The folllowing electrochemical cell was constructed, S.C.E. II cell solution I P
ID: 930599 • Letter: T
Question
The folllowing electrochemical cell was constructed, S.C.E. II cell solution I Pt(s) with a cell voltage of 0.353 V. With a total volume of 1 L, the cell solution (pH 6.85) was comprised of 0.0025 M Ce(SO4)2, 0.0010 M CeCl3, and 0.0045 M Na2EDTA.
Please do not answer this if you do not know EXACTLY what you are doing. This is my THIRD time asking this question. Also please answer all parts. Thank you
The folllowing electrochemical cell was constructed, S.C.E. Il cell solution IPt(s) with a cell voltage of 0.353 V. With a total volume of 1 L, the cell solution (pH 6.85) was comprised of 0.0025 M Ce(SO4)2,0.0010 M CeCl3, and 0.0045 M Na2EDTA A) Write the reaction for the right half-cell as a reduction and provide its standard reduction potential (E°). Number 4+ 3 + E" = 1 1.61 B) Calculate the quotient for the uncomplexed ions, (Ce3TCe"], in the cell solution. Number 3+ Ce | | 0.4 = 4+ Ce C) Determine the formatign constant quotient, KCeEDTA)CeEDTA) Number K, (CeEDTA) K CeEDTA -2.5 IncorrectExplanation / Answer
a.
The reaction for the right half cell as a reduction is:
CeEDTA + e = CeEDTA- EoCeEDTA / CeEDTA- = ?
(EDTA is here the symbol for Y4- , a quadrivalent anion of EDTA)
Eo has to be find in a table or can be calculated.
Theoretical:
Ce4+ + e = Ce3+ EoCe4+/Ce3+ = 1.61 V (1)
CeEDTA + e = CeEDTA- EoCeEDTA / CeEDTA- = ? (2)
From Eq.1 – Eq.2 :
Ce4+ + CeEDTA- = Ce3+ + CeEDTA
For this cell write as usual
Eocell = EoCe4+/Ce3+ - EoCeEDTA / CeEDTA-and
Ecell = Eocell – 0.0592log([Ce3+ ][CeEDTA] / [Ce4+][CeEDTA-]) (3)
If the reaction is at equilibrium Ecell = 0 V and
[Ce3+ ][CeEDTA] / [Ce4+][CeEDTA-] is an equilibrium constant.
K = [Ce3+ ][CeEDTA] / [Ce4+][CeEDTA-] =
= [Ce3+ ][EDTA][CeEDTA] / [Ce4+][EDTA][CeEDTA-] =
= Kf,CeEDTA / Kf,CeEDTA- (4)
EoCe4+/Ce3+ - EoCeEDTA / CeEDTA- = 0.0592log(Kf,CeEDTA / Kf,CeEDTA-)
From reference tables I found that
logKf,CeEDTA = 24.20 Kf,CeEDTA = 1.585x1024
logKf,CeEDTA- = 15.98 Kf,CeEDTA- = 9.550x1015
K = 1.585x1024 /9.550x1015 = 1.66x108
(If the reaction is not at equilibrium [Ce3+ ][CeEDTA] / [Ce4+][CeEDTA-] is the reaction quotient Q)
At equilibrium, using eq. 3 and 4:
0 = Eocell – 0.0592logK
Eocell = 0.0592logK
= 0.0592x8.22 =
= 0.487 V
EoCe4+/Ce3+ - EoCeEDTA / CeEDTA- = 0.49 V
1.61 V- EoCeEDTA / CeEDTA- = 0.49 V
EoCeEDTA / CeEDTA- = 1.12 V
You have to find this value in your book or to use this value, calculated by me.
b.
Your electrochemical cell is not at equilibrium, use eq. (3):
Ecell = Eocell – 0.0592logQ
(If the reaction is not at equilibrium [Ce3+ ][CeEDTA] / [Ce4+][CeEDTA-] is the reaction quotient Q)
Ecell is measured vs SCE, correct the value as referenced to NHE
Ecell = 0.353 + 0.241= 0.594 V vs NHE
Eocell = 0.487 V (see a.)
0.594 V = 0.487 - 0.0592logQ
logQ = - 0.107/0.0592
= - 1.807
Q = 0.01558 = 0.0156
Q = ([Ce3+ ][CeEDTA] /([Ce4+][CeEDTA-])) = 0.0156
The initial concentrations used for the preparation of the cell solution were:
CCe3+ = 0.0010M
CCe4+ = 0.0025
CEDTA = 0.0045
The complexes [CeEDTA] and [CeEDTA-] are very stable, thus you may assume with negligible errors that
[CeEDTA] = 0.0025 M
[CeEDTA-] = 0.0010 M
Then:
Q = 0.0156 = ([Ce3+ ] x 0.0025 / ([Ce4+] x 0.0010])) =
= 2.5 x ([Ce3+ ]/ [Ce4+])
[Ce3+ ]/ [Ce4+] = 0.00624
c.
Eocell = 0.0592logK (see a.)
Where K = Kf,CeEDTA / Kf,CeEDTA-
Eocell = 0.487 V (see a.)
0.487 = 0.0592 log Kf,CeEDTA / Kf,CeEDTA-
log Kf,CeEDTA / Kf,CeEDTA- = 8.226
Kf,CeEDTA / Kf,CeEDTA- = 108.22 = 1.66 x108