Cell Potential

1
Cell Potential

• L.O.
• Construct redox equations using half-equations
  or oxidation numbers.
• Describe how to make an electrochemical cell.

2

• An oxidation number is a measure of the number of
  electrons that an atom uses to bond with atoms of
  another element.

3

• Each element in a compound is given an oxidation
  number.

4
Species Oxidation number examples
Uncombined element 0 C, Na, O2, H2, P4, Cl2
Combined oxygen -2 H2O, CaO
Combined Hydrogen 1 NH3, H2S
Simple ion Charge on ion Na, Mg2, Cl-1
Combined fluorine -1 NaF, CaF2
5
Species Oxidation number examples
Combined group 1 1 NaCl
Combined group 2 2 MgCl2
6

• The sum of the oxidation numbers must equal the
  overall charge.

7
Oxidation states.

• F -1
• Cl -1
• O -2
• H 1

Mg in MgCl2?
? - 2 0
? 2
Put the in!
8
(No Transcript)
9
BALANCING REDOX HALF EQUATIONS
1 Work out formulae of the species before and
after the change balance if required 2 Work
out oxidation state of the element before and
after the change 3 Add electrons to one side
of the equation so that the oxidation states
balance 4 If the charges on the species (ions
and electrons) on either side of the equation do
not balance then add sufficient H ions to
one of the sides to balance the charges 5 If
equation still doesnt balance, add sufficient
water molecules to one side
Example 2 MnO4 being reduced to Mn2 in
acidic solution Step 1 MnO4 gt
Mn2
No need to balance Mn equal numbers
10
BALANCING REDOX HALF EQUATIONS
1 Work out formulae of the species before and
after the change balance if required 2 Work
out oxidation state of the element before and
after the change 3 Add electrons to one side
of the equation so that the oxidation states
balance 4 If the charges on the species (ions
and electrons) on either side of the equation do
not balance then add sufficient H ions to
one of the sides to balance the charges 5 If
equation still doesnt balance, add sufficient
water molecules to one side
Example 2 MnO4 being reduced to Mn2 in
acidic solution Step 1 MnO4 gt
Mn2 Step 2 7 2
Overall charge on MnO4 is -1 sum of the OSs of
all atoms must add up to -1 Oxygen is in its
usual oxidation state of -2 four oxygen atoms
add up to -8 To make the overall charge -1, Mn
must be in oxidation state 7 ... 7 (4x -2)
-1
11
BALANCING REDOX HALF EQUATIONS
1 Work out formulae of the species before and
after the change balance if required 2 Work
out oxidation state of the element before and
after the change 3 Add electrons to one side
of the equation so that the oxidation states
balance 4 If the charges on the species (ions
and electrons) on either side of the equation do
not balance then add sufficient H ions to
one of the sides to balance the charges 5 If
equation still doesnt balance, add sufficient
water molecules to one side
Example 2 MnO4 being reduced to Mn2 in
acidic solution Step 1 MnO4 gt
Mn2 Step 2 7
2 Step 3 MnO4 5e gt
Mn2
The oxidation states on either side are
different 7 gt 2 (REDUCTION) To
balance add 5 negative charges to the LHS
7 (5 x -1) 2 You must ADD 5 ELECTRONS
to the LHS of the equation
12
BALANCING REDOX HALF EQUATIONS
1 Work out formulae of the species before and
after the change balance if required 2 Work
out oxidation state of the element before and
after the change 3 Add electrons to one side
of the equation so that the oxidation states
balance 4 If the charges on the species (ions
and electrons) on either side of the equation do
not balance then add sufficient H ions to
one of the sides to balance the charges 5 If
equation still doesnt balance, add sufficient
water molecules to one side
Example 2 MnO4 being reduced to Mn2 in
acidic solution Step 1 MnO4 gt
Mn2 Step 2 7
2 Step 3 MnO4 5e gt
Mn2 Step 4 MnO4 5e 8H gt
Mn2
