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 degree C. How much more negative than the standard reduction potential (E degree) of ferricyanide must the electrode potential be so that 99.9% of the ferricyanide, [Fe(CN)_6]^3-, is reduced to ferrocyanide, [Fe(CN)_6]^4-? The total charge passed through a 95.0 mL solution of unknown ferricyanide concentration is 86.51 C. Calculate the concentration of ferricyanide in the solution.

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 86.51

=> moles of ferricyanide, n = (86.51 / 96500) = 8.965x10-4 mol

Also  moles of ferricyanide, n = M x V(L) = 8.965x10-4 mol

=> M x 95.0 mL x (1L / 1000 mL) = 8.965x10-4 mol

=> M = 9.44x10-3 M (answer)

  

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