Say we were to have a reaction between silver sulfate and copper. What
charge would the copper take on? How do we know?
Silver sulfate is not very water-soluble. Let us say that we use a generic soluble Ag(I) salt and we wish to know what will be the resulting charge on the copper ion after the reaction. If we were discovering/ studying this reaction for the first time, we might do the following:
One is the pure observational way: We make a solution of a silver salt in water and place a copper plate in the solution. After some time, the solution turns blue. Using previous knowledge that Cu(II) solutions in water are typically blue, one can make an educated guess that the resulting ion is Cu(II) not Cu(I). Cu(I) is colorless.
Second option is pure analytical way: We let copper and silver nitrate react completely (i.e., excess copper), and after a reaction, we physically separate the silver particles and remaining copper by filtration. We evaporate the solution and analyze the blue crystals chemically. The resulting formula after analysis will suggest that copper must be in Cu(II) form, otherwise the elemental analysis data would not agree with Cu(I) salt.
Third approach is the electrochemical approach which you might not be familiar as yet: One can start with half-cells and look at electrode potential values. The question is, if Cu is given a choice to react with silver ions in water, what will be the most favorable product?
Ag(I)/ Ag half-cell has an electrode potential of +0.80 V
Cu(I)/Cu half-cell has an electrode potential of +0.52 V
Cu(II)/Cu half-cell has an electrode potential of +0.34 V
Which half-cell of copper will give a larger positive potential difference with silver half-cell? It is the Cu(II)/Cu half cell, which means that if Cu(s) reacts with Ag(I), the thermodynamically favored product will be Cu(II) not Cu(I).