Total charges on either side are not equal LHS
1- and 5- 6- RHS
2 Balance them by adding 8 positive charges to
the LHS 6- (8 x 1) 2 You must
ADD 8 PROTONS (H ions) to the LHS of the equation
13
BALANCING REDOX HALF EQUATIONS
1 Work out formulae of the species before and
after the change balance if required 2 Work
out oxidation state of the element before and
after the change 3 Add electrons to one side
of the equation so that the oxidation states
balance 4 If the charges on the species (ions
and electrons) on either side of the equation do
not balance then add sufficient H ions to
one of the sides to balance the charges 5 If
equation still doesnt balance, add sufficient
water molecules to one side
Example 2 MnO4 being reduced to Mn2 in
acidic solution Step 1 MnO4 gt
Mn2 Step 2 7
2 Step 3 MnO4 5e gt
Mn2 Step 4 MnO4 5e 8H gt
Mn2 Step 5 MnO4 5e 8H gt
Mn2 4H2O now balanced
Everything balances apart from oxygen and
hydrogen O LHS 4 RHS 0 H LHS
8 RHS 0 You must ADD 4 WATER MOLECULES to the
RHS the equation is now balanced
14
5.3 Exercise 1
15
BALANCING REDOX HALF EQUATIONS
Q. Balance the following half equations... Na
gt Na Fe2 gt Fe3 I2
gt I C2O42- gt CO2 H2O2
gt O2 H2O2 gt
H2O NO3- gt NO NO3- gt
NO2 SO42- gt SO2
REMINDER 1 Work out the formula of the species
before and after the change balance if
required 2 Work out the oxidation state of the
element before and after the change 3 Add
electrons to one side of the equation so that the
oxidation states balance 4 If the charges on
all the species (ions and electrons) on either
side of the equation do not balance then add
sufficient H ions to one of the sides to balance
the charges 5 If the equation still doesnt
balance, add sufficient water molecules to one
side
16
BALANCING REDOX HALF EQUATIONS
Q. Balance the following half equations... Na
gt Na e- Fe2 gt Fe3
e- I2 2e- gt 2I
C2O42- gt 2CO2 2e- H2O2
gt O2 2H 2e-
H2O2 2H 2e- gt 2H2O NO3-
4H 3e- gt NO 2H2O NO3-
2H e- gt NO2 H2O SO42-
4H 2e- gt SO2 2H2O
17
COMBINING HALF EQUATIONS
A combination of two ionic half equations, one
involving oxidation and the other reduction,
produces a REDOX equation. The equations are
balanced as follows... Step 1 Write out the
two half equations Step 2 Multiply the
equations so that the number of electrons in each
is the same Step 3 Add the two equations and
cancel out the electrons on either side Step 4
If necessary, cancel any other species which
appear on both sides
18
COMBINING HALF EQUATIONS
A combination of two ionic half equations, one
involving oxidation and the other reduction,
produces a REDOX equation. The equations are
balanced as follows... Step 1 Write out the
two half equations Step 2 Multiply the
equations so that the number of electrons in each
is the same Step 3 Add the two equations and
cancel out the electrons on either side Step 4
If necessary, cancel any other species which
appear on both sides
The reaction between manganate(VII) and iron(II)
19
COMBINING HALF EQUATIONS
A combination of two ionic half equations, one
involving oxidation and the other reduction,
produces a REDOX equation. The equations are
balanced as follows... Step 1 Write out the
two half equations Step 2 Multiply the
equations so that the number of electrons in each
is the same Step 3 Add the two equations and
cancel out the electrons on either side Step 4
If necessary, cancel any other species which
appear on both sides
The reaction between manganate(VII) and
iron(II) Step 1 Fe2 gt Fe3
e Oxidation MnO4 5e 8H gt
Mn2 4H2O Reduction
20
COMBINING HALF EQUATIONS
A combination of two ionic half equations, one
involving oxidation and the other reduction,
produces a REDOX equation. The equations are
balanced as follows... Step 1 Write out the
