# 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 ${\text{NAD}}^{+}$ to $\text{NADH}$? What is the expression for $Q$? b) If you have a ${\text{Pt"|"NADH, NAD"^(+), "H}}^{+}$ half-cell at $\text{pH}$ $7$, and ${E}_{red}^{\circ} = - \text{0.11 V}$ at $\text{298 K}$, find ${E}_{red}$ if the concentrations of everything else is $\text{1 M}$? c) $\text{NADH}$ reduces cytochrome c from the ${\text{Fe}}^{3 +}$ to the ${\text{Fe}}^{2 +}$ form. Write this reaction? d) Given a "Pt"|"cyt c"("Fe"^(2+)), "cyt c"("Fe"^(3+)) half-cell, with ${E}_{red}^{\circ} = \text{0.25 V}$ relative to SHE, write $Q$ and find ${E}_{c e l l}$ at $\text{298 K}$ for the reactions given for parts $\left(a\right)$ and $\left(c\right)$?

May 12, 2017

$\text{NAD"^(+)(aq) + "H"^(+)(aq) + 2e^(-) -> "NADH} \left(a q\right)$

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 ${E}_{\text{cell}}$ to the ${E}_{\text{cell}}^{\circ}$, i.e. at standard conditions of $\text{1 M}$ concentrations, $\text{1 atm}$ pressure, and ${25}^{\circ} \text{C}$:

${E}_{\text{cell" = E_"cell}}^{\circ} - \frac{R T}{n F} \ln Q$

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

$Q = \left(\left[{\text{NADH"])/(["NAD"^(+)]["H}}^{+}\right]\right)$

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 ${\text{Pt" | "NADH, NAD"^(+),"H}}^{+}$ half-cell at $\text{pH}$ $7$, we know:

• that the platinum electrode is inert and does not participate in the reaction.
• that the concentration of $\boldsymbol{{\text{H}}^{+}}$ is ["H"^(+)] = 10^(-"pH") = 10^(-7) "M".
• that from the half-cell reaction, the number of electrons transferred is TWO.

We are also given ${E}_{\text{red"^@ = -"0.11 V}}$, and that all other components are at $\text{1 M}$ and the solutions are all at $\text{298 K}$.

So, we just use the equation we wrote down in part $\left(a\right)$:

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"))

$= \textcolor{b l u e}{- \text{0.32 V}}$

This indicates that the reduction of $\text{NADH}$ is nonspontaneous using these concentrations in these conditions.

c) We know that cytochrome c with ${\text{Fe}}^{3 +}$ was reduced by $\text{NADH}$ to cytochrome c with ${\text{Fe}}^{2 +}$.

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 $\text{NADH}$ 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.

$2 \left({\text{Fe"^(3+)(aq) + cancel(e^(-)) -> "Fe}}^{2 +} \left(a q\right)\right)$
${\text{NADH"(aq) -> "NAD"^(+)(aq) + "H}}^{+} \left(a q\right) + \cancel{2 {e}^{-}}$
$\text{-------------------------------------------------------------}$
$\textcolor{b l u e}{2 {\text{Fe"^(3+)(aq) + "NADH"(aq) -> "NAD"^(+)(aq) + "H"^(+)(aq) + 2"Fe}}^{2 +} \left(a q\right)}$

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 ${E}_{\text{red"^@ = "0.25 V}}$ relative to the standard hydrogen electrode:

${\text{Fe"^(3+)(aq) + e^(-) -> "Fe}}^{2 +} \left(a q\right)$, ${E}_{\text{red"^@ = "0.25 V}}$

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

This is a pretty bulky reaction quotient expression...

$Q = \left(\left[\text{NAD"^(+)]["H"^(+)]["Fe"^(2+)]^2)/(["Fe"^(3+)]^2["NADH}\right]\right)$

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 ${E}_{\text{cell}}^{\circ}$. We were given in part $\left(b\right)$ that ${E}_{\text{red"^@ = -"0.11 V}}$ for the $\text{NAD"^(+) -> "NADH}$ reduction, and in part $\left(d\right)$ that ${E}_{\text{red"^@ = +"0.25 V}}$ for the ${\text{Fe"^(3+) -> "Fe}}^{2 +}$ reduction.

