# Electrochemistry conditions for voltage to increase in a galvanic cell

I understand that increasing the concentration of reactants in a galvanic cell generally causes the voltage of the cell to increase. But is that really the case always?

Consider a standard cell of zinc and copper, 1 M and 1 L each solution of copper nitrate and zinc nitrate in each half cell. Another cell, increased the concentration of copper nitrate to 4 M, but added 0.25 L of it and kept the same zinc nitrate 1 M and 1 L.

This way I increased the concentration, but the amount of substance is the same. Does the voltage increase in the new cell? I would really appreciate someone to explain to me the specifics of what causes the voltage of a cell to increase based on the reactants

• Hmm so are you referring to the Nernst equation then? en.wikipedia.org/wiki/Nernst_equation#Expression – ManRow Dec 6 '19 at 9:50
• Please don't use a zinc plate in a solution of zinc nitrate. Metallic Zinc reduces nitrate in neutral solutions and produce nitrous ions and then hydrazine.Better use zinc sulfate. – Maurice Feb 5 at 21:21
• @Maurice, zinc does not reduce nitrate in neutral condition. Nothing happens if you dip Zn plate in 1 M zinc nitrate. – M. Farooq Feb 5 at 21:39

In an electrochemical cell, increasing the concentration of reactants will increase the voltage difference, as you have indicated. A higher concentration of reactant allows more reactions in the forward direction so it reacts faster, and the result is observed as a higher voltage.

If you have adjusted the cell volume to keep the total amount of reactants the same, the rate of electron flow thru the cell will be higher at first. Then the cell will run down faster, but the total amount of electron flow will be the same.

Electrical energy is the product of the voltage times the total amount of current (total number of electrons flowing). Ordinarily we say volts x amps = power (which is the rate of energy produced or used), but by energy here I mean volts x amps x time. You might think that there is a free source of energy here: higher concentrations give higher voltage at the same current. Can we make this into a perpetual motion battery?

Well, no, because the increase in cell voltage is rather minuscule (0.0178 V vs 1.12 V, according to ManRow). One way to explain this is that the more dilute solution has more entropy, but we did not extract the work associated with that when we made up the dilute solution. So if you find a source of dilute reactants and try to concentrate them to get more energy out, you have to put in work to concentrate the solutions, and it will cost you more to do that than you gain in voltage.

• Did you by any chance mean voltage x current = power, not electrical energy? Also thanks for the thought-provoking idea regarding perpetual motion – Bandoo Jan 5 at 20:45
• Yes, thanks! I was thinking total electron flow (accumulated) and was imprecise. I will edit the original answer. – James Gaidis Jan 6 at 13:59

Yes, your voltage will change in accordance with the Nernst equation

$$\Delta E_\text{cell} = - \frac{RT}{zF}\ln Q_{\text{rxn}}$$

So, in your copper-zinc battery, where

$$\ce{Cu^2+(aq) + Zn(s) -> Cu(s) + Zn^2+(aq)}$$

with a standard $$E^\circ_\text{cell} \approx \pu{1.10 V},$$ setting the $$\left[\ce{Cu^2+}\right] = \pu{4 M}$$ should raise your cell potential by

