A 20.00-mL aqueous solution containing tin(II) ions with a concentration of 0.0850 M is titrated with...

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A 20.00-mL aqueous solution containing tin(II) ions with a concentration of 0.0850 M is titrated with...

A 20.00-mL aqueous solution containing tin(II) ions with a concentration of 0.0850 M is titrated with a standard aqueous solution of cerium(IV) ions having a concentration of 0.1100 M.

Write a balanced chemical equation for the titration reaction and

then find the volume, in mL, of the titrant that is required to

reach the stoichiometric point of the titration.

b. If the titration were followed potentiometrically using a

platinum electrode and a saturated Ag/AgCl electrode, find the

following voltages in the titration:

(1) at 10% titration

(2) at 50% titration

(3 at the stoichiometric point of the titration

(4) at 150% titration

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Answer 1

Answer #1

This is an example of red-ox titration, where Ce(IV) oxidises Sn(II) to Sn(IV) and itself gets reduced to Ce(III).

Half cell reactions (in reduction form) are: Sn2+ + Sn + + 2e

Cett + e + Ce3+

Therefore the cell reaction is: Sn2+ (aq) + 2 Cet+ (aq) + Sn++ (aq) + 2 Ce3+ (aq)

The reaction above is the balanced equation for the titration process.

(a)

110oo 20 ml of 0.0850M tin (11) - solution Contains = (20 L) x 0,0850 mol/l : R. - x 10 me/ in (11) Now, according to the ba

(b) Half cell Reaction: Sn++ (aq) +20 + Sn2+ (aq)

Now, according to Nerst equation, the reduction potential of this half cell or electrode can be written as:

RT Sn2+1 Esn: +Sn2+ = E$n*+18n2+ + log Sm+1

where, Esne+Sn2+ = electrode potential

En4+ Sn2+ = standard electrode potential = 0.15 V

n = number of electrons that are tranfered during the reaction = 2

Putting the values R=8.314 J K-1 mol-1 and T = 298 K and F=96500 C, we get

or, Esn4+|Sn2+ = ESne+| Sn2+ + 0.059 Sn2+1 logisn4+1

Now, Ecel = En4+ Sn2+ – EC4+10e3+ = 0.15V - (-1.61V) = 1.76V

and here n = 2

Therefore, 0.059 Ecell = 1.76 V + (Ce3+1 Sn4+1 Sn2+1 Ce4+12

Now, to follow this titration potetiometrically we have to couple this with another standard electrode, which is here the saturated Ag|AgCl electrode.

The reaction for this electrode is: AgCl(s) + + Ag(s) + C1-(aq)

The standard reduction potential of this electrode, EACH AC- = 0.2415V

Since, standard reductionpotential is higher for the AgCl|Ag|Cl- half cell, it will act as a cathode and the Sn2+|Sn half cell will act as anode.

Therefore cell potential will be:

0.059 [Sn2+1 Esnt+ Sn2+ = Ecathode-Eanode = 0.2415 V-(E9n4+1 Sn2+ + log: 9 Sn4+1

or, Esne+|Sn2+ = 0.2415 V - (0.15 V + - 0.059 [Sn2+] logisn4+1

or, $E_{Sn^{4+}|Sn^{2+}}= 0.0915\,V\,+\,\frac{0.059}{2}\,log \frac{[Sn^{4+}]}{[Sn^{2+}]})$

Now to calculate the values of voltage (or potential) we will use this equation:

(1) at 10 % titration:

at 10 % titration, 10% has been converted to Sn4+ and 90% is still Sn2+.

0.059 10 Ecell = 0.0915 V + - log = 0.0633V

2. at 50% titration: 50% has been converted to Sn4+ and 50% is still Sn2+.

$E_{cell}= 0.0915\,V\,+\,\frac{0.059}{2}\,log \frac{50}{50}\,= 0.0915\,V$

3. at the stoichiometric point of the titration:

$Potential,\,E = \frac {n_1E_1\,+\,n_2E_2}{n_1\,+\,n_2} = \frac {(2\times 0.15)+(1\times 1.61)}{2+1} = 1.173\,V$

where, $E_1\,and\,E_2\,are\,the\,redduction\,potentials\,of\,the\,analyte\,and\,titrant\,respectively$

Analyte= tin(II) and Titrant= Cerium (IV)

n and n are the number of electrons transferred in the concerned electrode reactions.

4. at 150% titration:

Now, there is no Sn2+ , therefore the half cell Sn4+|Sn2+ is not there. The operating cell has been changed to Ce4+|Ce3+ system. Now the cell potential is calculated from the equation:

$E_{Ce^{4+}|Ce^{3+}}= E_{cathode}-E_{anode}= (E^o_{Ce^{4+}|Ce^{3+}}\,+\,\frac{0.059}{n}\,log \frac{[Ce^{4+}]}{[Ce^{3+}]}) - 0.2415\,V$

Here, n=1

$E_{Ce^{4+}|Ce^{3+}}= (1.61\,+\,\frac{0.059}{n}\,log \frac{[Ce^{4+}]}{[Ce^{3+}]}) - 0.2415\,V$

or, $E_{Ce^{4+}|Ce^{3+}}= 1.3685\,V\,+\,0.059\,log \frac{[Ce^{4+}]}{[Ce^{3+}]}$

Now, since we have added 17* 10-4 mol of Ce(IV) to reach stoichiometric point which have converted to Ce(III) already, then another (1/2)* 17* 10-4 = 8.5* 10-4 (unreacted Ce(IV)) has been aded at 150% titration.

EC4+Ce3+ = 1.3685 V + 0.059 log 8.5 x 10-4 17 x 10-4 = 1.3507V

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Cett + e + Ce3+

We were unable to transcribe this image

110oo 20 ml of 0.0850M tin (11) - solution Contains = (20 L) x 0,0850 mol/l : R. - x 10" me/ in (11) Now, according to the balanced reaction, a moles of Cerium (IV) reacts with 1 mole of tin (11) ions. Therefore we have to calculate how much rokome of the 0.1000 m Cerium (IV) solution would contain 22x 8.5*10-4) mot Cerium(IV) Now, according to the definition of molarity, 0.1100 moles of Ce (IV) in 12 solution = 0.1100 M nd, we 1 yer volume - (2x 8.5*10") moles of ce (IV) is present in . 1 L x 2X8.5 x 10 0.1100 mot 15452= 15:45 mL of the titrant Iceland uired to reach the stoichiometric point, .. regu mot = 0.01545L solution. Icelav) 0.015454 ] is

Sn++ (aq) +2e + Sn2+ (aq)

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