two half equations Step 2 Multiply the
equations so that the number of electrons in each
is the same Step 3 Add the two equations and
cancel out the electrons on either side Step 4
If necessary, cancel any other species which
appear on both sides
The reaction between manganate(VII) and
iron(II) Step 1 Fe2 gt Fe3
e Oxidation MnO4 5e 8H gt
Mn2 4H2O Reduction Step 2
5Fe2 gt 5Fe3 5e multiplied by
5 MnO4 5e 8H gt Mn2
4H2O multiplied by 1
21
COMBINING HALF EQUATIONS
A combination of two ionic half equations, one
involving oxidation and the other reduction,
produces a REDOX equation. The equations are
balanced as follows... Step 1 Write out the
two half equations Step 2 Multiply the
equations so that the number of electrons in each
is the same Step 3 Add the two equations and
cancel out the electrons on either side Step 4
If necessary, cancel any other species which
appear on both sides
The reaction between manganate(VII) and
iron(II) Step 1 Fe2 gt Fe3
e Oxidation MnO4 5e 8H gt
Mn2 4H2O Reduction Step 2
5Fe2 gt 5Fe3 5e multiplied by
5 MnO4 5e 8H gt Mn2
4H2O multiplied by 1 Step 3 MnO4 5e 8H
5Fe2 gt Mn2 4H2O 5Fe3
5e
22
COMBINING HALF EQUATIONS
A combination of two ionic half equations, one
involving oxidation and the other reduction,
produces a REDOX equation. The equations are
balanced as follows... Step 1 Write out the
two half equations Step 2 Multiply the
equations so that the number of electrons in each
is the same Step 3 Add the two equations and
cancel out the electrons on either side Step 4
If necessary, cancel any other species which
appear on both sides
The reaction between manganate(VII) and
iron(II) Step 1 Fe2 gt Fe3
e Oxidation MnO4 5e 8H gt
Mn2 4H2O Reduction Step 2
5Fe2 gt 5Fe3 5e multiplied by
5 MnO4 5e 8H gt Mn2
4H2O multiplied by 1 Step 3 MnO4 5e 8H
5Fe2 gt Mn2 4H2O 5Fe3
5e Step 4 MnO4 8H 5Fe2
gt Mn2 4H2O 5Fe3
23
COMBINING HALF EQUATIONS
A combination of two ionic half equations, one
involving oxidation and the other reduction,
produces a REDOX equation. The equations are
balanced as follows... Step 1 Write out the
two half equations Step 2 Multiply the
equations so that the number of electrons in each
is the same Step 3 Add the two equations and
cancel out the electrons on either side Step 4
If necessary, cancel any other species which
appear on both sides
The reaction between manganate(VII) and
iron(II) Step 1 Fe2 gt Fe3
e Oxidation MnO4 5e 8H gt
Mn2 4H2O Reduction Step 2
5Fe2 gt 5Fe3 5e multiplied by
5 MnO4 5e 8H gt Mn2
4H2O multiplied by 1 Step 3 MnO4 5e 8H
5Fe2 gt Mn2 4H2O 5Fe3
5e Step 4 MnO4 8H 5Fe2
gt Mn2 4H2O 5Fe3
SUMMARY
24
COMBINING HALF EQUATIONS
A combination of two ionic half equations, one
involving oxidation and the other reduction,
produces a REDOX equation. The equations are
balanced as follows... Step 1 Write out the
two half equations Step 2 Multiply the
equations so that the number of electrons in each
is the same Step 3 Add the two equations and
cancel out the electrons on either side Step 4
If necessary, cancel any other species which
appear on both sides
Q. Construct balanced redox equations for the
reactions between... Mg and H
Cr2O72- and Fe2 H2O2 and MnO4
C2O42- and MnO4 S2O32- and I2
Cr2O72- and I
25
Mg gt Mg2 2e (x1) H e gt ½
H2 (x2) Mg 2H gt Mg2
H2 Cr2O72- 14H 6e gt 2Cr3
7H2O (x1) Fe2 gt Fe3 e
(x6) Cr2O72- 14H 6Fe2 gt 2Cr3
6Fe2 7H2O MnO4 5e 8H
gt Mn2 4H2O (x2) H2O2 gt O2
2H 2e (x5) 2MnO4 5H2O2
6H gt 2Mn2 5O2 8H2O MnO4
5e 8H gt Mn2 4H2O
(x2) C2O42- gt 2CO2 2e (x5) 2MnO4
5C2O42- 16H gt 2Mn2 10CO2
8H2O 2S2O32- gt S4O62- 2e
(x1) ½ I2 e gt I (x2) 2S2O32-
I2 gt S4O62- 2I Cr2O72- 14H
6e gt 2Cr3 7H2O (x1) ½ I2 e
gt I (x6) Cr2O72- 14H 3I2
gt 2Cr3 6I 7H2O
BALANCING REDOX EQUATIONS
ANSWERS
26
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.

