# Question #d0498

Jun 27, 2016

For every oxidation there is a reduction; and for every reduction there is an oxidation.

#### Explanation:

$2 N a + F e C {l}_{2} \rightarrow 2 N a C l + F e$

Sodium is oxidized; iron is reduced.

$H C \equiv C H + \frac{5}{2} {O}_{2} \rightarrow 2 C {O}_{2} + {H}_{2} O$

Carbon is oxidized; oxygen is reduced.

$2 P b S + 3 {O}_{2} \rightarrow 2 S {O}_{2} + 2 P b O$

Oxygen is reduced, sulfur is oxidized.

$2 {H}_{2} + {O}_{2} \rightarrow 2 {H}_{2} O$

Dihydrogen is oxidized; oxygen is reduced.

$A g N {O}_{3} + C u \rightarrow C u N {O}_{3} + A g$

Copper is oxidized; silver is reduced.

All the oxidations feature an increase in oxidation state of the species that is oxidized. Likewise, all reductions feature a decrease in oxidation number of the species that is reduced. You might want to look up the method of half equations for the balancing of redox reactions.