# Help out with the following questions about the Nernst equation and #"NAD"^+#, cytochrome c, and #"Fe"# ions?

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#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 **nonspontaneous** using these concentrations in these conditions.

*reduced by*

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 *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.

#"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 **spontaneous**.

#### Answer:

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 **most** +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: