A 30 mL solution of Tl + (0.45 M) is titrated with 0.60 M Co 3+ . The anode is a

Question

A 30 mL solution of Tl + (0.45 M) is titrated with 0.60 M Co 3+ . The anode is a platinum wire and the cathode is a saturated silver/silver chloride (Ag/AgCl) reference electrode; both are inserted into the titration beaker. Determine the cell potential at the titration volumes listed below:

a) 21.65 mL of titrant added

b) 45.00 mL of titrant added

c) 49.72 mL of titrant added

d) 53.09 mL of titrant added

2)

A 20 mL solution of Ag3+(52 mM) can be reduced by a basic sulfite (SO32-) solution. A titration is performed with a saturated calomel reference electrode as the anode and a platinum wire as a cathode. Determine the cell potential at various titration volumes, below, if the titrant is 0.1 M SO32- and both solutions are buffered to pH 9.5.

a) 3.49 mL of titrant added

b) 8.37 mL of titrant added

c) 10.40 mL of titrant added

d) 15.92mL of titrant added

Explanation / Answer

I'm not so sure, but I think this is the right procedure.

Frist let's calculate the standard potential of the cell, we already know that 1 electron is transfered, and the standard reduction potential of Tl and Co are the following:
E° Tl = -0.34 V; E° Co3+ = +1.92 V
Co3+ + 1e- -------> Co2+
Tl ----------> Tl+ + 1e-

E° = 1.92 + 0.34 = 2.26 V

From the nerst equation: E = E° - 0.059/n log K

The values of K can change with the concentrations of the reactants. In this case it will change when the titrant is added. So, let's calculate the moles of Tl and Co, and then, we'll calculate the concentrations with the volume added:

moles Tl+ = 0.030 * 0.45 = 0.0135 moles
moles Co3+ = 0.60 * 0.02165 = 0.013 moles

Total volume 1 = 51.65 mL
Total volume 2 = 75 mL
Total volume 3 = 79.72 mL
Total volume 4 = 83.09 mL

Now let's calculate the concentrations for each component at this volume:
[Tl]1 = 0.45 * (30/51.65) = 0.2614 M; [Co]1 = 0.60 * (21.65/51.65) = 0.2515 M
[Tl]2 = 0.45 * (30/75) = 0.18 M; [Co]2 = 0.60 * (45/75) = 0.36 M
[Tl]3 = 0.45 * (30/79.72) = 0.1693 M; [Co]3 = 0.60 * (49.72/79.72) = 0.3742 M
[Tl]4 = 0.45 * (30/83.09) = 0.1625 M; [Co]4 = 0.60 * (53.09/83.09) = 0.3833 M

Now let's calculate E for each of these concentrations:
E1 = 2.26 - 0.059 log (0.2614/0.2515) = 2.2590 V
E2 = 2.26 - 0.059 log (0.18/0.36) = 2.2778 V
E3 = 2.26 - 0.059 log (0.1693/0.3742) = 2.2803 V
E4 = 2.26 - 0.059 log (0.1625/0.3833) = 2.28199 V

Question 2, post them in another question thread. Hope this helps

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