EOCE 20.114

[Zellmer, Robert]

I've received 2-3 questions about this question.

You're given two electrodes and their E^o values:

1) Ag^+ (aq) + e-  --> Ag(s)                            E^o = +0.80 V

2) AgSCN(s) + e- --> Ag(s) + SCN- (aq)        E^o = +0.09 V

You want to calculate the Ksp for AgSCN, which is the following rxn:

AgSCN(s) --> Ag^+(aq)  +  SCN- (aq)

You need to use the electrode reactions above to get this rxn.
You need to figure out which one is the anode (oxidation) and
which is the cathode (reduction) so they add up to give the Ksp
rxn.  Look at rxns 1 and 2 and figure out which one you'll have
to reverse so you can add them up to give the Ksp rxn.  When
you do this that rxn will become the ox. half reaction.  Remember
though, you don't change the sign of the E^o for that reaction.
To determine E^o(cell) you use

E^o(cell) = E^o(reduction, cathode) - E^o(oxidation, anode)

where these are the reduction potentials given (w/o changing the
sign of the rxn you must reverse).

That doesn't mean the E^o(cell) will be positive and it will be a voltaic
cell.  It can't be.  How do I know this?  Ksp values are small, much
less than 1.  That means it's nonspont. to product and that in turn
means E^o(cell) < 0.


Dr. Zellmer
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