Why put zinc solution in a galvanic cell?

Question

In a galvanic cell with Cu and Zn electrodes, I was taught that each electrode is immersed in a solution containing it's own positive ions. I understand why you need copper ions, so they can gain electrons, but why do you need zinc ions - the reaction is going to produce zinc ions anyway, and the negative ions from the salt bridge will migrate to neutralize them, so I don't see why you need zinc ions at the first place. Is there anything I'm missing in how the cell works?

Answer 1

You're right: you don't need a zinc solution to start with. In fact the lemon battery experiment (http://en.wikipedia.org/wiki/Lemon_battery) shows you don't even need the copper solution to generate a current. (The reaction at the copper end in that case is 2H$^+$ + 2e$^-$ = H$_2$ and doesn't involve reduction of copper.)

The reason the textbook description is the way it is has to do with reproducibility and simplicity. I'll address the question of reproducibility later on, but first consider: What's the simplest thing the zinc could be immersed in? If you put the zinc electrode in the same copper solution as the other electrode, you'll get copper-plated zinc instead of a battery. If you put it in distilled water you won't have a circuit. There are other things you could put it into, such as salt water, but in any case two things are true: It needs to be a solution different from the one where the copper electrode is sitting; and it will have zinc ions as soon as you have a closed circuit, as the zinc electrode starts to dissolve and the copper electrode starts acquiring more copper. So in a way the simplest answer is, stick it in a solution of zinc ions.

Now about reproducibility. The potential of the cell depends on (among other things) the concentration of Zn$^{2+}$ in solution near the Zn electrode, and the concentration of Cu$^{2+}$ in solution near the Cu electrode. Other things being equal, if you increase the concentration of Zn$^{2+}$ by a factor of 10, the potential will drop by about 0.118 volts (Nernst equation). So in a way it's actually good to have a low concentration of Zn$^{2+}$; but that concentration will never be zero, at least after you close the circuit, and when the concentration is very very low it will be growing very fast, relatively speaking (the first micromole of Zn to dissolve might increase the concentration a million times, say), which means the potential will drop quickly in the initial moments. You might as well start with a concentration of Zn$^{2+}$ that is not going to change so fast (in relative terms).

Answer 2

It's tempting to look at the whole cell here and ask: hmm what's going on? But try taking a reductionist approach and look at each electrode system one at a time, in isolation.

Let's consider a basic model.If we immerse an electrode of $\ce{Cu}$ metal into a solution of $\ce{Cu^2+}$, the following can be thought to take place:

$$\ce{Cu(s) <=> Cu^2+(aq) + 2e-}$$

Where the ions in the solution are bombarding the electrode surface, and are being taken up by gaining two electrons from the metal, whilst the metal is giving up ions from its surface, keeping hold of two electrons. At equilibrium, this forward and backward bombardment occurs at the same rate.

You might say to me: okay but we can't actually probe this system with a voltmeter to say that this is going on, because this would require a circuit to be made with an external system, and hence would perturb any effect we wish to observe. I'd then say to you: you're correct but bear with my argument, just noting what I've said.

By a similar token, in our other isolated beaker we have:

$$\ce{Zn(s) <=> Zn^2+(aq) + 2e-}$$

Now, the whole idea behind an electrochemical cell is to extract chemical energy by systems which are not at equilibrium. If a system is at equilibrium then no energy will be available. The equation which gives the free energy available from an electrochemical cell is (derivation not given):

$$\Delta G=−RT\ln K_\mathrm{eq}=−nFE_\text{cell}$$

Where $R$ is the ideal gas constant, $T$ is the parameter of temperature, $F$ is the Faraday constant (the amount of charge carried by a mole of electrons) and $E_\text{cell}$ is your available electrochemical energy.

So for these two individual single-electrode systems at equilibrium, can we extract chemical energy? Well no, because Gibbs at equilibrium is zero, and hence so is $E_\text{cell}$.

What then do we do?

Complete the circuit between the two, and if the potential between two half-reactions is sufficient to warrant the moving of electrons, a dis-equilibrium will be created until the reaction has completed (which is indeed what we get for a system of zinc/copper). Once the electron transfer between zinc and copped has been achieved i.e. equilibrium has been achieved, then we can no longer extract any useful chemical energy. Our cell is dead!

So in summary: the dual metal electrode/same-metal ion solution is needed for your initial half-reactions, which you then perturb by completing your circuit. This perturbation of the individual half-reactions results in our being able to extract useful chemical energy by shifting the equilibrium until a new equilibrium is established.