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The information below describes a redox reaction.
Cr3+(aq)+2Cl-(aq)---->Cr(s)+Cl2(s)
2Cl-(aq)--->Cl2(g)+2e-
Cr3+(aq)+3e- ---->Cr(s)

What is the final, balanced equation for this reaction?
1.) 2cr3+(aq)+6Cl-(aq) ------> 2Cr(s)+3Cl2(g)
2.) 2Cr3(aq)+2Cl-(aq)+6e- --->Cl2(g)+2Cr(s)
3.) Cr3+(aq)+6Cl-(aq)+3e- ---->2Cr(g)+3Cl2(g)
4.) Cr3+(aq)+2Cl-(aq)------>Cr(s)+Cl2(g)

Respuesta :

Edufirst
Answer: option 1)  2Cr3+(aq)+6Cl-(aq) ------> 2Cr(s)+3Cl2(g)

Explanation:

1) Write the oxidation half-reaction:

[tex]2Cl^-(aq)---\ \textgreater \ Cl_2(g)+2e^-[/tex]

2) Write the reduction half-raction:

[tex]Cr^{3+}(aq)+3e^{-}---\ \textgreater \ Cr(s)[/tex]

3) Multiply each half-reaction by the appropiate coefficient to equal the number of electrons of both half-reactions.

[tex]6Cl^{-}(aq)---\ \textgreater \ 3Cl_2(g)+6e^{-} 2Cr^{3+}(aq)+6e^{-}---\ \textgreater \ 2Cr(s)[/tex]

4) Add both half-reactions

[tex]2Cr^{3+}+6Cl^{-}(aq)---\ \textgreater \ 2Cr(s) +3Cl_2(g)[/tex]

And that is the answer. You can count the atoms and charges on every side and check they are equal.

Answer:

A

Explanation:

got it correct on edge

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