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
May 12, 2017

DISCLAIMER: LONG ANSWER!

#"NAD"^(+)(aq) + "H"^(+)(aq) + 2e^(-) -> "NADH"(aq)#

#a)# This part is just asking you to write out your starting equation. It turns out that the Nernst equation is also for half-cells.

The general Nernst equation relates the non-standard #E_"cell"# to the #E_"cell"^@#, i.e. at standard conditions of #"1 M"# concentrations, #"1 atm"# pressure, and #25^@ "C"#:

#E_"cell" = E_"cell"^@ - (RT)/(nF)lnQ#

We now know how to write #Q#, the reaction quotient:

#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"^(+)]))#

#b)# Given a #"Pt" | "NADH, NAD"^(+),"H"^(+)# half-cell at #"pH"# #7#, we know:

  • 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 #E_"red"^@ = -"0.11 V"#, and that all other components are at #"1 M"# and the solutions are all at #"298 K"#.

So, we just use the equation we wrote down in part #(a)#:

#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 #"NADH"# is nonspontaneous using these concentrations in these conditions.

#c)# We know that cytochrome c with #"Fe"^(3+)# was reduced by #"NADH"# to cytochrome c with #"Fe"^(2+)#.

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 #"NADH"# was the reducing agent, it gets oxidized.

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.

#d)# Now, we are given the #"Pt" | "cyt c"("Fe"^(2+)), "cyt c"("Fe"^(3+))# half-cell, with an #E_"red"^@ = "0.25 V"# relative to the standard hydrogen electrode:

#"Fe"^(3+)(aq) + e^(-) -> "Fe"^(2+)(aq)#, #E_"red"^@ = "0.25 V"#

We now essentially repeat what we did in parts #(a,b)# with the reaction we wrote in part #(c)#.

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 #E_"cell"^@#. We were given in part #(b)# that #E_"red"^@ = -"0.11 V"# for the #"NAD"^(+) -> "NADH"# reduction, and in part #(d)# that #E_"red"^@ = +"0.25 V"# for the #"Fe"^(3+) -> "Fe"^(2+)# reduction.

Since we know from part #(c)# that #"NADH"# was oxidized, we can flip its reduction potential sign to obtain #E_"ox"^@ = +"0.11 V"#. Therefore, the standard cell potential is:

#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 #"pH"# #7# and (presumably) #"298 K"# again. The same number of electrons were transferred (two!), and the concentration of #"H"^(+)# is #10^(-7) "M"# again.

#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 #"NADH"# at these concentrations in these conditions is spontaneous.

May 14, 2017

Answer:

See below:

Explanation:

#sf(NAD^(+)+H^++2erightleftharpoonsNaDH" "E^@=-0.11color(white)(x)V)#

#sf((a))#

The potential E for a 1/2 cell is given by:

#sf(E=E^@-(RT)/(zF)ln([["reduced form"]]/(["oxidised form"]))#

At 298K this can be simplified to:

#sf(E=E^@-0.0591/(z)log([[NaDH]]/([NaD^+][H^+])))#

Where z is the no. of moles of electrons transferred which, in this case =2.

#sf((b))#

Putting in the numbers:

#sf(E=-0.11-0.591/2log((1)/((1)xx(10^(-7))))#

#sf(E=-0.11-0.2068=color(red)(-0.32color(white)(x)V))#

#sf((c))#

It helps to consider the #sf(E^@)# values:

#stackrel(color(white)(xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx))(color(blue)(larr)#

#sf(NAD^(+)+H^++2erightleftharpoonsNaDH" "E^@=-0.11color(white)(x)V)#

#sf(cFe^(3+)+erightleftharpoonscFe^(2+)" "E^@=+0.25color(white)(x)color(white)(x)V)#

#stackrel(color(white)(xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx))(color(red)(rarr)#

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:

#sf(NaDHrarrNaD^++H^++2e)#

#sf(cFe^(3+)+erarrcFe^(2+))#

To get the electrons to balance we X the 2nd 1/2 cell by 2 then add:

#sf(NaDH+2cFe^(3+)+cancel(2e)rarrNaD^(+)+H^++cancel(2e)+2cFe^(2+))#

#sf((d))#

Now we have to use the full Nernst Equation:

#sf(E_(cell)=E_(cell)^@-(RT)/(zF)lnQ)#

This can be simplified at 298K to:

#sf(E_(cell)=E_(cell)^@-0.0591/(z)logQ)#

Where z = 2.

To find #sf(E_(cell)^@)# there are different conventions around but they all rely on finding the arithmetic difference between the two #sf(E^@)# values for the 1/2 cells.

The clearest way is to subtract the least +ve #sf(E^@)# value from the most +ve.

Using the values in (c):

#sf(E_(cell)^@=+0.25-(-0.11)=+0.36color(white)(x)V)#

From the complete equation the reaction quotient Q is written:

#sf(Q=([NaD][H^+][cFe^(2+)]^2)/([cFe^(3+)]^2[NaDH]))#

pH = 7 #:.##sf([H^+]=10^(-7)color(white)(x)"mol/l")#

Putting the numbers into The Nernst Equation:

#sf(E_(cell)=+0.36-0.0591/2log((1xx10^(-7)xx0.004^2)/(0.01^2xx1)))#

#sf(E_(cell)=+0.36+0.22885 color(white)(x)V)#

#sf(E_(cell)=color(red)(+0.59color(white)(x)V)#