Question

When iron ( Fe
Fe
) is placed in a copper sulfate ( CuSO4
CuSO
4
) solution, the blue solution slowly turns light green and the iron is coated with copper.

This equation shows the transfer of electrons during the reaction.

Cu2++Fe→Cu+Fe2+
Cu
2
+
+
Fe
→
Cu
+
Fe
2
+


Which best describes what happens during this oxidation-reduction reaction?

(1 point)
Responses

Iron gives up two electrons and copper accepts two electrons.
Iron gives up two electrons and copper accepts two electrons.

Copper gives up one electron and iron accepts one electron.
Copper gives up one electron and iron accepts one electron.

Iron gives up one electron and copper accepts one electron.
Iron gives up one electron and copper accepts one electron.

Copper gives up two electrons and iron accepts two electrons.
Copper gives up two electrons and iron accepts two electrons.

Answer 1

In the given oxidation-reduction reaction:

\[ \text{Cu}^{2+} + \text{Fe} \rightarrow \text{Cu} + \text{Fe}^{2+} \]

Iron (Fe) is oxidized and loses two electrons to form \(\text{Fe}^{2+}\), while copper ions (\(\text{Cu}^{2+}\)) are reduced and gain those two electrons to form solid copper (Cu).

Therefore, the correct description of what happens during this oxidation-reduction reaction is:

**Iron gives up two electrons and copper accepts two electrons.**