Answer:
Chromium (Cr)
Explanation:
We need to find the element which decreases its oxidation number in the reaction given below.
\(\rm 3Mg(s) + 2Cr(NO_3)_3 (aq)\rightarrow 3Mg(NO_3)_2 (aq) + 2Cr (s)\)
First let's determine the oxidation number of reactants:
Oxidation number of Mg = 0
Oxidation number of Chromium (Cr) in \(\rm 2Cr(NO_3)_3\) :
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 \(\rm NO_3^-\) :
We know that each nitrate ion has charge of -1. There are 3 nitrate ions, the total charge from nitrate ions is -3.
So,
\(Cr + (-3) = 0 \\ \\ Cr - 3 = 0 \\ \\ Cr = +3 \\\)
Hence, The oxidation number of Chromium is +3.
Now determine the oxidation number of products:
Oxidation number of Mg in \(\rm3Mg(NO_3)_2\) :
Each nitrate ion \(\rm({NO}_3^-)\) 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.
\(Mg + (-2) = 0 \\ \\ Mg = +2\)
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.
