Question

The information below describes a redox reaction.

Answer 1

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

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.