Question

When iron is places in a copper sulfate solution, the blue solution slowly turns light green and the iron is coated with copper.

Wihch best describes what occurs during this oxidation-reduction reaction?

Copper gives up two electrons and iron accepts two electrons
Copper gives up one electron and iron accepts one electron
Iron gives up one electron and copper accepts two
Iron gives up on electron and copper accepts one

Answer 1

The correct description of what occurs during the oxidation-reduction reaction when iron is placed in a copper sulfate solution is:

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

Here’s a breakdown of the reaction:

1. Iron (Fe) is oxidized. It loses electrons, specifically two electrons:
\[
\text{Fe} \rightarrow \text{Fe}^{2+} + 2e^-
\]

2. Copper ions (\(\text{Cu}^{2+}\)) in the copper sulfate solution are reduced. They gain electrons to form copper metal:
\[
\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}
\]

Overall, the process involves iron oxidizing to iron ions and copper ions reducing to copper metal. The transformation seen - a blue solution gradually turning light green and copper being deposited on the iron - indicates the transfer of electrons between iron and copper ions. Thus, iron gives up electrons, and copper, in turn, accepts them.

So, it is most accurate to say:
**Iron gives up two electrons and copper accepts two.**