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

Jan 11, 2017

The oxidation half-reaction is

$M g \left(s\right) \rightarrow M {g}^{2 +} + 2 {e}^{-}$

#### Explanation:

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:

$M g \left(s\right) + Z {n}^{2 +} \left(a q\right) + 2 C {l}^{-} 1 \left(a q\right) \rightarrow Z n \left(s\right) + M {g}^{2 +} \left(a q\right) + 2 C {l}^{-} 1 \left(a q\right)$

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

$M g \left(s\right) + Z {n}^{2 +} \left(a q\right) \rightarrow Z n \left(s\right) + M {g}^{2 +} \left(a q\right)$

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

$Z {n}^{2 +}$ has been reduced to $Z n \left(s\right)$

$M g \left(s\right)$ has been oxidized to $M {g}^{2 +}$

Jan 12, 2017

Anode: Mg → $M {g}^{2 +}$ + 2e- 2.38V

#### Explanation:

From the equation we see that Mg metal (oxidation state 0) changes to the ion $M {g}^{2 +}$. 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, $Z {n}^{2 +}$ 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:

$E = {E}^{o} \left(red\right)$ (cathode) - ${E}^{o} \left(red\right)$ (anode)

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

$E = {E}^{o} \left(red\right)$(Zn) - ${E}^{o} \left(red\right)$(Mg) = -0.76 - (-2.38) = 1.62V if the solutions are 1.0 M.

The two “standard cell” cathode (reduction) half reactions are:
$M {g}^{2 +}$ + 2e- → Mg - 2.38V
$Z {n}^{2 +}$ + 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 + $Z {n}^{2 +}$$M {g}^{2 +}$ + Zn
-0.76 - (-2.38) = 1.62V