For the titration of 25.0 mL of 0.0100 M Sn2+ by 0.0500 TI3+ in 1 M HCl, using P

Question

For the titration of 25.0 mL of 0.0100 M Sn2+ by 0.0500 TI3+ in 1 M HCl, using Pt and Ag|AgCl electrodes. Write the balanced titration reaction? _____________ ? _______________ Complete the two half reactions for the Pt electrode. Sn + e- ? E0 = _________ TI + e- ? E0 = __________ From the list select the two correct Nernst equation for the cell E = 0.77- 0.05916log(Sn2+]/[Sn3+]-0.0.197 E = 0.139- 0.05916log([Sn2+]/[Sn3+]-0.197 E = 0.77- 0.05916log([Sn2+]/[Sn4+]-0.197 E = 00.139- 0.05916log([Sn2+]/[Sn4+]-0.197 E = 0.77- 0.05916log([TI2+]/[TI3+]-0.197 E = 0.139- 0.05916log([TI2+]/[TI3+]-0.197 E = 0.77- 0.05916log([TI+]/[TI3+]-0.197 E = 0.139- 0.05916log([TI+]/[TI3+]-0.197 What is the value of E at the following volumes of added TI? 1.00 mL 2.50 mL 4.90 mL 5.00 mL 5.10 mL 10 mL

Explanation / Answer

0.05916log([Sn2+]/[Sn4+]-0.197 E = 0.77- 0.05916log([TI2+]/[TI3+]-0.197 E = 0.139-

5.10 mL 10 mL

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