If we have a solution of S2O8-2 (+2.01V) ion and Ag+ (+0.80V)
ion then which wil be discharged at cathode? Assuming both have 1 M
solution.
An excellent trick question you can use to test your students!
Although the standard reduction potential of peroxodisulfate(VI)
aka peroxydisulfate(VI) to sulfate(VI) is more positive than the
standard reduction potential of silver(I) to silver(0), and hence
peroxodisulfate(VI) aka peroxydisulfate(VI) will seem to be more
thermodynamically favored to be reduced to sulfate(VI) at the
cathode, but that does NOT occur at all.
(Note that the term "discharged" can only be used
specifically if the species loses a charge, ie. Ag+ discharged to
Ag, but when S2O8 2- is reduced to SO4 2-, the negative charge
remains, so advise your students to write "oxidized at anode" and
"reduced at cathode" instead of loosely (and often erroneously)
misusing the term "discharge").
The reason for this, is because the peroxodisulfate(VI) aka
peroxydisulfate(VI) ion is ANIONIC, and hence electrostatically
migrates to the ANODE (instead of cathode), but cannot be further
oxidized (so some other anion present, or H2O, will have to be
oxidized instead, exactly which, depends on their oxidation
potentials).
The Ag+ CATION will migrate to the CATHODE and be reduced to
Ag(s) instead. So in this setup, to answer your question, Ag+ is
reduced instead of the peroxodisulfate(VI) aka
peroxydisulfate(VI).