current Zero (use high resistance voltmeter)

S tandard H ydrogen E lectrode

Emf = E = Eright - Eleft

H2 at 100 kPa

o

o

o

o

o

o

o

o

o

o

o

o

salt bridge

1.0 M H+(aq)

Pt

temperature= 298 K

1.0 M Cu2+(aq)

V

Cu

high resistancevoltmeter

E = Eright

H2 at 100 kPa

o

o

o

o

o

o

o

o

o

o

o

o

salt bridge

1.0 M H+(aq)

Pt

temperature= 298 K

1.0 M Cu2+(aq)

V

Cu

high resistancevoltmeter

Pt(s) | H2(g) | H+(aq) || Cu2+(aq) | Cu(s)

Ni(s) | Ni2+(aq) || Sn4+(aq), Sn2+(aq) | Pt(s)

K(s) | K+(aq) || Mg2+(aq) | Mg(s)

ROOR

Emf = Eright - Eleft

ELECTRODE POTENTIALS – Q1

- 2.71 = Eright - 0

Eright = - 2.71 V

Emf = Eright - Eleft

ELECTRODE POTENTIALS – Q2

Emf = - 0.44 - 0.22

Emf = - 0.66 V

Emf = Eright - Eleft

ELECTRODE POTENTIALS – Q3

Emf = - 0.13 - (-0.76)

Emf = + 0.63 V

Emf = Eright - Eleft

ELECTRODE POTENTIALS – Q4

+1.02 = +1.36 - Eleft

Eleft = + 1.36 - 1.02 = +0.34 V

Emf = Eright - Eleft

ELECTRODE POTENTIALS – Q5

a) Emf = + 0.15 - (-0.25) = +0.40 Vb) Emf = + 0.80 - 0.54 = +0.26 Vc) Emf = + 1.07 - 1.36 = - 0.29 V

Emf = Eright - Eleft

ELECTRODE POTENTIALS – Q6

a) Eright = +2.00 - 2.38 = - 0.38 V

Ti3+(aq) + e- Ti2+(aq)

b) Eleft = -2.38 - 0.54 = - 2.92 V

K+(aq) + e- K(aq)c) Eright = - 3.19 + 0.27 = - 2.92 V Ti3+(aq) + e- Ti2+(aq)

ELECTRODE POTENTIALS – Q7

Emf = -0.76 - (-0.91) = +0.15 V

a) Cr(s) | Cr2+(aq) || Zn2+(aq) | Zn(s)

Emf = +0.77 - 0.34 = +0.43 V

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

Emf = +1.51 – 1.36 = +0.15 V

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

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

The more +ve electrode gains electrons

(+ charge attracts electrons)

– + 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+

– + 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–

Be + Cu2+ → Be2+ + Cu

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 SHE

E = 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+

Electrochemical cells

i appreciate that electrochemical cells can be used as a commercial source of electrical energy

j appreciate that cells can be non-rechargeable (irreversible), rechargeable and fuel cells

k be able to use given electrode data to deduce the reactions occurring in non-rechargeable and rechargeable cells and to deduce the e.m.f. of a cell

l understand the electrode reactions of a hydrogen-oxygen fuel cell and appreciate that a fuel cell does not need to be electrically recharged

m appreciate the benefits and risks to society associated with the use of these cells

Non-rechargeable (primary) cells – Zinc-carbon

-0.80 V Zn(NH3)22+ + 2 e- Zn + 2 NH3

+0.70 V 2 MnO2 + 2 H+ + 2 e- Mn2O3 + H2O

• Standard cell

• Short lifeDetermine: a) cell emf

b) overall reaction during discharge

Non-rechargeable (primary) cells – alkaline

-0.76 V Zn2+ + 2 e- Zn

+0.84 V MnO2 + H2O + e- MnO(OH) + OH-

• Longer lifeDetermine: a) cell emf

b) overall reaction during discharge

Non-rechargeable (primary) cells – lithium

• Very long life

• High voltageDetermine: a) cell emf b) overall reaction during discharge

Rechargeable (secondary) cells

• In non-rechargeable (primary) cells, the chemicals are used up so the voltage drops

• In rechargeable (secondary) cells the reactions are reversible – they are reversed by applying an external current.

• It is important that the products from the forward reaction stick to the electrodes and are not dispersed into the electrolyte.

Rechargeable (secondary) cells – Li ion

+0.60 V Li+ + CoO2 + e- LiCoO2

-3.00 V Li+ + e- Li

Determine: a) cell emf b) overall reaction during dischargec) overall reaction during re-charge

• Rechargeable

• Most common rechargeable cell

Rechargeable (secondary) cells – lead-acid

+1.68 V PbO2 + 3 H+ + HSO4- + 2 e- PbSO4 + 2 H2O

-0.36 V PbSO4 + H+ + 2 e- Pb + HSO4

-

• Used in sealed car batteries (6 cells giving about 12 V overall)

Determine: a) cell emf b) overall reaction during dischargec) overall reaction during re-charge

Rechargeable (secondary) cells – nickel-cadmium

+0.52 V NiO(OH) + 2 H2O + 2 e- Ni(OH)2 + 2 OH-

-0.88 V Cd(OH)2 + 2 e- Cd + 2 OH-

Determine: a) cell emf b) overall reaction during dischargec) overall reaction during re-charge

FUEL CELLS

+0.40 V O2 + 2 H2O + 4 e- 4 OH-

-0.83 V 2 H2O + 2 e- H2 + 2 OH-

Determine: a) cell emf b) overall reaction

• High efficiency (more efficient than burning hydrogen)

• How is H2 made?

• Input of H2/O2 to replenish so no need to recharge

From wikipedia (public domain)

Pros & cons of cells

+ portable source of electricity

Pros & cons of non-rechargeable cells

+ cheap, small

– waste issues

Pros & cons of rechargeable cells

+ less waste, cheaper in long run

– still some waste issues

Pros & cons of fuel cells

+ water is only product

– most H2 is made using fossil fuels, fuels cells expensive