Half-Reaction Explanations in a Simple Battery

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In summary: The voltmeter I used is rather old.I attempted to find the half-reactions of this scenario, in which the full redox is...Zn(s) + Cu2+(aq) -> Zn2+(aq) + Cu (s)I found the half reactions to be...Zn(s) -> Zn2+(aq) + 2e-Cu2+(aq) + 2e- -> Cu(s)These half-reactions had to be arranged in such a way as to combine to form the full redox. This can be explained by activity series, where Zn is more easily oxidized than Cu. This leaves the Zn half-reaction as the oxidation and Cu as the reduction
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I created a simple battery. I used a copper cathode and zinc anode resting in a saturated solution of NaCl (about 25% NaCl by mass), connected by alligator clips to a voltmeter. I received a voltage of .85V, whereas the theoretical voltage should have been about 1.10V. That's not of too much concern, considering my setup wasn't top-notch by any means, and the voltmeter I used is rather old.

I attempted to find the half-reactions of this scenario, in which the full redox is...

Zn(s) + Cu2+(aq) -> Zn2+(aq) + Cu (s)

I found the half reactions to be...

Zn(s) -> Zn2+(aq) + 2e-
Cu2+(aq) + 2e- -> Cu(s)

These half-reactions had to be arranged in such a way as to combine to form the full redox. This can be explained by activity series, where Zn is more easily oxidized than Cu. This leaves the Zn half-reaction as the oxidation and Cu as the reduction.

My question... can the determination of which half-reaction is to be an oxidation or reduction be explained through the spontaneity of the reaction as well? I went through, using the formula...

ΔG = ΔH - TΔS

...and found the change in free energy to be approximately -211 kJ, which I also confirmed using ΔG = -nFE. Thus, the redox is spontaneous and should occur on its own, as it did in the battery. If my logic is correct, would that also imply that free energy and spontaneity generally explains the activity series? For example, Li is higher than Zn on the activity series, and has a much lower ΔG than Zn. Zn is higher than Cu on the series, and has a much lower ΔG than Cu.

Thank you in advance!
 
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... can the determination of which half-reaction is to be an oxidation or reduction be explained through the spontaneity of the reaction ... ?

Yes ... Using standard reduction potentials for the half-reactions, Eocell must be > 0 for the combination of half-reactions used. That is, EoZn/Cu = EoCu - E oZn will give a positive value for the net cell potential and is therefore defined as a spontaneous combination. Generally, at standard conditions, if one calculates the net cell potential using Eocell = Eoreduction - Eooxidation one will always get a positive value. Reversing the values in the equation will result in a negative value and would be contrary to defining the combination as spontaneous. Only a combination that will produce Eocell > 0 can one say the combination is spontaneous.

If using ΔGo values, one can still define spontaneity as the same idea applies but ΔGo must be negative for the differential between the ΔGo(Reduction half-reaction) and ΔGo(Oxidation half-reaction). However, one must use the Hess's Law Equation ΔGoRxn = Σn⋅ΔGof(Products) - Σn⋅ΔGof(Reactants) which must be applied so that the calculated value is negative for the reaction of interest. For the Zn/Cu cell, the net reaction must be Zno(s) + Cu+2(aq) ↔ Zn+2(aq) + Cuo(s) showing Zn undergoing oxidation and Copper undergoing reduction in order to obtain a negative ΔGoRxn. Reversing the rolls of the ions; i.e., ... + Zn+2(aq) ↔ Cu+2(aq) + ... would give a positive ΔGoRxn and would not define the process as spontaneous. Hope this answers your question. Doc :-)
 
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James Pelezo said:
Hope this answers your question. Doc :-)
This helps a lot!

Just to further my understanding, can spontaneity properly explain the activity series as I mentioned at the end of my post?
 
  • #4
Yes, but in terms of the cell potentials being always positive and reactive free energy being negative as noted in my reply post. For example, Lithium would always undergo oxidation with respect to all elements listed below it in the activity series, sodium would always undergo oxidation with respect to all elements below it in the activity series but not with respect to those above it in the series. Again, such is confirmed when Eocell > 0 &/or ΔGoNet Cell Rxn < 0 for the combination of half-reactions chosen as these conditions specify/define the spontaneity of process.
 
  • #5
James Pelezo said:
Yes, but in terms of the cell potentials being always positive and reactive free energy being negative as noted in my reply post. For example, Lithium would always undergo oxidation with respect to all elements listed below it in the activity series, sodium would always undergo oxidation with respect to all elements below it in the activity series but not with respect to those above it in the series. Again, such is confirmed when Eocell > 0 &/or ΔGoNet Cell Rxn < 0 for the combination of half-reactions chosen as these conditions specify/define the spontaneity of process.
Makes sense... thank you for the help!
 
  • #6
Comeback City said:
I created a simple battery. I used a copper cathode and zinc anode resting in a saturated solution of NaCl (about 25% NaCl by mass), connected by alligator clips to a voltmeter. I received a voltage of .85V, whereas the theoretical voltage should have been about 1.10V.
Also take in mind that the theoretical value refers to a combination of standard half-cells where each clip is in a solution containing copper and zinc ions, respectively, at an activity of 1 mol/L.
 
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  • #7
DrDu said:
Also take in mind that the theoretical value refers to a combination of standard half-cells where each clip is in a solution containing copper and zinc ions, respectively, at an activity of 1 mol/L.
I'd imagine the voltage would also increase if I used a strong acid rather than an electrolyte such as NaCl, yes?
 
  • #8
The voltage is quite undefined. Take the reaction ##\mathrm{Cu^{++} + 2e^- \rightarrow Cu}##: How should it proceed if there are no Copper ions present? More probable is that on the cathode, Hydrogen is being produced.
 
  • #9
DrDu said:
The voltage is quite undefined. Take the reaction ##\mathrm{Cu^{++} + 2e^- \rightarrow Cu}##: How should it proceed if there are no Copper ions present? More probable is that on the cathode, Hydrogen is being produced.
So the cathode reaction should look more like this...

Cu(s) + 2H+ → Cu2+ + H2(g)
 
  • #10
No, simply ##\mathrm{2H^+ + 2 e^- \rightarrow H_2}##.
 
  • #11
I'd imagine the voltage would also increase if I used a strong acid rather than an electrolyte such as NaCl, yes?

Not exactly in that context...
Since it hasn’t been mentioned, for voltages outside the ‘Standard Cell Potentials’, one should apply the Nernst Equation. This equation defines voltages at non-standard conditions; i.e., where ion concentrations are not equal to 1.0M.

Enon-std = Eo std – (0.0592/n)log([Anodic Ions]/[Cathodic Ions]

The sustained discharge of a Voltaic Cell is due to separation of the oxidation and reduction processes called a controlled Galvanic Process that can sustain a charge flow over a long period of time. In contrast, an uncontrolled Galvanic Process would be like dropping a zinc bar into a solution of Cu+2ions and allowing the redox process to proceed until it naturally stops due to the Cuo(s) coating on the zinc bar surface preventing further oxidation of the Zn to Zn+2.

In a controlled Galvanic process, the change in voltage (Δv) as a function of time would follow an exponential decay trend like a 1st order decay process and is dependent on the ion concentration ratios at specified times. That's the log (A/C) term at the end of the Nernst Equation. The following graphic illustrates such a trend for a Galvanic Cell like the Zn/Cu cell system.
241338

Such explains the need for a high Cu+2 ion concentration at the cathodic cell and a low concentration of Zn+2 ions at the anodic cell in order to optimize the net cell voltage at the ‘beginning’ of discharge (i.e.; new battery). As time passes, the Zn+2 ion concentration increases with oxidation of the anodic metal and Cu+2 ion concentration decreases with reduction to standard state Cuo(s) from the cathodic cell solution. The concentration of ions will momentarily be equal and the net cell voltage is equal to the standard cell voltage. Beyond this, the Zn+2 ion concentration continues to increase and Cu+2 ion concentration decreases until the anodic Zno(s) metal is consumed by the oxidation process and the net cell voltage reaches zero, dead battery.

So, simply changing the electrolyte to a strong acid in the anodic cell wouldn't necessarily increase voltage, but might increase the oxidation rate of the metal and result in a faster battery discharge rate and dead battery sooner that if a common ion solution were used.

 

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  • #12
James Pelezo said:
Such explains the need for a high Cu+2 ion concentration at the cathodic cell and a low concentration of Zn+2 ions at the anodic cell in order to optimize the net cell voltage at the ‘beginning’ of discharge (i.e.; new battery). As time passes, the Zn+2 ion concentration increases with oxidation of the anodic metal and Cu+2 ion concentration decreases with reduction to standard state Cuo(s) from the cathodic cell solution. The concentration of ions will momentarily be equal and the net cell voltage is equal to the standard cell voltage. Beyond this, the Zn+2 ion concentration continues to increase and Cu+2 ion concentration decreases until the anodic Zno(s) metal is consumed by the oxidation process and the net cell voltage reaches zero, dead battery.
This makes a lot more sense now! I now understand how voltage would be at highest when the reaction first begins, given (as you mentioned) the ideal conditions of high Cu2+(aq) and Zn(s).

James Pelezo said:
So, simply changing the electrolyte to a strong acid in the anodic cell wouldn't necessarily increase voltage, but might increase the oxidation rate of the metal and result in a faster battery discharge rate and dead battery sooner that if a common ion solution were used.
Can I conclude this would simply increase the current rather than the voltage, then? It makes sense that the faster oxidation rate would mean a faster transfer of electrons through the electrolyte solution... would this also imply an increase in travel rate through the wire?
 
  • #13
No ... the net cell potential is due to the combination of anode and cathode materials chosen, not the rate of oxidation. Oxidation is a natural product of the chemistry of the combination chosen as is the reduction side of the cell. Boosting oxidation rates to achieve higher charge flow would only come with changing electrode materials having a higher cell potential not adjusting pH.
 
  • #14
Why is it then that sulfuric acid is used in batteries, considering its low pH does not have an effect on cell potential? Why not use something like table salt?
 
  • #15
Sulfuric acid is used in lead-acid batteries. During the reaction Pb is reduced and PbO2 oxidized both to Pb2+. The concentration of Pb2+ in both half reactions is reduced by the formation of insoluble PbSO4. Therefore H2SO4 increases the voltage of the battery.
 
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The sulfuric acid in the lead storage battery is there to react with the Pb(II) ions forming the lead(II) sulfate coatings at the electrodes. This does enhance the voltage but is a unique case. I haven't tried the effect on other systems but I would not expect that to be a universal truth for other material combinations; such as the Zn/Cu system in the original question.
 
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Related to Half-Reaction Explanations in a Simple Battery

1. What is a half-reaction in a simple battery?

A half-reaction in a simple battery is a chemical reaction that occurs at one of the electrodes. It involves the transfer of electrons from one chemical species to another, resulting in the production of an electric current.

2. How does a half-reaction contribute to the overall functioning of a battery?

A half-reaction is essential for the functioning of a battery because it creates a flow of electrons, which is necessary for the production of an electric current. Without half-reactions, a battery would not be able to generate the energy needed to power electronic devices.

3. What are the two types of half-reactions in a simple battery?

The two types of half-reactions in a simple battery are oxidation and reduction. Oxidation involves the loss of electrons, while reduction involves the gain of electrons. These two reactions occur simultaneously at the anode and cathode, respectively.

4. How are half-reactions balanced in a simple battery?

Half-reactions in a simple battery are balanced by ensuring that the number of electrons lost in the oxidation half-reaction is equal to the number of electrons gained in the reduction half-reaction. This balance is necessary for the proper functioning of the battery.

5. Can half-reactions be reversed in a simple battery?

Yes, half-reactions in a simple battery can be reversed. This occurs when the battery is being charged, and the flow of electrons is reversed. In this case, the anode becomes the cathode, and the cathode becomes the anode, and the half-reactions are reversed.

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