I was recently teaching some electrochemistry to some students, now before we get going it is important to note that the conventions on writing electrode potentials changed years ago. So if thou gets one’s ye olde text book out or ye olde almanack of chemie be ready for some possible problems.
For example in the 1950s paper on plutonium redox chemistry by Sherman W. Rabideau and Joe F. Lemon, Journal of the American Chemical Society, 1951, 73, 2895-2899 the redox couple for Pu(IV)/Pu(III) is listed as being -0.953 volts vs the standard hydrogen electrode. Now in modern text books it is listed as about + 0.95 volts. This modern use is an example of the “European” convention while the example I gave as an example of the “American” convention. I would say that both are equally valid but if you want to do a redox calculation be careful that you do not mix unwittingly mix data from both conventions up.
I have not seen the american convention being used much in modern text books, but please be aware that it does exist.
OK Health warning over lets get on with some chemistry. Now as a brain teaser (or brain expanded) I asked my students to consider the question of will a solution of iron(II) tend to reduce a solution of plutonium(IV) to plutonium(III) thus forming iron(III) in the process.
Now the redox couple of iron(II) / iron (III) is +0.77 volts (European convention). So we can calculate the cell voltage for our cell under “standard” conditions (1 mole per litre of everything, at 25 ºC and 1 bar).
We can combine the following two half equations
Fe2+ → Fe3+ + e–
e– + Pu4+ → Pu3+
To give us
Fe2++ Pu4+ → Fe3+ + Pu3+
The emf of the cell will be under standard conditions equal to 0.18 volts, but which thing will be reduced and what will be oxidized. The plutonium couple is higher (more positive) than the iron one in our European type text. So the plutonium will tend to oxidize the iron to form plutonium(III).
We get the emf of the cell from the difference between the two redox couples. All redox couples are expressed relative to hydrogen gas / hydrogen ions in 1 M acid. This choice of standard is simply a convention. Like the Greenwich meridian we need some arbitrary point to call zero. Hydrogen is a good choice as it features in so many reactions.
From the value of the emf of our cell we can get the ΔG of the reaction, to do this we use the following equation.
ΔG = -nFE
n is the number of electrons which are transferred in the cell reaction (1) and F is Faraday’s constant (charge on a mole of electrons) which is equal to 96485 C. Using this we can get a value for the Gibbs free energy of the reaction. Keep in mind that a cell with a positive emf is a cell which is able to do work, thus ΔG for the reaction in the cell must be negative.
This works out as -17367.3 joules per mole.
We can now get the equilibrium constant for this reaction, when a cell has a emf of zero then it has reached equilibrium. Now we need to use a different equation.
ΔG = ΔGº + RT Ln Q
In our case
Q = aPu3+ aFe3+ / aPu4+ aFe2+
Now in the ideal world (nice place I wish I lived there) the activity coefficient is equal to one, the closest we get to an ideal world is a dilute solution. So for nice and dilute solutions we can write
Q = [Pu3+] [Fe3+] / [Pu4+] [Fe2+]
If ΔG = 0 (zero) then
ΔGº = -RT ln ([Pu3+] [Fe3+] / [Pu4+] [Fe2+])
-ΔGº / RT = ln ([Pu3+] [Fe3+] / [Pu4+] [Fe2+])
exp (-ΔGº / RT) = ([Pu3+] [Fe3+] / [Pu4+] [Fe2+])
Now do the maths
exp (7) = ([Pu3+] [Fe3+] / [Pu4+] [Fe2+]) = k
Now k = 1107
we can rewrite the equations to give us
ΔGº = -RT ln k
Now it should be clear to my readers that a solution of iron(II) will reduce tetravalent plutonium into trivalent plutonium. You might be interested to read that the classic method of adjusting the oxidation state of plutonium from +4 to +3 in the PUREX process is to use ferrous sulfamate.