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?

1 Answer
Sep 7, 2017

Answer:

#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), #"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(color(blue)(-n))("A")#in the oxidation half-reaction.

Now, notice that the reaction consumes #2# moles of #stackrel(color(blue)(-n))("A")# for every #1# mole of dichromate anions

#(3.36 * color(red)(cancel(color(black)(10^(-3))))color(white)(.)"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-))#

This means that the total of #6# electrons needed for the reduction half-reaction will be delivered by #2# #stackrel(color(blue)(-n))("A")# atoms, which implies that every #stackrel(color(blue)(-n))("A")# atom will lose #3# electrons.

Consequently, you can say that the oxidation number of #"A"# will increase by #3# and go from #-n# on the reactants' side to

#color(darkgreen)(ul(color(black)("new oxidation state" = -n + 3 = 3 - n)))#

on the products' side.