question on 6M.5 part c

[Elizabeth Bowen 1J]

The question ins to write a cell diagram, determine the standard cell potential, and calculate delta G for the reaction:
Cr2O7 2-(aq) + 14H+ (aq) + 6 Pu 3+ (aq) --> 6Pu 4+ (aq) + 2Cr 3+ (aq) + 7H2O (l)

How do we find the standard reduction potential for the oxidation half reaction (6Pu3+ (aq)->6Pu4+ (aq) + e-)--> because in the appendex 2B they have the have the reverse half reaction (Divided by 6) with a standard potential of 0.97, but would that be the same as the reaction given? In the solutions manual they used 0.97, but I don't understand where they got that from because in the appendix it's the opposite reaction. Also, in the solutions they say that n=6, which is confusing to me because when I wrote out the reduction reaction i got
9e- + Cr2O7 2- (aq) + 14H+ (aq) --> 2 Cr 3+ (aq) + 7H2O (l), so how would they get that 6 electrons are transferred in this redox reaction, when it seems like 9 are needed to even out the charge on both sides?

Thanks

[asannajust_1J]

the reduction of cr2o7 2- goes from oxidation number (6+-->3+) which is a transfer of 3 electrons, but since there are two, it would transfer 6 electrons. The coefficients of the reaction do not influence the cell potential so you would not divide the cell potential by this value. 1 electron is transferred in the oxidation of Pu3+ --> Pu4+, and since you need to balance each half reaction this would result in multiplying the half reaction by 6 thus. n=6 for the overall reaction since 6 electrons are transferred in the redox rxn. Use Ecell=Er-Eo to find the standard cell potential for the overall reaction.