1

Instrumental AnalysisInstrumental Analysis

Electrogravimetry , CoulometryElectrogravimetry , Coulometry

and Amperometry and Amperometry

Instrumental AnalysisInstrumental Analysis

Electrogravimetry , CoulometryElectrogravimetry , Coulometry

and Amperometry and Amperometry

Tutorial 6

2

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 overpotential for O2 evolution at a Ni surface at a current of 0.20 A is 0.519 V. The reactions are:

Cathode: [Zn(OH)4]2- + 2e Zn(s) + 4OH E° = –1.199 VAnode: H2O 1/2O2 + 2H+ + 2e E° = 1.229 V

V179.2

519.0)35.0)(20.0(450.0140.1

ialoverpotentIRanodeEcathodeEappliedE

V450.02]1310[.2/1)2.0(log2

05916.0229.1

2]H[.2/1OP

1log

2

05916.0EanodeE

V140.1]01.0[

4]1.0[log

2

05916.0199.1

]24)OH(Zn[

4]OH[log

2

05916.0EcathodeE

2

Example 1:

Solution

1/2O2 + 2H+ + 2e H2O

Written as reduction:

NaOH]= 0.1, pOH = 1pH = 13, [H+] = 10-13

3

Ions that react with Ag+ can be determined electrogravimetrically by deposition on a silver working anode:

Ag(s) + X AgX(s) + e-

What 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

Example 2:

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

44

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).

Theoretical Basis for coulometric Titration

t.Iq

Coulombs Amperes seconds

F

t.IeofMoles

Fn

t.IreactedMoles

If a reaction requires n electrons per mole of reactant, the quantity reacting of chemical species in time t is

Faraday’s Law

)massmolar(.Fn

t.IMass

5

H2S(aq) can be analyzed by titration with coulometrically generated I2. IH)s(SISH 2222

FWFn

tImass

g7525102

34

96500

10x6.52812

n

FW

F

ItSHofMass 6

3

2

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 I = current in ampst = time in secondsFW = formula weight n = number of electrons transferred per 1 mole of species

Concentration of H2S = 7525/50 = 150.5 g/mL

Example 3:

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Example 4:

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+ + 2e-

M3+ + 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, Mn metal is the anode

b) .

(Since 1 mol of M gives 3 e-)

Atomic mass of M = 0.504 g / 0.0097 mol = 52.0 g/mol

c) In the electrolysis 0.0291/2 = 0.01455 mol of Mn2+ were produced.

[Mn2+] = 0.0250 + 0.01455 = 0.0396 M

Mofmol.eofmol.mol/C

)s.)(s/C.(0097002910

96500

60018602

8

Example 5:

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) Ag+ + e- (Ag+ generated anodically)

X-(aq) + Ag+ AgX(s) (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 is chlorine, express the TOX as g Cl/L

9

The corrected coulometric titration time is 387 – 6 = 381 s

Number of moles of e- = 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).

Solution

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Amperometry is the electroanalytical technique in which:

• the potential of the working electrode is adjusted at a fixed

value on the plateau of i-E curve and

• the current is measured as a function of concentration.

Amperometry is a measure of limiting current at a fixed applied potential as a function of concentration

Amperometric titrations

• Amperometry can be used to estimate the equivalence point of

titrations, provided at least one of the participants or products

of the reaction involved is oxidized or reduced at a

microelectrode.

• In this case, the current at fixed potential (at the plateau of

the i-E curve) is measured as a function of the titrant volume

added.

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reactant reacts at electrode

XneA

ACi

VT

iEPCA 0

Titrant reacts at electrode

XneT

TCi

VT

iEP

CT 0

Both reactant and titrant react

VT

i EP

Predict the shape of the amperometric titration

curve if the product is the only electroactive species

?

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Exercise 1

Suppose we wish to electrolyze I- to I3- in a 0.10 M KI

solution containing 3.0x10-5 M I3- at pH 10.00 with PH2

fixed at

1.00 bar.

3I- + 2H2O I3- + H2(g) + 2OH-

a) Find the cell potential if no current is flowing .

b) Suppose that electrolysis increases I3- to 3.0x10-4 M,

but other concentrations are unaffected. If the cell

resistance is 2.0 , the current is 63 mA, the cathode

overpotential is 0.382 V and the anode overpotential is

0.025 V, what voltage is needed to drive the reaction?

Answer: a) –1.081 V

b) –1.644 V

Hint:

Cathode : 2H2O + 2e- H2(g ) + 2OH Eo = – 0.828 V

Anode : I3– + 2e– 3I Eo = 0.535 V

13Try to solve Exercise 17-B and problems 17-1, 17-2, 17-4, 17-8 17-Try to solve Exercise 17-B and problems 17-1, 17-2, 17-4, 17-8 17-11, 17-12, 17-13, 17-1911, 17-12, 17-13, 17-19 (Harris text book, p400-403)(Harris text book, p400-403)

reacted with the 8-hydroxyquinoline (HOC9H6N) that was released when the indium(III) compound was dissolved:

HOC9H6N + 2Br2 HOC9H4NBr2 + 2HBr

A heated buffer solution containing indium(III), In3+, was treated with an excess of 8-hydroxyquinoline (HOC9H6N) to precipitate quantitatively an insoluble indium(III) compound, according to the following reaction:

In3+ + 3HOC9H6N In(OC9H6N)3(s) + 3H+

After the precipitate was separated and washed, it was dissolved in dilute hydrochloric acid solution (which caused the above reaction to proceed in the right-to-left direction). Then an excess of bromide ion (Br–) was added, and the solution was transferred to an electrochemical cell equipped with a platinum generator anode and a platinum auxiliary cathode. Elemental bromine (Br2), which was electrogenerated by oxidation of bromide ion,

2Br– Br2 + 2e–

An end point was reached after a titration time of 186.6 seconds at a constant current of 125.3 mA. Calculate the mass of In3+ in milligrams in the original sample solution.

Faraday constant (F) = 96,500 coulombs/mol e-. Atomic mass of In = 114.8)

(Answer: 2.318 mg of In3+)

Exercise 2