# Question #2179b

##### 1 Answer

#### Answer:

#### Explanation:

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 **Increasing** from *losing* **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 **increasing** from *losing* **electron**.

Now, you know that the two steps required a **lost** by element

Since you know that element **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 **electrons** to get to a

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

You can thus say that the cation carried a

#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,

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+)#