Question #39ff1

For simplicity, I will write Cytochrome C as Cc

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

#2H^+ +2e^(-)-> H_2#
#Cc(Fe^(3+))+5e^(-)->Cc(Fe^(2-))#

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:

#10H^+ +10e^(-)-> 5H_2#
#2Cc(Fe^(3+))+10e^(-)->2Cc(Fe^(2-))#

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:

#5H_2 ->10H^+ +10e^(-)#
#2Cc(Fe^(3+))+10e^(-)->2Cc(Fe^(2-))#

Add the two together and cancel out the electrons:

#2Cc(Fe^(3+))+5H_2 ->10H^+ 2Cc(Fe^(2-))#

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:

#Pt[H_(2(g))]|H_(aq)^+||Cc(Fe^(2-))|Pt|Cc(Fe^(3+))#

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)
...F = faradays constant (96500)

Hence:

#21.22=-10*96500*E_(cell)^°#

#21.22=-965000*E_(cell)^°#

#E_(cell)^°=-2.199*10^(-5)V#