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`