When Cu(NO3?)2(aq) reacts with Zn(s), Cu(s) is a product. Cu(NO3?)2(aq) + Zn(s)

Question

When Cu(NO3?)2(aq) reacts with Zn(s), Cu(s) is a product.

Cu(NO3?)2(aq) + Zn(s) ? Cu(s) + Zn(NO3?)2(aq)

Which statement best describes what happened in this redox reaction?


A) Zn(s) is reduced to Zn2+?(aq) by the oxidizing agent, Cu2+.
B) Cu2+?(aq) is oxidized (to Cu(s)) by the reducing agent, Zn(s).
C) Cu2+, as an oxidizing agent, causes Zn(s) to be oxidized to Zn2+?(aq). Zn(s), as a reducing agent, causes Cu2+?(aq) to be reduced to Cu(s).
D) Zn(s) bonds with Cu(NO3?)2 to form Zn(NO3?)2(aq), releasing copper metal.

Explanation / Answer

It's called an oxidation-reduction reaction. One reactant donates electrons (becoming "oxidized") while the other reactant accepts electrons (becoming "reduced"). It makes it easier to look at the ionic equation for this: Cu (2+) + 2(NO3) (1-) + Zn ---> Cu + Zn (2+) + 2(NO3) (1-) By looking at the reaction you can see that Cu went from a 2+ charge to a +0 charge where it became a solid (meaning it gained 2 electrons). Therefore, Cu is reduced. Zn went from a solid to having a 2+ charge indicating that it lost 2 electrons. This means that Zn is oxidized. To make it easier to remember which one is oxidized and which one is reduced, use this: LEO the lion says GER. L(ose)E(lectrons)O(xidized) the lion says G(ain)-E(lectrons)-R(educed) So the answer is that copper was reduced (gained 2 electrons) and became a solid.

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