You are performing a titration of 25.0 mL of 0.0100 M Sn4 in 1 M HCl with 0.0500

Question

You are performing a titration of 25.0 mL of 0.0100 M Sn4 in 1 M HCl with 0.0500 M Ag to give Sn2 and Ag3 using a Pt indicator electrode and a saturated calomel electrode (SCE) as the reference electrode.

Map You are performing a titration of 25.0 mL of 0.0100 M Sn4+in 1 M HCI with 0.0500 M Ag to give Sn2 ana Ag3* using a Pt indicator electrode and a saturated calomel electrode (SCE) as the reference electrode A) Write the balanced titration reaction B) Complete the two half-reactions that occur at the indicator electrode (shown is their corresponding reduction potential) Sn4++ 2e- Sn2+ Eo=0.139 V Eo 1.90 V C) Choose the two different expressions for the potential of the cell, consisting of the Nernst equation for each half-reaction minus the potential of the SCE (0.241 V) [A] E=10.139-0.05916 log| 2+ [B] E=10.139-0.05916 looLSn1 A E- 0.139 0.241 BE- [0.139 0.241 Scroll down to answer part D of this question

Explanation / Answer

For the given titration reaction

From the half cell reactions

C) The two Nernst equation for the cell would be,

A

E

D) Cell voltage

(a) 1 ml of Ag+ added

moles of Sn4+ present = 0.01 x 25 = 0.25 mmol

moles of Ag+ added = 0.05 M x 1 ml = 0.05 mmol

[Sn4+] remained = 0.2/26 = 0.0077 M

[Sn2+] formed = 0.05/26 = 0.0019 M

E = [0.139 - 0.0592/2 log(0.0019/0.0077)] - 0.241 = -0.084 V

(b) 2.5 ml Ag+ added

moles of Sn4+ present = 0.01 x 25 = 0.25 mmol

moles of Ag+ added = 0.05 M x 2.5 ml = 0.125 mmol

[Sn4+] = [Sn2+]

E = 0.139 - 0.241 = -0.102 V

(c) 4.8 ml Ag+ added

moles of Sn4+ present = 0.01 x 25 = 0.25 mmol

moles of Ag+ added = 0.05 M x 4.8 ml = 0.24 mmol

[Sn4+] remained = 0.01/29.8 = 0.000335 M

[Sn2+] formed = 0.24/29.8 = 0.00805 M

E = [0.139 - 0.0592/2 log(0.00805/0.000335) = -0.143 V

(d) 5 ml Ag+ added

moles of Sn4+ present = 0.01 x 25 = 0.25 mmol

moles of Ag+ added = 0.05 M x 5 ml = 0.25 mmol

E = [(0.139 + 1.90)/2] - 0.241 = 0.7785 V

(e) 5.10 ml Ag+ added

moles of Sn4+ present = 0.01 x 25 = 0.25 mmol

moles of Ag+ added = 0.05 M x 5.1 ml = 0.255 mmol

[Ag+] remained = 0.005/30.1 = 0.00017 M

[Ag3+] formed = 0.25/30.1 = 0.0083 M

E = [1.90 - 0.0592/2 log(0.00017/0.0083)] - 0.241 = 1.76 V

(f) 10 ml Ag+ added

moles of Sn4+ present = 0.01 x 25 = 0.25 mmol

moles of Ag+ added = 0.05 M x 10 ml = 0.5 mmol

E = 1.90 - 0.241 = 1.659 V

Related Questions

Navigate

• Browse (All)
• Browse Y
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.