Download Tutorial 7

Instrumental Analysis
Electrogravimetry , Coulometry
Tutorial 7
1
Objectives
By the end of this session, the student should be able to:
To calculate potential of a cell when current passes.
Differentiate between electrogravimetry, coulometry.
 Correlate concentration to the electrochemical phenomena measured.
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Example 1:Electrolytic cell
Calculate the initial voltage that should be applied to electrolyze
0.010 M [Zn(OH)4]2 in 0.10 M NaOH, using Ni electrodes. Assume that
the current is 0.20 A, the cell resistance is 0.35 Ω, and O2 is
evolved at 0.20 bar. The over-potential for O2 evolution at a Ni
surface at a current of 0.20 A is 0.519 V. The reactions are:
Cathode:
Anode:
[Zn(OH)4]2- + 2e
Zn(s) + 4OH
H2O
1/2O2 + 2H+ + 2e
E° = –1.199 V
E° = 1.229 V
Ecell = Ecathode  Eanode  IR  OVERPOTENTIALS
Solution
0.05916
[OH  ]4
Ecathode  E 
log
2
[Zn(OH)24 ]

0.05916
[0.1]4
  1.199 
log
  1.140 V
2
[0.01]
Written as reduction:
1/2O2 + 2H+ + 2e
H2O
[NaOH]= 0.1, pOH = 1
pH = 13, [H+] = 10-13
Eanode  E 
0.05916
1
log
1/ 2
2
PO . [H  ]2
2
 1.229 
0.05916
log (0.2)1/ 2 . [10 13 ]2  0.450 V
2
Eapplied  Ecathode  Eanode  IR  overpotential
  1.140  0.450  (0.20 )( 0.35)  0.519
  2.179 V
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Example 2:Electrogravimetry
Ions that react with Ag+ can be determined electro-gravimetrically
by deposition on a silver working anode:
Ag(s) + X  AgX(s) + eWhat will be the final mass of a silver anode used to electrolyze
75.00 mL of 0.0238 M KSCN if the initial mass of the anode is
12.463 g.
Solution
75.00 mL of 0.0238 M KSCN = 1.785 mmol of SCN - which gives
1.785 mmol of AgSCN, CONTAINING 0.1037 g of SCN-.
Mass of SCN- = 1.785 mmol SCN- x 10-3 x 58.09 (g/mol SCN-)
= 0.1037 g SCN-
Final mass of silver anode = 12.4638 + 0.1037 = 12.5675 g
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Theoretical Basis for Electrogravimetry and Coulometry
Faraday’s law:
The amount of chemical reaction at an
electrode (mass of copper deposited on the
cathode surface) is proportional to the
quantity of electricity passed in the circuit.
Faradaic current: The current passes in the circuit as a result
of actual electrolysis (oxidation and
reduction at electrode surface).
q  I . t
Coulombs Amperes seconds
Moles of e 

I.t
F
If a reaction requires n electrons per mole of
reactant, the quantity reacting of chemical species
in time t is
Moles reacted 
Mass 
I.t
n F
I.t
. ( molar mass )
n F
Faraday’s
Law
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Example 3:
A 1.00-L electrolysis cell initially containing 0.0250 M Mn2+ and
another metal ion, M3+, is fitted with Mn and Pt electrodes. The
reactions are:
Mn(s) → Mn2+ + 2eM3+ + 3e- → M(s)
a) Is the Mn electrode the anode or the cathode?
b)
A constant current of 2.60 A was passed through the cell for
18.0 min, causing 0.504 g of the metal M to plate out on the
Pt electrode. What is the atomic mass of M?
c)
What will the concentration of Mn2+ in the cell be at the end
of the experiment?
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Solution
a) Sine Mn is oxidized, it is the anode.
b)
( 2.60 C / s )(18.0  60 s )
 0.0291 mol of e  / 3  0.0097 mol of M
96500 C / mol
Since 1 mol of M gives 3 e-
Atomic mass of M = 0.504 g / 0.0097 mol = 52.0 g/mol
One step solution:
Wt. of M3+ Electrolyzed = It/nF X atomic.wt = 2.6X18X60/3X96500Xatomic.wt
0.504 = 0.0097 X atomic.wt
atomic.wt = 52.0 g/mol
c) In the electrolysis :
no. of moles of Mn2+ produced = It/nF =2.6 X 18 X 60/2 X 96500 =0.01455mol
Total moles of [Mn2+] = 0.0250 + 0.01455 =
conc. Of [Mn2+] = 0.0396 moles in 1 Liter = 0.0396 M
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Example 4: Coulometric titration
H2S(aq) can be analyzed by titration with coulometrically
generated I2.


H 2 S  I 2  S(s)  2 H
 2I
To 50.0 mL of sample were added 4 g KI. Electrolysis required
812 s at 52.6 mA. Calculate the concentration of H2S (g/mL) in
the sample.
Solution:
Faraday's Law
The number of grams reduced at the cathode or oxidized at the
anode is given by:
Where
Mass of H2S 
It
mass 
 FW
nF
I = current in amps
t = time in seconds
FW = formula weight
n = number of electrons transferred per 1 mole
of species
It FW
812  52.6 34



 10 3  7525 g
F
n
96500
2
Concentration of H2S= Wt of sample /volume = 7525/50 = 150.5 g/mL
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Example 5: Coulometric titration
Chlorine has been used for decades to disinfect drinking water. An
undesirable side effect of this treatment is the reaction of
chlorine with organic impurities to create organochlorine compounds,
some of which could be toxic. Monitoring total organic halide is now
required for many water providers. A standard procedure is to pass
water through activated charcoal that adsorbs organic compounds.
Then charcoal is combusted to liberate hydrogen halide:
Organic halide (RX)
CO2 + H2O + HX
The HX is absorbed into aqueous solution and measured by automatic
coulometric titration with a silver anode:
Ag(s)

X-(aq) +
Ag+ + eAg+ 
AgX(s)
(Ag+ generated anodically)
(formed AgX is deposited on anode)
When 1.00 L of drinking water was analyzed, a current of 4.23 mA was
required for 387 s.
A blank prepared by oxidizing charcoal required 6 s at 4.23 mA.
Express TOX of the drinking water as mol halogen/L. If all halogen
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is chlorine, express the TOX as g/L
Solution
- The corrected coulometric titration time is 387 – 6 = 381 s
q = It/F = [(4.23 mA)(381 s)] / (96500 C/mol) = 0.0167 mmol e= 16.7 mol e-.
- Because 1e- is equivalent to one x-, the concentration of
organohalide is 16.7 M.
- If all halogen is Cl, this corresponds to 592 g Cl/L
(16.7 mol/L x 35.45 g/mol).
One Step solution
Wt of Cl = It/nf X M.Wt = 4.23x 10-3 x 381/1x96500 x 35.45=
0.000592 gms = 592 g
Conc of Cl= 592 per 1 liter sample = 592 g Cl/L
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