Question #2179b

Start by writing the oxidation half-reaction for the first oxidation step

#stackrel(color(blue)(+n))("X")""^(n+) -> stackrel(color(blue)(+6))("X") stackrel(color(blue)(-2))("O")_ 4""^(2-) + x"e"^(-)#

In the first oxidation step, the oxidation number of #"X"# is Increasing from #color(blue)(+n)# on the reactants' side to #color(blue)(+6)# on the products' side because every atom of #"X"# is losing #x# electrons.

In the second oxidation step, you have

#stackrel(color(blue)(+6))("X")stackrel(color(blue)(-2))("O")_ 4""^(2-) -> stackrel(color(blue)(+7))("X")stackrel(color(blue)(-2))("O")_ 4""^(-) + "e"^(-)#

This time, the oxidation number of #"X"# is increasing from #color(blue)(+6)# on the reactants' side to #color(blue)(+7)# on the products' side because every atom of #"X"# is losing #1# electron.

Now, you know that the two steps required a #4:1# ratio of oxidizing agent. This implies that the numbers of electrons lost by element #"X"# in the two steps are in a #4:1# ratio as well.

Since you know that element #"X"# lost #1# electron in the second step, you can say that it lost

#1 color(red)(cancel(color(black)("e"^(-)color(white)(.)"in 2nd step"))) * ("4 e"^(-)color(white)(.)"in 1st step")/(1color(red)(cancel(color(black)("e"^(-)color(white)(.)"in 2nd step")))) = 4 "e"^(-)# #"in 1st step"#

Therefore, you can say that you have

#stackrel(color(blue)(+n))("X")""^(n+) -> stackrel(color(blue)(+6))("X") stackrel(color(blue)(-2))("O")_ 4""^(2-) + 4"e"^(-)#

Since every atom of #"X"# loses #4# electrons to get to a #color(blue)(+6)# oxidation state, you can say that its initial oxidation state was

#(+n) = (+6) + 4 xx (1-) implies n= +2#

You can thus say that the cation carried a #2+# charge.

#stackrel(color(blue)(+2))("X")""^(2+) -> stackrel(color(blue)(+6))("X") stackrel(color(blue)(-2))("O")_ 4""^(2-) + 4"e"^(-)#

Notice that this oxidation half-reaction is not balanced. Assuming that the reaction takes place in acidic solution, you can balance the oxygen atoms by adding water molecules to the side that needs oxygen and hydrogen cations, #"H"^(+)#, to the side that needs hydrogen.

You will have

#4"H"_ 2"O" + stackrel(color(blue)(+2))("X")""^(2+) -> stackrel(color(blue)(+6))("X") stackrel(color(blue)(-2))("O")_ 4""^(2-) + 4"e"^(-) + 8"H"^(+)#

The charge is balanced becuase you have

#(2+) = (2-) + 4 xx (1-) + 8 xx (1+)#