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Say we have a simple single replacement reaction between a salt and a metal $$\ce{2 AgNO3 + Cu -> Cu(NO3)2 + 2 Ag}$$

We know in normal circumstances, silver always has a $+1$ charge, and in normal circumstances for copper, it typically varies between a $+1$ or a $+2$.

Now I have been told that in copper nitrate, the copper ion will (almost?) always be in a $+2$ state, but this raises the question of why we would know that. Why can't the reaction be

$$\ce{AgNO3 + Cu -> CuNO3 + Ag}$$

instead?

Say we were to have a reaction between silver sulfate and copper. What charge would the copper take on? How do we know?

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  • $\begingroup$ There are a few ways to make an educated guess if you don't know it already, from textbook - that's a generic reaction, you'd stumble on it sooner or later. 1) You should know that Cu is pretty easily oxidised to +2, so if you use pretty strong oxidiser like Ag+... 2) Another option is that Cu+ actually tends to disproportionate in aqueous solutions to Cu and Cu 2+ $\endgroup$
    – Mithoron
    Commented Jun 2, 2022 at 0:14
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    $\begingroup$ Hmm, now that I think of it you mean actual experimental reinventing the wheel? I.e. we just ignore few hundred years of chemistry and find it out? $\endgroup$
    – Mithoron
    Commented Jun 2, 2022 at 0:19
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    $\begingroup$ You cannot know from basic principles. You must have background knowledge of element chemistry, eventually involving also concept of redox potentials. It can be also verified via element molar masses and experimentally determined equivalent mass ratios. $\endgroup$
    – Poutnik
    Commented Jun 2, 2022 at 7:24

1 Answer 1

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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).

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