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I am confused as to why galvanic cells have electrolytes. For example, a zinc-hydrogen galvanic cell has a zinc sulfate anode electrolyte and hydrochloric acid cathode electrolyte. The overall reaction of this may be written as:

$$ \ce{ZnSO4 + 2HCl -> ZnCl2 + H2SO4} $$

But rather, it is written as:

$$ \ce{Zn(s) + 2H+(aq) -> Zn^2+(aq) + H2(g)} $$

Why is this the overall reaction? What is then the purpose of an electrolyte, if only the anodes and cathodes are written in the overall equation?

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  • $\begingroup$ I know to write the overall cell reaction, you need to eliminate spectator ions. Is it just assumed that sulfate ions and chloride ions are spectators? $\endgroup$
    – user510
    Commented Feb 9, 2016 at 4:01
  • $\begingroup$ Yes they are spectators in this reaction. There should be 2HCl in the first step of the balanced equation. The electrolyte carries the charge between the electrodes. $\endgroup$
    – Technetium
    Commented Feb 9, 2016 at 4:11
  • $\begingroup$ Then why are Zn and H counted as spectator ions? They also cancel when you write ZnSO4 + 2HCl --> ZnCl2 + H2SO4 in total ionic form. $\endgroup$
    – user510
    Commented Feb 9, 2016 at 4:15
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    $\begingroup$ Because they are involved in the electron transfer. The oxidation state of the Cl and SO4 remains the same therefore there not directly involved in the transfer of electrons and are considered spectator ions. $\endgroup$
    – Technetium
    Commented Feb 10, 2016 at 2:50
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    $\begingroup$ The electrolyte may still be a variable as it carries the current across the electrodes and conductivity varies significantly between electrolytes meaning this will occur at different rates resulting in different potentials. This is going off-topic from the OP. I hope I could help you out. $\endgroup$
    – Technetium
    Commented Feb 10, 2016 at 5:29

2 Answers 2

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Unless I'm badly misunderstanding your question, I think your confusion arises from an error in the balanced chemical equation you wrote. As you have it, a salt is reacting with a strong acid to make another strong acid and a new salt. This reaction may occur spontaneously as written (I haven't figured its Gibb's free energy) , but it's not the oxidation-reduction reaction that drives a galvanic cell. In fact, it's not an oxidation reduction at all, it's some sort of acid-base reaction, and all of the reactants and products remain in solution anyway.

The actual chemistry that occurs in this system is between the zinc anode and the acid.

$\ce{Zn + 2 HCl -> ZnCl2 + H2}$

From this it is much clearer why the net ionic equation is as you wrote. The zinc sulfate could be replaced with zinc chloride with almost no effect on the battery's potential. In fact, the zinc solution exists only to create standard conditions for the cell and to carry electrical current through the solution - if it were replaced by a nonreactive electrolyte solution the cell would still function, albeit at a different (higher) potential.

You might then ask what the purpose of the hydrogen cathode is. Again, it's to create standard state. If you did this in reality, you would be using some sort of inert electrode at the hydrogen end to carry the electrical current into the acid solution and likely would just vent the hydrogen to the surroundings.

In short, this galvanic cell does not function because of a reaction between its electrolytes, nor because of a reaction between its electrodes. Instead it is ultimately a reaction between the anode and the cathode's electrolyte. The anode's electrolyte and cathode are required for electrical continuity and are used in examples of this sort to put the cell in standard state. Beyond that they don't serve much of a purpose.

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  • $\begingroup$ zinc is the anode. So is it that in very galvanic cell, the anode and the cathode electrolyte are the parts that cause the actual chemistry? $\endgroup$
    – user510
    Commented Feb 27, 2016 at 0:04
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    $\begingroup$ You're right, I've switched cathode and anode throughout, thanks for the correction. I almost wrote something like what you asked in your question, but there are sooooo many different galvanic cells that you can build that I really don't feel comfortable making that broad a statement. Certainly in the ones that you're going to see in an introductory course this is true. $\endgroup$
    – Jason Patterson
    Commented Feb 27, 2016 at 2:32
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    $\begingroup$ @user510 I do feel comfortable with a more general statement like: In a galvanic cells, electrons are removed from the anode, generating a substance with a higher oxidation state, and transferred to the cathode, where they generate a substance with a lower oxidation state. $\endgroup$
    – Jason Patterson
    Commented Feb 27, 2016 at 2:37
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The function of electrolyte is to carry the charge in the liquid phase and close the circuit in order to maintain the charge flux. In the external part of the cell (copper wires, some conductor) it is the electrons which move and carry the charge. In the solution it is the movement of ions, being the potential difference between anode and cathode the driving force for moving them (anions and cations move in opposite directions).

The 'spectator' ions may be called that way because they are not electroactive, meaning with this that they do not participate on electrode processes. As pointed out by @lcnittl, the first reaction is incomplete and does not describe the electrochemistry of reaction. For thermodynamic properties, it is mainly needed the redox reaction, reaction N°2, and potential difference associated (that can be found on Standard Redox Potentials Table). It will also depend on other variables such as valence of ions, number of electrons transfered, pH, temperature, etc.

The second reaction can be decomposed into two semi reactions, one occurring on the anode (oxidation) and the other on the cathode (reduction)

$$ \ce{2e⁻ + 2H⁺(aq) -> H_2(g)} ---(Cathode) -- E°=+0.7618 V $$ $$ \ce{Zn(s) -> Zn^2+(aq) + 2e⁻} --- (Anode) -- E°=0 V$$

The sum of the standard potentials of each semi-reaction gives a positive value (dE=0.7618V + 0V); a positive value means a spontaneous reaction based on the relation dG=-n.F.dE, where dG is the Gibbs free energy change, n the number of electrons transfered (n=2 in this case), F the Faraday constant and dE the potential difference between two semi-reactions.

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  • $\begingroup$ What do you mean by 'Both reactions are OK'? $\endgroup$
    – lcnittl
    Commented Feb 24, 2016 at 17:09
  • $\begingroup$ The first one is the net reaction of the cell -it does not contain practical information, only species involved and mass balance. The second one is the oxidation semi reaction $\endgroup$
    – Orr22
    Commented Feb 26, 2016 at 10:35
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    $\begingroup$ Yeah, but the first misses an HCl molecule to be balanced, and furthermore cannot be the net equation of the cell, since it's an acid-base reaction, and not a redox reaction. $\endgroup$
    – lcnittl
    Commented Feb 26, 2016 at 10:44
  • $\begingroup$ @lcnittl The first of the two reactions is a reduction, which it has to be to pair with the oxidation step shown in the second reaction. You need both for either to occur. $\endgroup$
    – Jason Patterson
    Commented Feb 26, 2016 at 11:30
  • $\begingroup$ @JasonPatterson How can it be a reduction, if there is no change on any oxidation state of any participating atom? (Not to talk at cross purposes, we were talking about the OPs first equation) $\endgroup$
    – lcnittl
    Commented Feb 26, 2016 at 11:33

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