Why put zinc solution in a galvanic cell?
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:

$$\ce{Cu(s) &lt;=&gt; Cu^2+(aq) + 2e-}$$

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:

$$\ce{Zn(s) &lt;=&gt; Zn^2+(aq) + 2e-}$$

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.