# Question 39ff1

Apr 12, 2017

Check it out

#### Explanation:

For simplicity, I will write Cytochrome C as Cc

Hence, we are dealing with the following two half reactions:

$2 {H}^{+} + 2 {e}^{-} \to {H}_{2}$
$C c \left(F {e}^{3 +}\right) + 5 {e}^{-} \to C c \left(F {e}^{2 -}\right)$

Note: I balanced the individual reactions. There are plenty of videos on balancing redox reactions, so I will not get into too much detail here.

Now that we have the two half-reactions, we have to get the electrons to have the same coefficient; i.e. multiply the hydrogen half-reaction by 5 and the second half-reaction by 2:

$10 {H}^{+} + 10 {e}^{-} \to 5 {H}_{2}$
$2 C c \left(F {e}^{3 +}\right) + 10 {e}^{-} \to 2 C c \left(F {e}^{2 -}\right)$

The question tells you that "cytochrome c (Fe +) is reduced by hydrogen", which means that Hydrogen has to be oxidized. Hence, you have to flip the hydrogen-half-reaction-equation:

$5 {H}_{2} \to 10 {H}^{+} + 10 {e}^{-}$
$2 C c \left(F {e}^{3 +}\right) + 10 {e}^{-} \to 2 C c \left(F {e}^{2 -}\right)$

Add the two together and cancel out the electrons:

$2 C c \left(F {e}^{3 +}\right) + 5 {H}_{2} \to 10 {H}^{+} 2 C c \left(F {e}^{2 -}\right)$

You got your balanced half-reaction Yey

For the cell notation, I am assuming that the hydrogen half cell composed of a standard hydrogen electrode, containing a platinum black electrode with a constant ${H}_{2}$ flow. I do not know what kind of an electrode is used for the Cytochrome C, but I will assume that it is an inert platinum electrode:

$P t \left[{H}_{2 \left(g\right)}\right] | {H}_{a q}^{+} | | C c \left(F {e}^{2 -}\right) | P t | C c \left(F {e}^{3 +}\right)$

For the standard cell potential you have to know the following equation:

∆G=-nFE_(cell)^°

Where...
...n = number of electrons flowing (in this case there are 10)

Hence:

21.22=-10*96500*E_(cell)^°

21.22=-965000*E_(cell)^°

E_(cell)^°=-2.199*10^(-5)V