What is the oxidation half reaction for `Mg(s) + ZnCl_2(ag) -> MgCl_2(ag) + Zn(s)`?

If you write what is known as the net ionic equation, it is a simpler matter to identify the oxidation (and the reduction form that matter).

First, write the two salts in aqueous ion form:

`Mg(s)+Zn^(2+)(aq) +2Cl^-1 (aq) rarr Zn(s)+Mg^(2+)(aq) +2Cl^-1 (aq)`

Since no change occurs in the chloride ion, these are "spectators" and can be omitted:

`Mg(s)+Zn^(2+)(aq) rarr Zn(s)+Mg^(2+)(aq)`

Now, we have a more clear description of what has taken place.

`Zn^(2+)` has been reduced to `Zn(s)`

`Mg(s)` has been oxidized to `Mg^(2+)`

From the equation we see that Mg metal (oxidation state 0) changes to the ion `Mg^(2+)`. It is thus “oxidized” and is the reducing agent. It is thus the anode of the cell.

Zn is changed from it’s ionic state, `Zn^(2+)` to its metallic state of 0. It is thus “reduced” and is the oxidizing agent. It is thus the cathode of the cell.

By convention, all half-cell emf's are compared to the emf of the standard hydrogen electrode. The emf of a half-cell, with respect to the standard hydrogen electrode, is called the reduction potential.
In an electrochemical cell, the general equation is:

`E = E^o(red)` (cathode) - `E^o(red)` (anode)

The emf for the Mg-Zn cell described would be:

`E = E^o(red)`(Zn) - `E^o(red)`(Mg) = -0.76 - (-2.38) = 1.62V if the solutions are 1.0 M.

The two “standard cell” cathode (reduction) half reactions are:
`Mg^(2+)` + 2e- → Mg - 2.38V
`Zn^(2+)` + 2e- → Zn - 0.76V
http://hyperphysics.phy-astr.gsu.edu/hbase/Tables/electpot.html

Put in the general equation for the actual cell, the full reaction is:
Mg + `Zn^(2+)`→`Mg^(2+)` + Zn
-0.76 - (-2.38) = 1.62V