Spontaneity is relating to the value the E cell takes. If E cell > 0 it is spontaneous and if E cell < 0 it is non-spontaneous
These types of questions you do the following things:
1. Write half equations (if they give you a full equation you need to break it apart and get the oxidation and reduction equations out from that)
2. Find the cell potentials for the half equations on the standard reduction potentials data sheet page (noting that if the equation we have is the reverse of what is shown on the table we need to reverse the sign from + to - or - to +)
3. Calculate E cell using E cell = E oxd + E red
1)
Cu(s) --> Cu^2+(aq) + 2e^- (oxidation) E oxd = -0.34 V (had to reverse the sign of this)
H^+(aq) + e^- --> 1/2 H2(g) (reduction) E red = 0.00 V
Therefore, E cell = -0.34 + 0 = -0.34 V
As E cell < 0, non-spontaneous
2)
1/2 H2(g) --> H^+(aq) + e^- (oxidation) E oxd = 0.00 V
Cu^2+(aq) + 2e^- --> Cu(s) (reduction) E red = 0.34 V
Therefore, E cell = 0.34 + 0 = 0.34 V
As E cell > 0, spontaneous
