I was wondering how to calculate cell spontaneity for a galvanic cell. Do you just look up the enthalpy values of electrode reactions? Does the electrolyte solution you choose have an effect on cell spontaneity?
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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:
$$\ce{Zn_{(s)} + Cu^2+_{(aq)} -> Zn^2+_{(aq)} + Cu_{(s)}}$$
Your cathode half reaction is:
$$\ce{Cu^2+_{(aq)} + 2e- -> Cu_{(s)}}$$
and your anode half reactions is:
$$\ce{Zn_{(s)} -> Zn^2+_{(aq)} + 2e-}$$
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.
