Question #2179b
1 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
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
Now, you know that the two steps required a
Since you know that element
#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
#(+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+)#