Controlled potential coulometry at a large platinum electrode is used to determi

Question

Controlled potential coulometry at a large platinum electrode is used to determine the amount of ferricyanide in a solution at 25 °C. How much more negative than the standard reduction potential (E°) of ferricyanide must the electrode potential be so that 99.9% of the ferricyanide, [Fe(CN)13-, is reduced to ferrocyanide, [Fe(CN)147 Number The total charge passed through a 80.0 mL solution of unknown ferricyanide concentration is 52.14 C Calculate the concentration of ferricyanide in the solution. Number

Explanation / Answer

The standard electrode potential for the reduction of [Fe(CN)6]3- to [Fe(CN)6]4- is +0.36 V

[Fe(CN)6]3- + 1e- ------> [Fe(CN)6]4- Eored = +0.36 V

Appluing Nernst equation

Ered = Eored - (0.0591 / 1)xlog[[Fe(CN)6]4-] / [[Fe(CN)6]3-]

Given [[Fe(CN)6]3-] = 100 - 99.9 = 0.1

and [[Fe(CN)6]4-] = 99.9

=> Ered = 0.36V - 0.0591xlog(99.9 / 0.1)

=> Ered = 0.36V - 0.1773V

Hence Ered should be 0.1773 V or 0.18 V negative than the standard reduction potential for 99.9% conversion.

Hence the answer is 0.1773 V or 0.18 V

Q.2: Since 1 electron is received during the redcution of ferricyanide to ferrocyanide, molar mass (Mw) = equivalent mass (E)

According to Faraday's first law of electrolysis

W = ZxQ

=> (moles of ferricyanide, n) x Mw = (E / 96500) x Q [As Z = E / 96500]

=> (moles of ferricyanide, n) x Mw = (Mw / 96500) x 52.14

=> moles of ferricyanide, n = (52.14/ 96500) = 5.40x10-4 mol

Also  moles of ferricyanide, n = M x V(L) = 5.40x10-4 mol

=> M x 80.0 mL x (1L / 1000 mL) = 5.40x10-4 mol

=> M = 6.75x10-3 M (answer)

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

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