Why is a Solution of ZnSO4 Necessary for Zn & Cu Half Cells?

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A solution of ZnSO4 is necessary in a zinc half cell connected to a copper half cell to provide the Zn2+ ions needed for the oxidation process, even though these ions are produced during oxidation. The presence of copper ions is essential for reduction to occur, allowing for current flow in the cell. Oxidation in water can occur due to H+ ions from water autodissociation or dissolved oxygen, despite copper's low reactivity. The potential of the half cells is influenced by the concentration of ions present, as described by the Nernst equation, which shows that ion concentration affects the cell's potential. Understanding these dynamics is crucial for grasping electrochemical cell behavior.
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I would like to know why is a solution of ZnSO4 necessary in a Zinc half cell when connected to a Cu half cell. The ions Zn2+ are produced by the zinc electrode so a priori no solution of ZnSO4 is needed and plain water would be enough.
More generally, why a half cell always needs both the oxidized and reduced form in it?
The Cu half cell when connected to a Zn half cell does not need a Cu electrode, or does it?
Thank you
 
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Zn2+ are not necessary, they will be created when zinc gets oxidized. Copper ions are necessary, if they are not present there is nothing that can be reduced, so you can't draw any current from the cell.

But to see potential difference you don't need ions, just electrodes are enough - some of the metal on both electrodes will get immediately oxidized and you will have both metal and ions present. I believe cells are usually presented the way they are as they are easier to understand conceptually then.
 
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Thank you for your clear answer.

If I understand well, whether I put zinc or copper in water some of the metal is going to oxidize and form ions in the water. What is causing this oxidation in water? Copper is low in the reactivity series. Why should it oxidize at all in water?

Another question: Does it affect the potential of Zn or Cu half cell if the metal is place in a solution of its ions instead of water?

I have with me several advanced chemistry textbooks and none of them come even close to give an answer to these (maybe too trivial) questions.

Thank you.
 
hushai1 said:
If I understand well, whether I put zinc or copper in water some of the metal is going to oxidize and form ions in the water. What is causing this oxidation in water?

Either just H+ from water autodissociation, or dissolved oxygen, or both.

Copper is low in the reactivity series. Why should it oxidize at all in water?

Reactivity series is only a simplified approximation of the reality. Potential of the cell depends on the ion concentrations. Metals in reactivity series are sorted according to their standard potentials, that is, potential of the metal electrode in contact with 1M solution of the ions. If the ion concentration is substantially different, reactivity series order can fail. See below.

Another question: Does it affect the potential of Zn or Cu half cell if the metal is place in a solution of its ions instead of water?

Potential can be calculated using Nernst equation, and one of the parameters is ion concentration. For example:

E = E_0 + \frac {RT} {nF} \ln [Cu^{2+}]

E0 is standard potential (determined experimentally - you can get it from tables), n is number of electrons exchanged in the electrode reaction (2 for copper and zinc), T is temperature, R & F are constants.

Obviously equation has no sense for zero concentration, but that's not a problem - metal gets oxidized immediately, so concentration is never really zero. Concentration will grow till the cell potential equals potential of whatever oxidizing agent is present (be it H+ mentioned earlier). In the case of copper in pure water it means concentration of metal ion is in the 10-26 M range.
 
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Thank you very much. Your answers are excellent.
 
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