# Which is the correct reduction half reaction for the following equation? Ca + CuO = Cu + CaO?

Oct 3, 2016

$C u O + 2 {H}^{+} + 2 {e}^{-} \rightarrow C u + {H}_{2} O$

#### Explanation:

$\text{Reduction}$

$C u O + 2 {H}^{+} + 2 {e}^{-} \rightarrow C u + {H}_{2} O$

$\text{Oxidation}$

$C a + {H}_{2} O \rightarrow C a O + 2 {e}^{-} + 2 {H}^{+}$

$\text{Overall}$

$C u O + \cancel{2 {H}^{+}} + C a + \cancel{{H}_{2} O} \rightarrow C a O + C u + \cancel{{H}_{2} O} + \cancel{2 {H}^{+}}$

i.e. $C u O + C a \rightarrow C a O + C u$, as required.

Here I added oxygen as water rather than representing the reduction of dioxygen:

$\frac{1}{2} {O}_{2} + 2 {e}^{-} \rightarrow {O}^{2 -}$

Of course, this is really making a meal out of a simple metathesis reaction. Oxygen transfer between the metals is simpler and more intuitive. The reaction of calcium with water is abstract (ordinarily calcium would reduce water), but electron transfer and redox reactions are abstract.