Post on 16-Dec-2015
transcript
Electrode Potentials
know the IUPAC convention for writing half-equations for electrode reactions.
Know and be able to use the conventional representation of cells.
Know that standard electrode potential, E , refers to conditions of 298 K, 100 kPa and 1.00 mol dm−3 solution of ions.
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)
These give a diagrammatic representation of what is happening in a cell.
• Place the cell with the more positive E° value on the RHS of the diagram.
Cu2+(aq) + 2e¯ Cu(s) E° = + 0.34V put on the RHS
Zn2+(aq) + 2e¯ Zn(s) E° = - 0.76V put on the LHS
ZINC IS IN CONTACT THE SOLUTIONS A SOLUTION OFWITH A SOLUTION ARE JOINED VIA A COPPER IONS IN OF ZINC IONS SALT BRIDGE TO CONTACT WITH COPPER
CELL DIAGRAMSCELL DIAGRAMS
Zn Zn2+ Cu2+ Cu
These give a diagrammatic representation of what is happening in a cell.
• Place the cell with the more positive E° value on the RHS of the diagram.
Cu2+(aq) + 2e¯ Cu(s) E° = + 0.34V put on the RHS
Zn2+(aq) + 2e¯ Zn(s) E° = - 0.76V put on the LHS
• Draw as shown… the cell reaction goes from left to right
• the zinc metal dissolves Zn(s) ——> Zn2+(aq) + 2e¯ OXIDATION
• copper is deposited Cu2+(aq) + 2e¯ ——> Cu(s) REDUCTION
• oxidation takes place at the anode
• reduction at the cathode
CELL DIAGRAMSCELL DIAGRAMS
Zn Zn2+ Cu2+ Cu
+_
CELL DIAGRAMSCELL DIAGRAMS
These give a diagrammatic representation of what is happening in a cell.
• Place the cell with the more positive E° value on the RHS of the diagram.
Cu2+(aq) + 2e¯ Cu(s) E° = + 0.34V put on the RHS
Zn2+(aq) + 2e¯ Zn(s) E° = - 0.76V put on the LHS
• Draw as shown… the electrons go round the external circuit from left to right
• electrons are released when zinc turns into zinc ions
• the electrons produced go round the external circuit to the copper
• electrons are picked up by copper ions and copper is deposited
Zn Zn2+ Cu2+ Cu
V
+_
CELL DIAGRAMSCELL DIAGRAMS
These give a diagrammatic representation of what is happening in a cell.
• Place the cell with the more positive E° value on the RHS of the diagram.
Cu2+(aq) + 2e¯ Cu(s) E° = + 0.34V put on the RHS
Zn2+(aq) + 2e¯ Zn(s) E° = - 0.76V put on the LHS
• Draw as shown… the cell voltage is E°(RHS) - E°(LHS) - it must be positive
cell voltage = +0.34V - (-0.76V) = +1.10V
Zn Zn2+ Cu2+ Cu
V
+_
Ni(s) | Ni2+(aq) || Sn4+(aq), Sn2+(aq) | Pt(s)
K(s) | K+(aq) || Mg2+(aq) | Mg(s)
ROOR
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).
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)