Question

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)`?

Answer 1

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.

Answer 2

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)`