# Element A in a compound ABD has oxidation number A^(n-) .It is oxidised by Cr_2O_7^(2-) in acid medium. In an experiment 1.68*10^-3 moles of K_2Cr_2O_7 were used for 3.36*10^(-3) moles of ABD. the new oxidation number of A after oxidation is?

Sep 7, 2017

$3 - n$

#### Explanation:

For starters, you should know that chromium has an oxidation number equal to $+ 6$ in the dichromate anion.

The two most common oxidation states for chromium are $+ 6$ and $+ 3$, so you can assume that in this reaction, the dichromate anion will be reduced to chromium(III), ${\text{Cr}}^{3 +}$.

This means that a single atom of chromium will gain $3$ electrons. Since the dichromate anion contains $2$ atoms of chromium, it follows that a total of $6$ electrons will be gained in the reduction half-reaction.

As you know, in a redox reaction, the number of electrons gained in the reduction half-reaction must be equal to the number of electrons lost in the oxidation half-reaction.

This implies that a total of $6$ electrons must be lost by $\stackrel{\textcolor{b l u e}{- n}}{\text{A}}$in the oxidation half-reaction.

Now, notice that the reaction consumes $2$ moles of $\stackrel{\textcolor{b l u e}{- n}}{\text{A}}$ for every $1$ mole of dichromate anions

$\left(3.36 \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{{10}^{- 3}}}} \textcolor{w h i t e}{.} {\text{moles"color(white)(.) stackrel(color(blue)(-n))("A"))/(1.68 * color(red)(cancel(color(black)(10^(-3)))) color(white)(.)"moles Cr"_2"O"_7^(2-)) = ("2 moles"color(white)(.) stackrel(color(blue)(-n))("A"))/("1 mole Cr"_2"O}}_{7}^{2 -}\right)$

This means that the total of $6$ electrons needed for the reduction half-reaction will be delivered by $2$ $\stackrel{\textcolor{b l u e}{- n}}{\text{A}}$ atoms, which implies that every $\stackrel{\textcolor{b l u e}{- n}}{\text{A}}$ atom will lose $3$ electrons.

Consequently, you can say that the oxidation number of $\text{A}$ will increase by $3$ and go from $- n$ on the reactants' side to

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{new oxidation state} = - n + 3 = 3 - n}}}$

on the products' side.