Answer :
Answer:
Chromium (Cr)
Explanation:
We need to find the element which decreases its oxidation number in the reaction given below.
[tex] \rm 3Mg(s) + 2Cr(NO_3)_3 (aq)\rightarrow 3Mg(NO_3)_2 (aq) + 2Cr (s) [/tex]
First let's determine the oxidation number of reactants:
Oxidation number of Mg = 0
Oxidation number of Chromium (Cr) in [tex] \rm 2Cr(NO_3)_3[/tex] :
As we know that the total oxidation number of all atoms in a neutral molecule must add up to zero. So the oxidation number of Cr and Mg will be zero.
Oxidation number of [tex] \rm NO_3^-[/tex] :
We know that each nitrate ion has charge of -1. There are 3 nitrate ions, the total charge from nitrate ions is -3.
So,
[tex] Cr + (-3) = 0 \\ \\ Cr - 3 = 0 \\ \\ Cr = +3 \\ [/tex]
Hence, The oxidation number of Chromium is +3.
Now determine the oxidation number of products:
Oxidation number of Mg in [tex]\rm3Mg(NO_3)_2[/tex] :
Each nitrate ion [tex] \rm({NO}_3^-)[/tex] has a charge of -1. There are 2 nitrate ions, the total charge from nitrate ions is -2.
Now, Sum of oxidation number of Mg and Cr Will be 0.
[tex] Mg + (-2) = 0 \\ \\ Mg = +2 [/tex]
Chromium has an oxidation number of zero.
Since, In Mg the oxidation number increases from 0 to +2 and in Cr, the oxidation number of Cr is decreased from +3 to 0.
Therefore, the element that decreases its oxidation number in this reaction is chromium.