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

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In summary, the use of a solution of ZnSO4 in a Zinc half cell when connected to a Cu half cell is not necessary because the Zn2+ ions are produced by the zinc electrode. However, in order for a cell to have a potential difference, both the oxidized and reduced forms of the metal must be present. This explains why a Cu half cell connected to a Zn half cell still needs a Cu electrode. The oxidation of the metal in water is caused by either H+ ions or dissolved oxygen. The reactivity series is only a simplified approximation and the potential of the cell depends on the ion concentrations. The potential can be calculated using the Nernst equation, where the ion concentration is a factor.
  • #1
hushai1
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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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  • #2
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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  • #3
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.
 
  • #4
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:

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

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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  • #5
Thank you very much. Your answers are excellent.
 

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

Why is a solution of ZnSO4 necessary for Zn & Cu half cells?

A solution of ZnSO4 is necessary for Zn & Cu half cells because it acts as an electrolyte, which allows for the movement of ions between the two half cells. This movement of ions is essential for the electrochemical reactions to occur and for the generation of electricity.

Can other electrolytes be used instead of ZnSO4 in Zn & Cu half cells?

Yes, other electrolytes such as NaCl or KCl can be used in place of ZnSO4 in Zn & Cu half cells. However, ZnSO4 is commonly used because it is readily available, inexpensive, and has a high solubility, making it an effective electrolyte for this system.

How does ZnSO4 affect the redox reactions in Zn & Cu half cells?

ZnSO4 acts as a medium for the redox reactions to occur in Zn & Cu half cells. It provides a source of Zn2+ ions, which are involved in the oxidation of Zn to Zn2+. Additionally, the SO42- ions present in ZnSO4 help to maintain electrical neutrality in the solution by balancing out the positive Zn2+ ions.

What happens if the concentration of ZnSO4 is too high or too low in Zn & Cu half cells?

If the concentration of ZnSO4 is too high, it can increase the conductivity of the solution and potentially alter the rate of the redox reactions. If the concentration is too low, it can slow down the rate of the reactions and affect the overall efficiency of the half cells. It is important to maintain a proper concentration of ZnSO4 for optimal performance.

Is the use of ZnSO4 limited to Zn & Cu half cells?

No, ZnSO4 can also be used as an electrolyte in other types of electrochemical cells, such as Zn & Ag or Zn & Fe half cells. However, the specific type of electrolyte used may vary depending on the specific redox reactions and materials being used in the cell.

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