The oxidation and reduction equations for when zinc is placed in water?

Jul 10, 2017

$\text{Oxidation:}$ $Z n \left(s\right) \rightarrow Z {n}^{2 +} + 2 {e}^{-}$ $\left(i\right)$

Explanation:

Water COULD be reduced.........

${H}_{2} O + {e}^{-} \rightarrow \frac{1}{2} {H}_{2} \left(g\right) \uparrow + H {O}^{-}$ $\left(i i\right)$

We add (i) and (ii) such that the electrons are eliminated from the reaction.......

$\left(i\right) + 2 \times \left(i i\right)$

$Z n \left(s\right) + 2 {H}_{2} O \rightarrow Z {n}^{2 +} + {H}_{2} \left(g\right) \uparrow + 2 H {O}^{-}$

We would need exceptionally clean and pure zinc metal for this reaction to occur, but in the presence of acid.......we could simply write.....

$Z n \left(s\right) + 2 H C l \left(a q\right) \rightarrow Z n C {l}_{2} \left(a q\right) + {H}_{2} \left(g\right) \uparrow$