Question

What will happen when a piece of magnesium metal is dropped into a beaker containing a 1 M solution of copper(II) chloride?
Mg2+ + 2e- Mg(s) Eo = - 2.37 V
Cu2+ + 2e- Cu(s) Eo = + 0.34 V

A. The magnesium will dissolve and the copper will be deposited.
B. The copper will dissolve and the magnesium will be deposited.
C. The magnesium will dissolve but nothing else will happen.
D. The magnesium will dissolve and chloride gas will be produced.
E. There will be no reaction.

Answer 1

@JFraser

Answer 2

Mg2+ + 2e- -> Mg(s) Eo = - 2.37 V
Cu2+ + 2e- -> Cu(s) Eo = + 0.34 V

So we would have the following if a reaction were to occur:
Cu2+ + 2e- -> Cu(s) -> Eo = + 0.34 V
Mg(s) -> Mg2+ + 2e- -> Eo = -(- 2.37 V) = +2.37V

So we have that:
-Mg reduces because it has higher Eo
-Cu oxidizes because it has a lower Eo

So your assumption is correct no reaction would occur, as copper is already oxidized