The molar solubility for Ag(Fe(CN)6 is 2.0x10^-9 M. An electrochemical cell is r

Question

The molar solubility for Ag(Fe(CN)6 is 2.0x10^-9 M. An electrochemical cell is represented by the following redox reaction: AgFe(CN)6 ---> AG + Fe(CN)6 1. Balance above reaction 2. Determine the Ksp for this reaction 3. If the reduction Ecell for the half reaction Ag+ + 1e --> Ag is 0.800 V, determine the Ecell for AgFe(CN)6 half reaction. 4. Calculate delta G The molar solubility for Ag(Fe(CN)6 is 2.0x10^-9 M. An electrochemical cell is represented by the following redox reaction: AgFe(CN)6 ---> AG + Fe(CN)6 1. Balance above reaction 2. Determine the Ksp for this reaction 3. If the reduction Ecell for the half reaction Ag+ + 1e --> Ag is 0.800 V, determine the Ecell for AgFe(CN)6 half reaction. 4. Calculate delta G 1. Balance above reaction 2. Determine the Ksp for this reaction 3. If the reduction Ecell for the half reaction Ag+ + 1e --> Ag is 0.800 V, determine the Ecell for AgFe(CN)6 half reaction. 4. Calculate delta G

Explanation / Answer

ANSWER:

Dear candidate The Reactant should be actually like this Ag3Fe(CN)6

(1) Ag3Fe(CN)6 ---> 3Ag + (Fe(CN)6)-2

(2)AgFe(CN)6 ---> Ag + Fe(CN)6

   s s

Ksp = s X s = (s)2 = 8.4 X 10-18

(3) Eocell = Eocathod - Eoanode

   = 0.8 - 0.3557 = 0.4443V

(3) delta Go = -nFEocell

n = no of electrons transfered in balanced chemical equation. = 3

F = Faraday constant = 96500

= -3X96500X0.4443 = -85.9KJ

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