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2815

Electrochemistry

ELECTROCHEMISTRY
REDOX

HALF CELLS

Reduction
Oxidation

gain of electrons
removal of electrons

Cu2+(aq) + 2e¯ ——> Cu(s)
Zn(s) ——> Zn2+(aq) + 2e¯

• these are systems involving oxidation or reduction
• there are several types
METALS IN CONTACT WITH SOLUTIONS OF THEIR IONS

C
O
P
P
E
R

Z
I
N
C

Reaction
Electrode
Solution
Potential

Cu2+(aq) + 2e¯
Cu(s)
copper
Cu2+(aq) (1M) - 1M copper sulphate solution
+ 0
...
76V

GASES IN CONTACT WITH SOLUTIONS OF THEIR IONS

Reaction
Electrode
Solution
Gas
Potential

2H+(aq) + 2e¯
H2(g)
platinum
H+(aq) (1M) - 1M hydrochloric acid or 0
...
00V

SOLUTIONS OF IONS IN TWO DIFFERENT OXIDATION STATES
P
L
A
T
I
N
U
M

Reaction
Electrode
Solution
Potential

Fe3+(aq) + e¯
Fe2+(aq)
platinum
Fe3+(aq) (1M) and Fe2+(aq) (1M)
+ 0
...
52 V

2

2815

CELL
POTENTIAL

Electrochemistry

• each electrode / electrolyte combination has its own half-reaction

Measurement

• it is impossible to measure the potential of a single electrode BUT
...


• temperature
• pressure of any gases
• solution concentration

The ultimate reference is the STANDARD HYDROGEN ELECTRODE
...

In the diagram below the standard hydrogen electrode is shown coupled up to a zinc half
cell
...


V
Salt Bridge (KCl)

Hydrogen
(1 atm
...
00V
2+

Zn (aq) (1M)

H + (1M)
(aq)

conditions

salt bridge

Secondary
standards

Calomel

temperature
solution conc
...
H
...
) is difficult to set up so it is easier to choose a
more convenient secondary standard which has been calibrated against the S
...
E
...
27V
is used as the left hand electrode to determine the electrode potential of an unknown
to obtain the E° value of the unknown half cell ADD 0
...

All equations are written as reduction processes
...
e
...
g
...
66V
E° = +1
...
e
...

E° / V

F2(g) + 2e¯
H2O2(aq) +

2F¯(aq

+2
...
77

+ 5e¯

Mn2+(aq)

2H+(aq)
+

MnO4¯(aq) + 8H

(aq)

PbO2(s) + 4H+(aq) + 2e¯
4+

Ce

(aq)

+ e¯

Cl2(g) + 2e¯
Cr2O72-(aq)

+

+ I4H

MnO2(s) +

Pb2+(aq) + 2H2O(l)
Ce

3+

2Cl¯(aq)

+ 6e¯
+ 2e¯

Mn2+(aq)

Br2(l) + 2e¯

+1
...
47
+1
...
36
+ 7H2O(l)
+ 2H2O(l)

+1
...
23

2Br¯(aq)

+ e¯

Ag(s)

+0
...
77

2H+(aq)

+ 2e¯

H2O2(l)

+0
...
54

Cu+(aq) + e¯

Cu(s)

+0
...
34

Cu2+(aq) + e¯

Cu+(aq)

+0
...
15

2H+(aq) + 2e¯

H2(g)

LH species better
oxidising agents

+1
...
00

+

Ag
O2(g) +

reaction is more
likely to go right

(aq)

2+

Pb

2e¯

(aq)

+ 2e¯

Pb(s)

+ 2e¯

Sn(s)
Ni(s)

-0
...
41

Fe(s)

-0
...
76

Al3+(aq) + 3e¯

Al(s)

-1
...
38

Na+(aq)

+ e¯

Na(s)

-2
...
87

K(s)

-2
...
14

Ni2+(aq) + 2e¯

RH species are
harder to oxidise

-0
...

• In theory ANY REDOX REACTION WITH A POSITIVE E° VALUE WILL WORK
...
40V
...


Example

What will happen if an Sn(s) / Sn2+(aq) cell and a Cu(s) Cu2+(aq) cell are connected?
• Write out the appropriate equations

Cu2+(aq) + 2e¯
2+

Sn
•
•
•
•
•

Q
...
34V
Sn(s) ; E° = -0
...
(+0
...
14) = + 0
...


Q
...


Q
...


5

2815

Electrochemistry

Combining
half-cells

In the cell shown, copper has a more positive E° value (+0
...
76V)
...
10V
...

• Place the cell with the more positive E° value on the RHS of the diagram
...

•
•
•
•

Conclusion

the cell reaction goes from left to right
the electrons go round the external circuit from left to right
the cell voltage is E°(RHS) - E°(LHS)
...
4

E°(RHS) - E°(LHS)

Electrochemistry

=

0
...
76V)

= 1
...

Work out the overall reaction and calculate the potential difference of the cell
...
36V

Cl2(g) + 2e¯

2Cl¯(aq)

E° = + 1
...
34V

as reductions with their E° values

Sn2+(aq) + 2e¯

Sn(s) ; E° = -0
...
e
...
(+0
...
14)
• If this is the equation you want then it will be spontaneous
• If it is the opposite equation (i
...
going the other way) it will not be spontaneous

Method 2

• 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
...
14V

and the usual equations
• Reverse one equation and its sign

Sn(s) ——> Sn2+(aq) + 2e¯

• Combine the two half equations

Sn(s) + Cu2+(aq)

• Add the two numerical values

(+0
...
14V) = +0
...
14V

Sn2+(aq) + Cu(s)

2815

Electrochemistry

• if the value is positive the reaction will be spontaneous

Q
...
6

Which of the following reactions occur spontaneously ?
• Cl2(g) + 2Br¯(aq)
• I2(g) + 2Br¯(aq)
• 2H+(aq) + Zn(s)

Q
...

The following half equations will help
...

Cu+(aq) + e¯

Cu(s)

E° = + 0
...
15V

+ e¯

7



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