3. G8 pts) Consider the titration of a 5000 mL aliquot containing iballows ion,T

Question

3. G8 pts) Consider the titration of a 5000 mL aliquot containing iballows ion,TI in IFHCJ with 0.01000 M potassium iodate, KIO, in 1F HCL monitored electrochemically with a platinum clectrode measured against a saturated calomel clectrode as the anode. a) Write the balanced chemical equation which describes the titration reaction. b) If 37.31 ml of iodate solution is required to reach the equivalence point, then what is the Ihallous ion concentration in the original aliquot? Write two different Nernst equations for the electrochemical monitoring of the cell reaction. d) Calculate the cell potential after 20.000 mLof iodate solution has been added.

Explanation / Answer

a)

Write the half reactions

Tl3+ + 2e-    <------> TI+

IO-3 + 2Cl-   6H+ + 4e- <------> ICl2-   + 3H2O

To blance the electron multiply 2 with the first equation

2Tl3+ + 4e-    <------> 2TI+

IO-3 + 2Cl-   6H+ + 4e- <------> ICl2-   + 3H2O

-------------------------------------------------------------------------

2TI++ IO-3 + 2Cl- <-----> 2Tl3+ + ICl2-   + 3H2O

b)

Number of moles of iodate is

= 37.31 mL x 0.01000 mmol / mL

= 0.0003731 mol

2TI++ IO-3 + 2Cl- <-----> 2Tl3+ + ICl2-   + 3H2O

Form the equation one mole iodate oxidizes 2 moles of TI+

So number of moles of TI+ is 2 x 0.0003731

=0.0007462 mol

The TI+ is present in 50 mL of solution

so the concentration is

0.0007462 mol / 50 mL

Concentration of TI+ is 0.014924 M

c)

Nernst equation

Ecell = Eright - Eleft

Here Eleft is calomel electrode for which Standard electrode potential is 0.241

Nernst equation is

aA + bB + ... + ne   <------> cC + dD + ...

E = Eo - RT/nF ln([C]c[D]d/[A]a[B]b)

Eo - Standard electrode potential

R - Gas constant

T - Temperature

n - number of electrons that appears in the half reaction.

F - Faraday 96,485 C

If we substitute numerical values for the constants, convert to base 10 logarithms,and specify 25°C for the temperature, we get

E = Eo - 0.0592/n log ([C]c[D]d/[A]a[B]b)

for

Tl3+ + 2e-    <------> TI+   Eo = 0.770V

Eleft = 0.770- 0.0592/2 log ([TI+]/[TI3+])

Ecell = 0.770- 0.0592/2 log ([TI+]/[TI3+]) - 0.241

for

IO-3 + 2Cl-   + 6H+ + 4e- <------> ICl2-   + 3H2O Eo = 1.242 V

Eleft = 1.242- 0.0592/4 log ([ICl2- ]/IO-3][Cl-]2[H+]6)

Ecell = 1.242- 0.0592/4 log ([ICl2-]/IO-3][Cl-]2[H+]6) - 0.241

d)

Cell potential after adding 20mL iodate

20 mL x 0.01000 mmol/mL = 0.0002

initial concentration of TI+ is 0.0007462

after adding 0.0002 moles of iodate TI+ is oxidized to TI3+

We know that 1 mol of iodate oxidizes 2 moles of TI+

So 0.0002 moles of iodate oxidizes 0.0002 x 2 = 0.0004 moles of TI+

Now the number of moles of TI+ is 0.0007462 - 0.0004 = 0.0003462

and the concentration of TI3+ is 0.0004

Substituting this in the following equation

Ecell = 0.770- 0.0592/2 log ([TI+]/[TI3+]) - 0.241

we have

Ecell = 0.770- 0.0592/2 log (0.0003462/ 0.0004) - 0.241

Ecell = 0.5308 V ~ =0.531V

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