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