### Solution

```Instrumental Analysis
Fundamentals of Electrochemistry
Tutorial 4
Electricity from the Ocean Floor
Bacteria oxidizes organic matter
using SO42- ions available in the
sediment layer. As a result, HS
is released and acts as one of the
reactant in the electrochemical
cell. The other reactant is
dissolved O2 molecules available
at the sediment-water interface.
Oceanographic
instruments can be
battery between the
water and sediment
layers.
Summary of Symbols, Units, and Relationships
Relation between
charge and moles
q
Work
**
Electric Power
.
F
moles
of e
=
Joules, J
C/mol
E
.
Volts, V
q
coulombs
G = ─ W = ─ n F E
J/mol
I
Ohm’s Law
z
Charge
(coulombs, C)
Relation between
work and voltage
Relation between free
energy and potential
difference
=
Current
A
=
E
/
Volts
V
P = work/s
Power
J/s
(Watt, W)
R
resistance
Ohm, 
= E . I
V
A
(96500 C/mol).
z is the total no. of moles of
electrons passes in the circuit
E is the electric potential
difference, E is the work (J)
needed to move a coulomb of
positive charge from one point to
the other.
n is the moles of charge moved
under a potential difference, E.
I is the electric current,
measured in ampere (A). It is the
coulombs per second moving
pas a point in the circuit.
P is the work per unit time done
by electricity moving through a
circuit.
** the minus sign in the equation indicates that the free energy of a system decreases
when the work is done on the surrounding
Relation between
Current and rate
of chemical
reaction
I
A
(C/s)
=
n is the number of moles of e- per
1 mole of reactant and rate is the
rate of electrochemical reaction in
mol/s
n . F . Rate
C/mol
mol/s
Nernst equation
b
Relation between
potential of the
electrode and
activity of ions
Relation between
activity and
concentration
0.05916 V
aB

EE 
log a
n
aA
a c
E is the reduction electrode
potential and E is the standard
reduction electrode potential when
the activities of all reactants and
products are unity. n is the
number of electrons in the half-cell
reaction. a is the activity of
reactant or product of the
electrochemical reaction of the
half-cell when written as reduction
Note: Nernst Equation can be
applied to one half cell as well
as to the reaction of the whole
cell.
a is the activity of a chemical
species,  is the activity coefficient,
c is the concentration of chemical
species
Example (Charge)
If 3.2 g of O2 were reduced in the overall reaction with HS:
HS   S  H   2e 
1
2
O 2  2H   2e   H 2O
HS   12 O 2  H   S  H 2O
how many coulombs have been transferred from HS to O2 or how many
charges pass in the circuit?
Solution
3.2 g O 2 /32 g m ol-1  0.1 m ol O 2
S in ce 1 2 m ole sof O 2 gain 2 m ole sof e 
No. of m ole sof e le ctron s 0.1 x 4  0.4 m ole s
q  z F  (0.4 mol)(96500 C/mole- )  38,600 C
Example (Current)
If electrons are forced into a wire which acts as anode to oxidize HS at a rate
of 4.24 mmol/hr, how much current passes through?
Note: The electrode at which
oxidation occurs is the anode
while the electrode at which
reduction occurs is the cathode
Solution
HS   S  H   2e 
No. of mole sof e  / s 
n
4.24 mmol HS
1 mol HS
2 mol e 
1h
 3


 2.356 10 6 mole se  / s


1h
10 mmol HS
1 molHS 3600 s
ch arge cou lom bs m ole s cou lom b


.
tim e
se cond
se cond m ol

cou lom s
 6 m ole se
 2.356 10
 96500
 0.227A
s
m ole
C u rre nt
Generally, the current may be related to the rate of electrochemical reaction by:
I  n . F . rate
Rate of consumption of reactants in
a unit of mol/s
n is the number of moles of e- per 1 mole of reactant
Example (Work)
How much work can be done if 2.4 mmol of electrons go through a potential
difference of 0.70 V in the ocean-floor battery?