27

• Half-cell an elements in two oxidation states.

28
Zn2(aq) 2 e ? Zn(s)
29

• Zn2(aq) 2e -gt Zn(s) E - 0.76V
• The electrode potential of the half cell
  indicates its tendency to lose or gain electrons.

30

• The standard electrode potential of a half-cell,
  is the e.m.f of a cell compared with a standard
  hydrogen half-cell, measured at 298 K with a
  solution concentration of 1 mol dm-3 and a gas
  pressure of 100KPa

31
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)
32
S tandard H ydrogen E lectrode
33
(No Transcript)
34
- electrode anode oxidation
electrode cathode reduction
electron 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.
Zn
Cu
Zn ? Zn2 2 e- oxidation
Cu2 2 e- ? Cu reduction
35
(No Transcript)
36
(No Transcript)
37
(No Transcript)
38
Emf E E(positive terminal) -
E(negative terminal)
39
(No Transcript)
40
(No Transcript)
41
Pt(s) H2(g) H(aq) Cu2(aq) Cu(s)
42
(No Transcript)
43
GOLDEN RULE

• The more ve electrode gains electrons
• ( charge attracts electrons)

44

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

45

• A2.CHEM5.3.003

5.3 EXERCISE 2 - electrochemical cells
46
ELECTRODE POTENTIALS Q1
Emf E?right - E?left
- 2.71 E?right - 0
E?right - 2.71 V
47
ELECTRODE POTENTIALS Q2
Emf E?right - E?left
Emf - 0.44 - 0.22
Emf - 0.66 V
48
ELECTRODE POTENTIALS Q3
Emf E?right - E?left
Emf - 0.13 - (-0.76)
Emf 0.63 V
49
ELECTRODE POTENTIALS Q4
Emf E?right - E?left
1.02 1.36 - E?left
E?left 1.36 - 1.02 0.34 V
50
ELECTRODE POTENTIALS Q5
Emf E?right - E?left
a) Emf 0.15 - (-0.25) 0.40 V
b) Emf 0.80 - 0.54 0.26 V
c) Emf 1.07 - 1.36 - 0.29 V
51
ELECTRODE POTENTIALS Q6
Emf E?right - E?left
a) E?right 2.00 - 2.38 - 0.38 V
Ti3(aq) e- ? Ti2(aq)
b) E?left -2.38 - 0.54 - 2.92 V
K(aq) e- ? K(aq)
c) E?right - 3.19 0.27 - 2.92 V
Ti3(aq) e- ? Ti2(aq)
52
ELECTRODE POTENTIALS Q7
a) Cr(s) Cr2(aq) Zn2(aq) Zn(s)
Emf -0.76 - (-0.91) 0.15 V
b) Cu(s) Cu2(aq) Fe3(aq),Fe2(aq) Pt(s)
Emf 0.77 - 0.34 0.43 V
c) Pt(s) Cl-(aq) Cl2(g)
MnO4-(aq),H(aq),Mn2(aq) Pt(s)
Emf 1.51 1.36 0.15 V