Question

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)

Answer 1

A

Step-by-step explanation:

got it correct on edge

Answer 2

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

Step-by-step explanation:

1) Write the oxidation half-reaction:


`2Cl^-(aq)---\ \textgreater \ Cl_2(g)+2e^-`

2) Write the reduction half-raction:


`Cr^(3+)(aq)+3e^(-)---\ \textgreater \ Cr(s)`

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


`6Cl^(-)(aq)---\ \textgreater \ 3Cl_2(g)+6e^(-) 2Cr^(3+)(aq)+6e^(-)---\ \textgreater \ 2Cr(s)`

4) Add both half-reactions


`2Cr^(3+)+6Cl^(-)(aq)---\ \textgreater \ 2Cr(s) +3Cl_2(g)`

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