Solution
q  z . F  ( 2.4  10  3 mol ) (96500 C/mol )  2.3  10 2 C
ElectricalWork  E . q  (0.70V)(2.3  102 C)  1.6 102 J
• The greater the difference in potential (V), the stronger the e will be
pushed around the circuit
• 12V battery “pushes” electrons through a circuit eight times harder than a
1.5V battery
Example (E from free energy change)
Calculate the voltage that will be measured by the potentiometer in the figure,
Knowing that the free energy change for the net reaction is 150 kJ/mol of Cd.
Solution
Re du ction:
O xidation:
2 AgC l(s )  2e   2 Ag(s )  2C l (aq)
C d(s )  C d2  (aq)  2e 
Ne tRe action: C d(s )  2 AgC l(s )  C d2  (aq)  2Ag(s )  2C l (aq)
G   n F E
J
G
m olC d
E

  0.777 V ( J / C)
m
ol
e

C
nF
(2
) (96500
)
m olC d
m ol e 
 150 103
A spontaneous chemical reaction of negative G produces a positive voltage.
Example (Ohm’s Law)
In the following circuit, a battery generates a potential difference of 3 V and
the resistor has a 100  resistance. How much current and power are
delivered by the battery?
Solution
E 3.0V
I 
 0.030 A
R 100 
P  E . I  (3.0 V)(0.030A)  0.090W
Nernst Equation
• The Nernst Equation relates the potential of the half cell to the activities of
the chemical species (their concentrations)
– For the reaction (written as reduction)
aA  ne-  bB
a Bb
RT

E E 
ln [
]
a
nF
aA
– We usually calculate half-reactions at 25º C, substituting that in with the
gas constant and to base 10 log gives:
b
a
0
.
05916
V
E  E 
log B
n
aAa
Where a is the activities of species A and B.
n is the no. of electrons
in either the electrode or
cell reaction
Example (Nernst Equation)
•
Write the Nernst equation for the reduction of O2 to water:
1
2
O 2  2H   2e-  H 2O
E  1.23 V
Note that:
asolid =1, agas = pressure
E  1.23 
0.05916
[ H 2 O]
log
1
2
PO 2 2 [H  ]2
E  1.23 
0.05916
1
log
2
[H  ]2
aH2O =1, aion = molarity
E  1.23 
0.05916
log [H  ]2
2
E  1.23  0.05916 log [H  ]
•
Note that multiplying the reaction by any factor does not affect either Eº or the
calculated E:
O2  4H   4e-  2H 2O
0.05916
[H 2O]2
E  1.23 
log
4
PO 2 [H  ]4
E  1.23 V
E  1.23  0.05916 log [H  ]
Example (Ecell)
• Find the voltage for the Ag-Cd cell and state if the reaction is spontaneous if
the right cell contained 0.50 M AgNO3(aq) and the left contained 0.010 M
Cd(NO3)2(aq).
For Ag/Ag+
electrode
Ag   e  Ag(s)
E  0.799 
For Cd/Cd2+
electrode
For the Cell
0.05916
1
log
 0.781 V
1
0.50
Cd 2   2e   Cd(s)
E  0.402 
E  0.799 V
E  0.402 V
0.05916
1
log
 0.461V
2
0.010
Ecell  E  E  0.781  (0.461)  1.242 V
 Note that you will get the same value of cell potential if you apply the
Nernst equation directly to the overall cell reaction.
2Ag  2e  2Ag(s)
Cd  Cd 2  2e
Cd(s)  2Ag  Cd 2  2Ag(s)
Ecell  Ecell
0.05916
[Cd 2 ][ Ag ]2

log
2
[Cd ][ Ag  ]2
Ecell  (EAg / Ag
Ecell
Ecell
2
0
.
05916
[
Cd
]
 ECd / Cd2 ) 
log
2
[ Ag  ]2
0.05916
[0.01]
 (0.799  ( 0.402)) 
log
2
[0.5]2
0.05916
[0.01]
 1.201 
log
2
[0.5]2
 1.242 V
Since Ecell is found to be +ve quantity, the reaction as written is spontaneous
Example (Ecell)
A cell was prepared by dipping a Cu wire and a saturated calomel electrode
(ESCE = 0.241 V) into 0.10 M CuSO4 solution (ECu/Cu2+ = 0.339 V). The Cu wire
was attached to the positive terminal and the calomel to the negative terminal of
the potentiometer.