$$\Delta E_\text{cell} = -\frac{RT}{2F}\ln\frac{1}{4} = \pu{+0.0178 V}$$

for a new cell potential (at $$\pu{25 °C})$$ of

$$E_{\text{cell}} = \pu{1.10 V} + \pu{0.0178 V} \approx \pu{1.12 V}$$

• So what is the reasoning for the increase in voltage? There is still the same number of moles of reactant. Why is it that only the concentration affects the cell potential rather than number of moles? Moreover, i want to know if there are any other factors that can affect voltage of a galvanic cell besides the reaction, and the reaction quotient. Finally, does the new cell last as long? Sorry I am self learning this subject and I don’t understand some concepts. Thanks. – Bandoo Dec 6 '19 at 14:43
• By varying $\left[\ce{Cu^2+}\right]$ you are shifting the system either closer to or farther away from chemical equilibrium, often expressed as a ratio of product & reactant concentrations. This principle also applies to galvanic cells like in your example as well, and reflects in the EMF that we can measure from them as well. Namely, the closer the system is to equilibrium, the smaller the magnitude of the EMF (and vica versa). – ManRow Dec 7 '19 at 3:45
• The new cell will last as long as the system has not reached equilibrium. However the closer the system is to equilibrium the lower the cell potential will be. Finally, yes, there are additional factors that can influence the cell potential as well, such as temperature, for example. In your example if we set $\left[\ce{Cu^2+}\right] = 4\text{M}$ and also raised the temperature (to something warmer than the standard 25°C), the new cell potential would above the calculated 1.12V as well. – ManRow Dec 7 '19 at 3:55
• I suppose it would benefit this answer to make clear at what point the concentration will effect the potential. At least for me it is not obvious, why Q = 1/4 with the change in concentration. – Martin - マーチン Feb 6 at 13:54
• At what point? The Nernst equation is technically relevant at "all points" and times of the reaction, not only initial conditions. The reaction quotient (see en.wikipedia.org/wiki/Reaction_quotient) I computed was for the initial conditions however (though, I thought that was obvious). An interesting side effect is that you can also use the Nernst equation to determine the equilibrium ("final") concentrations of the reactants, by finding the reaction quotient that would give a total, zero cell potential. – ManRow Feb 7 at 2:45

As others have answered, yes the potential of the cell will still change. It sounds like you're looking for a more "physics-y" answer than the ones provided so here goes.

The Nernst equation (and equilibrium equations in general) work based on concentration and not total moles. To understand why, start with a homogeneous reaction, like $$\ce{HCl + NaOH <-> NaCl + H2O}$$, or more usefully $$\ce{H+ + OH- <-> H2O}$$. In order for one molecule of $$\ce{H2O}$$ to be produced, we need a proton and a hydroxide ion to collide. If there are lots of them, this reaction will happen often. However, at the same time, some number of $$\ce{H2O}$$ molecules will dissociate into protons and hydroxide ions (i.e. the reverse reaction). If the forward reaction happens quickly compared to the reverse reaction, then the equilibrium will favor the products. If you double the volume but keep the concentration the same then your increasing the number of collisions of protons and hydroxide ions but also increasing the number of dissociations of water molecules so overall the equilibrium stays the same. Only if you increase the number of protons and hydroxides relative to the number of water molecules will the equilibrium shift. This is what you do when you increase the concentration.

Now let's switch to your battery example and consider the half-reaction of $$\ce{Cu^{2+}_{(aq)} + 2e- <-> Cu_{(s)}}$$. This is a heterogeneous reaction so we have to think about the interactions at the surface of the electrode. In order for the forward reaction to occur, a solvated copper ion must shed its solvation layer and adsorb to an active site on the surface of the metal then receive two electrons from the copper. At the same time, some number of atoms on the copper metal are releasing two electrons, solvating and entering the bulk electrolyte. Let's say you have a cell in equilibrium. This means that the rate of the forward reaction and the rate of the backward reaction are the same. Increasing the concentration of the electrolyte means that now every second more ions will be hitting the copper metal and it increases the likelihood of a conversion to copper metal. This increase in the rate of the forward reaction pushes the equilibrium farther towards the products. However, increasing the total volume of the electrolyte without changing the concentration doesn't affect the rates of adsorption of ions to the surface of the electrode, since that's an inherently surface phenomenon.

To use an analogy: the number of cars per minute that get through a toll-booth is only determined by how quickly the very front cars pay and drive off. Adding a bunch of cars in a queue behind the booth doesn't change anything. You could make the queue a mile long and still only the very first row of cars actually paying will set the rate of traffic. Of course, having a long line means it will take longer for all the cars in the queue to get through, just as in your case the additional of volume of electrolyte will increase the capacity of the cell.

• States of aggregation should not be subscripted, it is not wrong, but the recommendations (Sec. 2.1.) are different. What is wrong on the other hand is using mesomeric arrows ($\ce{<->}$) instead or equilibrium arrows ($\ce{<=>}$). – Martin - マーチン Feb 6 at 13:50
• Your link is broken, but sure I'll make sure to get the arrows right next time. Any thoughts on the actual chemistry, though? – Mark Wolfman Feb 6 at 15:00
• Sorry for the old link, fixed: recommendations (Sec. 2.1.). You might want to have a look at the Grotthuss Mechanism. In aqueous solution, there not really are any of the ions you suppose there to be, i.e. $\ce{H+}$, $\ce{OH-}$; collisions isn't the right word to use there. I can't really follow the parts after that. – Martin - マーチン Feb 6 at 15:24