[daveyavey]

Hey, sorry to bother you all, but I came across a question in my book, which i got a different answer to..
I just hope someone can clarify where I went wrong, or if there is an error in the book itself. Thanks

Q.Use the standard reduction potentials to calculate the EMF for the following cells.

Cu(s)|Cu2+(aq)||MnO4-(aq),Mn2+(aq)|Pt(s).

The standard reduction potentials for two half equations related are;
Cu2+ + e- ---> Cu     Eo= +0.34V
MnO4- + 8H+ + 5e- ----> Mn2+ + 4H2O  Eo=+1.51V

The answer I keep getting is +1.85V, however the answers say +1.17V,
Thanks for your help

[Borek]

Yous should subtract lower potential from higher potential.

[daveyavey]

Yes, I do that, but i have to reverse the first half equation, giving me -0.34V
So i minus a minus, adding them together to get +1.85V?
Or am I wrong?

[Borek]

You have to subtract lower from higher, if the lower has negative potential it is the same as changing sign and adding (-*- = +).

It is like calculating distance between two pints on the numerical axis.

[daveyavey]

Does that mean I am right and the text book is wrong?

[Sev]

Yes, I do that, but i have to reverse the first half equation, giving me -0.34V

Not nessesarily.  It is easier if you calculate: emf = E_cathode - E_anode (both as reduction potentials).  So like Borek said,

Yous should subtract lower potential from higher potential.

Does that mean I am right and the text book is wrong?

No, the textbook is right.

[daveyavey]

Yes, I do that, but i have to reverse the first half equation, giving me -0.34V

Not nessesarily.  It is easier if you calculate: emf = E_cathode - E_anode (both as reduction potentials).  So like Borek said,

Yous should subtract lower potential from higher potential.

Does that mean I am right and the text book is wrong?

No, the textbook is right.

Thanks, That just made me realise its standard REDUCTION potentials....
So I just use the reduction reaction E values which is why I would get +1.17V?
Cheers, was confusing me all day.