Since we know from part $\left(c\right)$ that $\text{NADH}$ was oxidized, we can flip its reduction potential sign to obtain ${E}_{\text{ox"^@ = +"0.11 V}}$. Therefore, the standard cell potential is:

${E}_{\text{cell"^@ = E_"red"^@ + E_"ox}}^{\circ}$

= "0.25 V" + [-stackrel(E_"red"^@)overbrace((-"0.11 V"))]

$=$ $+ \text{0.36 V}$

Now, we can calculate the non-standard cell potential at $\text{pH}$ $7$ and (presumably) $\text{298 K}$ again. The same number of electrons were transferred (two!), and the concentration of ${\text{H}}^{+}$ is ${10}^{- 7} \text{M}$ 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"))

$= \textcolor{b l u e}{+ \text{0.59 V}}$

Thus, the reduction of the iron center in cytochrome c by $\text{NADH}$ at these concentrations in these conditions is spontaneous.

May 14, 2017

See below:

#### Explanation:

$\textsf{N A {D}^{+} + {H}^{+} + 2 e r i g h t \le f t h a r p \infty n s N a D H \text{ } {E}^{\circ} = - 0.11 \textcolor{w h i t e}{x} V}$

$\textsf{\left(a\right)}$

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:

$\textsf{E = {E}^{\circ} - \frac{0.0591}{z} \log \left(\frac{\left[N a D H\right]}{\left[N a {D}^{+}\right] \left[{H}^{+}\right]}\right)}$

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

$\textsf{\left(b\right)}$

Putting in the numbers:

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

$\textsf{E = - 0.11 - 0.2068 = \textcolor{red}{- 0.32 \textcolor{w h i t e}{x} V}}$

$\textsf{\left(c\right)}$

It helps to consider the $\textsf{{E}^{\circ}}$ values:

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

$\textsf{N A {D}^{+} + {H}^{+} + 2 e r i g h t \le f t h a r p \infty n s N a D H \text{ } {E}^{\circ} = - 0.11 \textcolor{w h i t e}{x} V}$

$\textsf{c F {e}^{3 +} + e r i g h t \le f t h a r p \infty n s c F {e}^{2 +} \text{ } {E}^{\circ} = + 0.25 \textcolor{w h i t e}{x} \textcolor{w h i t e}{x} V}$

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:

$\textsf{N a D H \rightarrow N a {D}^{+} + {H}^{+} + 2 e}$

$\textsf{c F {e}^{3 +} + e \rightarrow c F {e}^{2 +}}$

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

$\textsf{N a D H + 2 c F {e}^{3 +} + \cancel{2 e} \rightarrow N a {D}^{+} + {H}^{+} + \cancel{2 e} + 2 c F {e}^{2 +}}$

$\textsf{\left(d\right)}$

Now we have to use the full Nernst Equation:

$\textsf{{E}_{c e l l} = {E}_{c e l l}^{\circ} - \frac{R T}{z F} \ln Q}$

This can be simplified at 298K to:

$\textsf{{E}_{c e l l} = {E}_{c e l l}^{\circ} - \frac{0.0591}{z} \log Q}$

Where z = 2.

To find $\textsf{{E}_{c e l l}^{\circ}}$ there are different conventions around but they all rely on finding the arithmetic difference between the two $\textsf{{E}^{\circ}}$ values for the 1/2 cells.

The clearest way is to subtract the least +ve $\textsf{{E}^{\circ}}$ value from the most +ve.

Using the values in (c):

$\textsf{{E}_{c e l l}^{\circ} = + 0.25 - \left(- 0.11\right) = + 0.36 \textcolor{w h i t e}{x} V}$

From the complete equation the reaction quotient Q is written:

$\textsf{Q = \frac{\left[N a D\right] \left[{H}^{+}\right] {\left[c F {e}^{2 +}\right]}^{2}}{{\left[c F {e}^{3 +}\right]}^{2} \left[N a D H\right]}}$

pH = 7 $\therefore$$\textsf{\left[{H}^{+}\right] = {10}^{- 7} \textcolor{w h i t e}{x} \text{mol/l}}$

Putting the numbers into The Nernst Equation:

$\textsf{{E}_{c e l l} = + 0.36 - \frac{0.0591}{2} \log \left(\frac{1 \times {10}^{- 7} \times {0.004}^{2}}{{0.01}^{2} \times 1}\right)}$

$\textsf{{E}_{c e l l} = + 0.36 + 0.22885 \textcolor{w h i t e}{x} V}$

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