## Hmi

4 see Sect. 9.5 for a discussion of the law of mass action hydrogen half-cell, the equilibrium condition for the overall reaction represented by Eq (10.30a) is:

To preserve the correct sign convention, the oxidized form of the species in half-cell A must appear on the right hand side of the overall reaction, as in reaction (10.30a).

Example Plutonium in aqueous solution in the IV oxidation state is to be reduced by adding a solution of ferrous ions. This is a step in the reprocessing of spent nuclear fuel. In a solvent extraction step, Pu4+ extracts along with U6+. Fe2+ reduces Pu4+ but not the U^. The reduced form of Pu is not extractable, and so separation of U and Pu is possible.

If 1 liter each of 1 M Fe2+ and 0.5 M Pu4+ solutions are mixed and equilibrated, what is the ratio of Pu4+ to Pu3+ ?

The overall reaction is:

This reaction can be broken up into half-cell reactions numbers 9 and 12 in Table 10.1. However, it is not necessary to write the Nernst equations for these half-cell reactions. Instead, the law of mass action based on the above overall reaction and of the equilibrium constant can be calculated from Eq (10.31). Component A in this equation is iron, since its oxidized form appears on the right hand side of the reaction as written. The result is:

K = cFe3+ cPu3+ = 1o(e-s^)/ 0.059 = 1°[-°.77-(-0.98)//0.059 = 