How to calculate galvanic cell spontaneity?
To determine spontaneity of a reaction, one looks at the sign of $\Delta G$ for the process. And you have the following relation between the change in standard Gibbs free energy and the (standard)EMF of the cell: 

$ \Delta G^0 = -n\,F\,E^0$

where, $n:=$ no. of electrons exchanged

$F:=$ Faraday constant (Charge on 1 mole of electrons. About $96,500\,\mathrm{C}$)

So, it is sufficient to compute the $E$ and look at it's sign. A positive sign for the EMF would ensure a negative value for $\Delta G^0$ and thus, you can conclude that your cell reaction is spontaneous.

The EMF of the cell is given as follows:

$$E^0_{\mathrm{cell}} = E^0_{\mathrm{cathode}} - E^0_{\mathrm{anode}} $$

Where, $E^0_{\mathrm{cathode}}$ and $E^0_{\mathrm{anode}}$ are the standard reduction potentials for your cathode and anode half reactions, and you can easily find tabulated values for most metals, ions.

An illustrative example: 

Consider the following cell reaction:

$$\ce{Zn_{(s)} + Cu^2+_{(aq)} -&gt; Zn^2+_{(aq)} + Cu_{(s)}}$$

Your cathode half reaction is:

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

and your anode half reactions is:

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

the tabulated reduction potential values for the two:

$$E^0_{\mathrm{anode}}= -0.763\,\mathrm{V} \qquad E^0_{\mathrm{cathode}}= +0.342\,\mathrm{V}$$

Plugging this into

$$E^0_{\mathrm{cell}} = E^0_{\mathrm{cathode}}-E^0_{\mathrm{anode}}$$

we obtain,

$$E^0_{\mathrm{cell}} = +1.103\,\mathrm{V}$$

Thus, this cell reaction is spontaneous.