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Chemistry, 20.11.2020 17:00, BraedynM2089
You are titrating 45.0 mL of 0.0100 M Sn2+ in 1 M HCl with 0.0500 M Tl3+ resulting in the formation of Sn4+ and Tl+. A Ptindicator electrode and a saturated Ag|AgCl reference electrode are used to monitor the titration.
What is the balanced titration reaction?
titration reaction:
βΆ
1. Complete the two halfβreactions that occur at the Pt indicator electrode. Write the half-reactions as reductions.
half-reaction:
Sn+eββ½βββ
β=0.139 V
half-reaction:
Tl+eββ½βββ
β=0.77 V
2. Select the two equations that can be used to determine the cell voltage at different points in the titration. of the saturated Ag|AgCl reference electrode is 0.197 V.
a. =0.139β0.05916log([Sn2+][Sn4+])β0.1 97
b. =0.139β0.05916log([Sn4+][Sn2+])β0.1 97
c. =0.139β0.059162log([Sn4+][Sn2+])β0. 197
d. =0.139β0.059162log([Sn2+][Sn4+])β0. 197
e. =0.77β0.05916log([Tl+][Tl3+])β0.197
f. =0.77β0.05916log([Tl3+][Tl+])β0.197
g. =0.77β0.059162log([Tl+][Tl3+])β0.19 7
h. =0.77β0.059162log([Tl3+][Tl+])β0.19 7
3. Calculate the cell potential () after each of the given volumes of the Tl3+ titrant have been added.
1.00 mL=
V
4.50 mL=
V
8.90 mL=
V
9.00 mL=
V
9.10 mL=
V
14.00 mL=
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Answers: 1
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You are titrating 45.0 mL of 0.0100 M Sn2+ in 1 M HCl with 0.0500 M Tl3+ resulting in the formation...
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