Consider the titration of 25.00 mL of 0.0100 M Sn2+ by 0.0500 M Tl3+ in 1.0 M HC
ID: 1028798 • Letter: C
Question
Consider the titration of 25.00 mL of 0.0100 M Sn2+ by 0.0500 M Tl3+ in 1.0 M HCl, using Pt and saturated calomel electrodes to find the end point. Read the textbook (Quantitative Chemical Analysis, 8th Ed. Daniel Harris, page 341-344). Sn2+ + Tl3+ Sn4+ + Tl+
Find the net cell voltage E (v) when 3.80 mL of Tl3+ is consumed. =-0.23 Incorrect.
Calculate the volume (mL) of Tl3+ used at the equivalence point. =5.00 You are correct.
Find the net cell voltage E (v) at the equivalence point.= 0.46 Incorrect.
Find the net cell voltage E (v) when 6.70 mL of Tl3+ is consumed.
Explanation / Answer
Sn2+ + Tl3+ -----------Sn4+ + Tl+
E 0(Tl+/Tl3+) = 0.77 V
E 0(Sn2+/Sn4+) = 0.139 V
1. At 3.80 mL of Tl3+ added
mmol Sn2+ = (25.0 mL)(0.0100 M) – (3.80 mL)(0.0500 M) = 0.06 mmol
mmol Sn4+ = (3.80 mL)(0.0500 M) = 0.19 mmol
Ecell = E0 (Sn2+/Sn4+) – E(SCE) – 0.05916/2 log([Sn2+ ]/[Sn4+])
= 0.139 – 0.241 - 0.05916/2 log(0.06/0.19) = -0.087 V
3. At 5.00 mL of Tl3+ added (at equiv. pt)
Ecell = {(E0 (Sn2+/Sn4+) + E0(Tl+/Tl3+)) / 2} – 0.241 V = (0.139 + 0.77)/2 – 0.241 V = 0.21 V
4. At 6.70 mL of Tl3+ added
mmol Tl3+ = (1.70 mL (excess))(0.0500 M) = 0.085 mmol
mmol Tl += (25.0 mL)(0.0100 M) = 0.250 mmol
Ecell = E0(Tl+/Tl3+) – E(SCE) – 0.05916/2 log([Tl+ ]/[Tl3+ ]) = 0.77 – 0.241 - 0.05916/2 log(0.250/0.085) = 0.515 V