Help out with the following questions about the Nernst equation and #"NAD"^+#, cytochrome c, and #"Fe"# ions?
#a)# Write out the Nernst equation for the reduction of #"NAD"^+# to #"NADH"# ? What is the expression for #Q# ?
#b)# If you have a #"Pt"|"NADH, NAD"^(+), "H"^(+)# half-cell at #"pH"# #7# , and #E_(red)^@ = -"0.11 V"# at #"298 K"# , find #E_(red)# if the concentrations of everything else is #"1 M"# ?
#c)# #"NADH"# reduces cytochrome c from the #"Fe"^(3+)# to the #"Fe"^(2+)# form. Write this reaction?
#d)# Given a #"Pt"|"cyt c"("Fe"^(2+)), "cyt c"("Fe"^(3+))# half-cell, with #E_(red)^@ = "0.25 V"# relative to SHE, write #Q# and find #E_(cell)# at #"298 K"# for the reactions given for parts #(a)# and #(c)# ?
2 Answers
DISCLAIMER: LONG ANSWER!
#"NAD"^(+)(aq) + "H"^(+)(aq) + 2e^(-) -> "NADH"(aq)#
The general Nernst equation relates the non-standard
#E_"cell" = E_"cell"^@ - (RT)/(nF)lnQ#
We now know how to write
#Q = (["NADH"])/(["NAD"^(+)]["H"^(+)])#
Hence, the Nernst equation for this problem, which is for a half-cell, is:
#bb(E_"red" = E_"red"^@ - (RT)/(nF)ln((["NADH"])/(["NAD"^(+)]["H"^(+)]))#
- that the platinum electrode is inert and does not participate in the reaction.
- that the concentration of
#bb("H"^(+))# is#["H"^(+)] = 10^(-"pH") = 10^(-7) "M"# . - that from the half-cell reaction, the number of electrons transferred is TWO.
We are also given
So, we just use the equation we wrote down in part
#color(blue)(E_"red") = -"0.11 V" - ("8.314472 J/mol"cdot"K"cdot"298 K")/("2 mol e"^(-)cdot"96485 C/mol e"^(-))ln(("1 M")/("1 M"cdot 10^(-7) "M"))#
#= color(blue)(-"0.32 V")#
This indicates that the reduction of
Since the important reaction is the reduction of the iron center, let's focus on that half-reaction. However, don't forget our other half-reaction. Since
Thus, we flip our first half reaction, which was a reduction to begin with, and turn it into an oxidation half-reaction.
#2("Fe"^(3+)(aq) + cancel(e^(-)) -> "Fe"^(2+)(aq))#
#"NADH"(aq) -> "NAD"^(+)(aq) + "H"^(+)(aq) + cancel(2e^(-))#
#"-------------------------------------------------------------"#
#color(blue)(2"Fe"^(3+)(aq) + "NADH"(aq) -> "NAD"^(+)(aq) + "H"^(+)(aq) + 2"Fe"^(2+)(aq))#
I'll leave you to check that the mass and charge are balanced.
#"Fe"^(3+)(aq) + e^(-) -> "Fe"^(2+)(aq)# ,#E_"red"^@ = "0.25 V"#
We now essentially repeat what we did in parts
This is a pretty bulky reaction quotient expression...
#Q = (["NAD"^(+)]["H"^(+)]["Fe"^(2+)]^2)/(["Fe"^(3+)]^2["NADH"])#
Thus, the Nernst equation for this problem is:
#E_"cell" = E_"cell"^@ - (RT)/(nF)ln((["NAD"^(+)]["H"^(+)]["Fe"^(2+)]^2)/(["Fe"^(3+)]^2["NADH"]))#
We first have to figure out
Since we know from part
#E_"cell"^@ = E_"red"^@ + E_"ox"^@#
#= "0.25 V" + [-stackrel(E_"red"^@)overbrace((-"0.11 V"))]#
#=# #+"0.36 V"#
Now, we can calculate the non-standard cell potential at
#color(blue)(E_"cell") = "0.36 V" - ("8.314472 J/mol"cdot"K"cdot"298 K")/("2 mol e"^(-)cdot"96485 C/mol e"^(-))ln(("1 M"cdot10^(-7)"M"cdot("0.004 M")^2)/(("0.01 M")^2cdot"1 M"))#
#= color(blue)(+"0.59 V")#
Thus, the reduction of the iron center in cytochrome c by
See below:
Explanation:
The potential E for a 1/2 cell is given by:
At 298K this can be simplified to:
Where z is the no. of moles of electrons transferred which, in this case =2.
Putting in the numbers:
It helps to consider the
The most +ve 1/2 cell is the one that will take in the electrons so you can see that the 1st 1/2 cell will move right to left and the 2nd 1/2 cell will move left to right in accordance with the arrows.
The two 1/2 equations are therefore:
To get the electrons to balance we X the 2nd 1/2 cell by 2 then add:
Now we have to use the full Nernst Equation:
This can be simplified at 298K to:
Where z = 2.
To find
The clearest way is to subtract the least +ve
Using the values in (c):
From the complete equation the reaction quotient Q is written:
pH = 7
Putting the numbers into The Nernst Equation: