What is the true voltage of a voltaic cell?

In summary, according to some sources, the voltage of a voltaic cell is 1.1 V, while others claim it is 0.76 V (the difference of potentials of zinc and hydrogen). The truth is probably somewhere in between these values.
  • #1
Seeit
5
0
TL;DR Summary
Hello,
What is the voltage of a voltaic cell? Should it be determined from the difference of the potentials of zinc and copper or zinc and hydrogen?
Hello,
What is the voltage of a voltaic cell? Some sources claim it's 1.1 V (the difference of potentials of zinc and copper). But I've also seen its being claimed to be 0.76 V (the difference of potentials of zinc and hydrogen). I know that zinc and hydrogen are the ones undergoing redox reactions and the copper only acts as a conductor, so to me, the value 0.76 V seems more logical. But what is it then with 1.1 V? What is the truth?
 
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  • #3
BvU said:
I meant the cell in the voltaic pile (https://en.m.wikipedia.org/wiki/Voltaic_pile). Wikipedia here itself says that the copper is only a conductor and hydrogen is the one reacting. The Wikipedia page mentions the voltage of 0.76 V.
Well, as voltaic cell itself can mean both the galvanic cell in general and the one created by Volta, this could cause the ambiguity.
But I asked about the voltage of the voltaic cell created by Volta containing sulfuric acid. I thought someone here would know.
 
  • #4
BvU said:
If you want that, you must use a solution of a copper salt, so copper metal can precipitate on the electrode. If you just use acid or brine, there's no way for the copper to participate in any reaction.
 
  • #5
Seeit said:
I meant the cell in the voltaic pile
My bad googling :rolleyes:. Voltaic cell gets galvanic cell. Voltaic pile gets, well, voltaic pile. Would have gone better if you had included thte link in post #1.

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  • #6
willem2 said:
If you just use acid or brine, there's no way for the copper to participate in any reaction.
Unless it gets oxidized.
 
  • #7
willem2 said:
If you want that, you must use a solution of a copper salt, so copper metal can precipitate on the electrode. If you just use acid or brine, there's no way for the copper to participate in any reaction.
Yes, I agree. So the voltage of a voltaic pile's one cell should be 0.76 V? Why do some pages mention 1.1 V? For example Czech Wikipedia mentions 1.1 V for this cell, should I edit it then?
 
  • #8
Borek said:
Unless it gets oxidized.
To get a battery we must have oxidation at one electrode and reduction at the other. The zinc is what gets oxidized, so copper must be reduced, so you'll have to start with Cu+.
 
  • #9
willem2 said:
To get a battery we must have oxidation at one electrode and reduction at the other. The zinc is what gets oxidized, so copper must be reduced, so you'll have to start with Cu+.

In Zn/Cu battery, yes, Zn is getting oxidized, copper is getting reduced, no doubt about it. But technically it is possible to make a battery which works by oxidizing copper, just a matter of selecting something that will be reduced at a potential high enough (Ag+/Ag will do the trick). Your original claim

willem2 said:
If you just use acid or brine, there's no way for the copper to participate in any reaction.

seems to be suggesting it is not possible at all.
 
  • #11
The Daniel Cell gives 1 volt approx and uses copper and zinc. I think the Volt was originally defined from this type of cell.
 

FAQ: What is the true voltage of a voltaic cell?

What is the true voltage of a voltaic cell?

The true voltage of a voltaic cell, also known as the electromotive force (EMF), is the potential difference between the two electrodes when no current is flowing. It is determined by the specific materials used for the electrodes and the electrolyte, and can be measured using a high-impedance voltmeter.

How is the voltage of a voltaic cell calculated?

The voltage of a voltaic cell can be calculated using the Nernst equation, which takes into account the standard electrode potentials of the half-cells, the temperature, and the concentrations of the reactants and products. The equation is: E_cell = E°_cell - (RT/nF) * ln(Q), where E°_cell is the standard cell potential, R is the gas constant, T is the temperature in Kelvin, n is the number of moles of electrons transferred, F is the Faraday constant, and Q is the reaction quotient.

What factors affect the voltage of a voltaic cell?

Several factors can affect the voltage of a voltaic cell, including the types of materials used for the electrodes, the concentration of the electrolyte, the temperature, and the presence of any impurities or side reactions. Changes in these factors can alter the cell potential and the overall efficiency of the cell.

Why does the voltage of a voltaic cell decrease over time?

The voltage of a voltaic cell decreases over time due to several reasons, including the depletion of reactants, the buildup of products, changes in electrolyte concentration, and internal resistance within the cell. These factors cause the cell to deviate from its initial conditions, leading to a drop in voltage.

How can the voltage of a voltaic cell be measured accurately?

The voltage of a voltaic cell can be measured accurately using a high-impedance voltmeter to ensure that no current flows through the cell during measurement. This prevents any additional voltage drops due to internal resistance. It is also important to ensure that the cell is at equilibrium and that the temperature and concentrations are stable.

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