This is an excellent question. One would normally expect the first ionization potentials to indicate how easily an element gives up the first electron. With its low first ionization energy, it would seem that Ag would easily react in solution. The trick is that the ionization energy is measured from the reaction
Ag --> Ag+ + electron
for an Ag atom in the gas phase. If the Ag were to react in solution, you have solid Ag metal that first needs to break up into atoms and then the atoms have to ionize to give an Ag+ in solution: A completely different picture!
So instead of ionization energies, what you really should use for solution reactions is the standard reduction potential as embodied in the so-called electrochemical series. This is related to the negative of the free energy of the reaction:
M(n+) + ne --> M
The values are expressed in volts, since these are measured in terms of the electric potential.
Here are some values:
Li+ + e --> LiE = -3.04 V
K+ + e --> K E = -2.93 V
Mn2+ + 2e --> Mn E = -1.19 V
Zn2+ + 2e --> Zn E = -0.76 V
2H+ + 2e --> H2 E = 0 V [This is used as the reference]
Cu2+ + 2e --> Cu E = +0.34 V
Ag+ + e --> Ag E = +0.80 V
So Cu and Ag will not react with water to give the ions, but clean Zn and Mn will give up their electrons and displace H from water. K and Li will do this violently. The large voltage associate with Li going to Li+ is why Li is used in batteries.
You see from this table that Ag is very inert despite its small ionization energy. So you should be testing a AgNO3 solution with a copper wire (the copper will replace Ag in solution and the wire will obtain a silver coating). Likewise Zn filings will displace Cu metal from a CuSO4 solution.
I hope this is useful.
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