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)6]3–, is reduced to ferrocyanide, [Fe(CN)6]4–?

___V

The total charge passed through a 90.0 mL solution of unknown ferricyanide concentration is 25.83 C. Calculate the concentration of ferricyanide in the solution.

__ M

Answer 1

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- E^o_red = +0.36 V

Appluing Nernst equation

E_red = E^o_red - (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

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

=> E_red = 0.36V - 0.1773V

Hence E_red 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 25.83

=> moles of ferricyanide, n = (25.83 / 96500) = 2.677x10^-4 mol

Also  moles of ferricyanide, n = M x V(L) = 2.677x10^-4 mol

=> M x 90.0 mL x (1L / 1000 mL) = 2.677x10^-4 mol

=> M = 2.974x10^-3 M (answer)