1

Instrumental AnalysisInstrumental Analysis

Electrogravimetry , coulometryElectrogravimetry , coulometry

and amperometry and amperometry

Instrumental AnalysisInstrumental Analysis

Electrogravimetry , coulometryElectrogravimetry , coulometry

and amperometry and amperometry

Tutorial 5

2

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.

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-.

Final mass of silver anode = 12.4638 + 0.1037 = 12.5675 g

Solution

Example 1:

3

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

nFW

FIt

mass

g.

n

FW

F

ItSHofMass

752510

2

34

96500

652812 32

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 species

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

Example 2:

4

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+ + 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?

5

Solution

a) Sine Mn is oxidized, it 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

6

Example 4:

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

7

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).

Solution

8Try 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 1