1- Write the half-cell reaction of Cu electrode.
2- Write the Nernst equation for the Cu electrode.
3- Calculate the cell voltage.
4- What would happen if the [Cu2+] were 4.864x10-4 M?
solution
Electrode connected to the positive terminal of the potentiometer is the
cathode and the other is the anode
1- Cu2+(aq) + 2e-  Cu(s)
0.05916
log [Cu 2 ]
2
0.05916
E  0.339 
log (0.10)  0.309V
2
2- E  E0 
3-
Ecell = ECu/Cu2+  ESCE = 0.309 V  0.241 V = 0.068 V
Example (Ecell)
A 100.0 ml solution containing 0.100 M NaCl was titrated
with 0.100 M AgNO3 and the voltage of the cell shown in
the figure was monitored.
a) Calculate the voltage after the addition of 65.0 ml
of AgNO3. (EAg/Ag+ = 0.799 V, ESCE = 0.241 V)
b) Explain why a silver electrode can be used as an
indicator electrode for both Ag+ and for halides.
Solution:
NaCl
a) The titration reaction is: Ag+(aq) + Cl(aq)  AgCl(s).
Equivalence point is reached upon addition of 100.0 ml titrant (in this case).
After addition of 65.0 ml of AgNO3, we can say:
(M V)reacted Cl- = (M V)Ag+ added
(note that Ag+ and Cl react in the ratio of 1:1)
(M V)Cl- remains unreacted = (M V)total Cl-  (M V)Ag+ = (0.1 x 100)  (0.1 x 65)
(M V)Cl- remains unreacted = 3.5 mmole
MCl- = 3.5 mmol/165 = 0.0212 mol/L
•
To find the cell voltage we need to know Ag+:
[ Ag  ] [Cl  ]  K sp
so  Ag+  = Ksp / [Cl]
where Ksp is the solubility product of insoluble AgCl
[Ag+] depends on the concentration of Cl ion
= 1.8 x 10-10 M / 0.0212 M
= 8.5 x 10-9 M
•
The potential of Ag electrode is given by:
Ag electrode reaction:
Ag+ + eAg+
E  E 
0.05916
[ Ag ]
1
log
 E  0.05916 log
 E  0.05916 log [ Ag  ]
1
[ Ag  ]
[ Ag  ]
EAg/Ag+ = 0.799 + 0.05916 log (8.5  10-9) = 0.322 V
•
The cell potential: Ecell = ESCE  EAg/Ag+ = 0.241  0.322 =  0.081 V
b) We can see from the example that the silver electrode can act indirectly as halide
indicator electrode if solid insoluble silver halide is present. The solubility of silver
halide will be affected by whatever halide ion is present and in turn the concentration of
Ag+ ion as well as the potential of the Ag electrode.
K sp
 ]  E
E Ag/Ag   E

0
.
05916
log
[
Ag

0
.
05916
log
Ag/Ag 
Ag/Ag 
[Cl  ]
Exercise 1
A mercury cell used to power heart pacemaker runs on the following reaction:
Zn(s) + HgO(s)
ZnO(s) + Hg()
E = 1.35 V
If the power required to operate the pace maker is 0.010 W, how many
kilogram of HgO (molar mass = 216.59) will be consumed in 365 days?
Exercise 2
Calculate the voltage of each of the following cells:
(a) Fe(s) / FeBr2(0.010 M) // NaBr(0.050 M) / Br2() / Pt(s)
(b) Hg() / Hg2Cl2(s) / KCl(0.060 M) // KCl(0.040 M) / Cl2(g, 0.50 bar) / Pt(s)
(EFe/Fe2+ = 0.44 V, EBr2/Br = 1.078 V, EHg/Hg2Cl2 = 0.268 V, ECl2/Cl = 1.360 V)
Try to solve problems 14-3, 14-4, 14-5, 14-7, 14-12, 14-14, 14-17, 14-18, and 14-19
Harris text book, p308-310
If you are unable to solve these problems or need to revise your answer, please refer to
the “Solution Manual for Quantitative Chemical Analysis”, by D.C. Harris (GUC library)
```