Calculate the cell potential in the redox titration when 10.0mL of 0.100M Ce(IV)

Question

Calculate the cell potential in the redox titration when 10.0mL of 0.100M Ce(IV) has been added to 50.0mL of 0.100M Fe(II) solution (in 1M HCl). The reference electrode is Ag-AgCl in saturated KCl. please explain how to solve this question, from book Answer: 0.499V

Explanation / Answer

HA ------> H+ + A- HA be given acid HC3H5O20.1-x x xK = 1.3 x10^ -5 = (x^2)/(0.1-x)x = [H+] = 0.001134H+ moles = 0.001134 x 50 /1000 = 0.000056725 ml of 0.1 M KOH meansOH moles = 25 x0.1/1000 = 0.0025net OH- moles after titration = 0.0025-0.0000567 = 0.0024433[OH-] = 0.0024433 x 1000/75 =0.0318pOH = 1.5299pH = 12.47

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

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