# Question #1650c

Nov 16, 2015

Here's what I got.

#### Explanation:

The balanced chemical equation for this redox reaction looks like this

${\text{Fe"_text((aq])^(2+) + "Cu"_text((aq])^(2+) -> "Cu"_text((aq])^(+) + "Fe}}_{\textrm{\left(a q\right]}}^{3 +}$

The thing to remember when dealing with ions is that their oxidation state is equal to their net charge.

In this case, you're dealing with the iron(II) and iron(III) cations, and the copper(II) and copper(I) cations.

The oxidation states for these cations will be equal to

• ${\text{Fe}}^{2 +} \to$ oxidation state: $+ 2$
• ${\text{Fe}}^{3 +} \to$ oxidation state; $+ 3$
• ${\text{Cu}}^{2 +} \to$ oxidation state: $+ 2$
• ${\text{Cu}}^{+} \to$ oxidation state: $+ 1$

Now, in order to identify which element is being oxidized and which element is being reduced, look at how the oxidation states change from one side of the reaction to the other.

Notice that iron's oxidation state goes from $+ 2$ on the reactants' side, to $+ 3$ on the products' side. This means that iron is being oxidized, since its oxidation number increased.

${\text{Fe"^(2+) -> "Fe"^(3+) + 1"e}}^{-}$

On the other hand, copper's oxidation state goes from $+ 2$ on the reactants' side, to $+ 1$ on the products' side. This means that copper is being reduced, since its oxidation number decreased.

${\text{Cu"^(2+) + 1"e"^(-) -> "Cu}}^{+}$

Now, copper is being reduced, which means that it's acting as an oxidizing agent. Iron is being oxidized by copper, which means that it's acting as a reducing agent.