Question #065fa
1 Answer
Here's what I got.
Explanation:
In the redox reaction given to you, the dichromate anions are being reduced to chromium(III) cations and the iron(II) cations are being oxidized to iron(III) cations.
Chromium has a
The fact that you have hydroxide anions present on the products' side tells you that this redox reaction takes place in basic medium.
So, the oxidation half-reaction looks like this
#stackrel(color(blue)(+2))("Fe") ""_ ((aq))^(2+) -> stackrel(color(blue)(+3))("Fe") ""_ ((aq))^(3+) + "e"^(-)#
Notice that the half-reaction is balanced in terms of charge because you have
#(2+) -> (3+) + (1-)#
The reduction half-reaction looks like this
#stackrel(color(blue)(+6))("Cr")_ 2"O"_ (7(aq))^(2-) + 6"e"^(-)-> 2stackrel(color(blue)(+3))("Cr")""_ ((aq))^(3+)#
Here each atom of chromium gains
In order to balance out the atoms of oxygen, add water to the side that needs oxygen and hydrogen cations,
You will have
#14"H"_ ((aq))^(+) + stackrel(color(blue)(+6))("Cr")_ 2"O"_ (7(aq))^(2-) + 6"e"^(-)-> 2stackrel(color(blue)(+3))("Cr")""_ ((aq))^(3+) + 7"H"_ 2"O"_ ((l))#
Now, because you're in basic medium, you need to add hydroxide anions to both sides of the equation in order to neutralize the added protons.
This means that you have
#overbrace(14"OH"_ ((aq))^(-) + 14"H"_ ((aq))^(+))^(color(red)(=14"H"_ 2"O")) + stackrel(color(blue)(+6))("Cr")_ 2"O"_ (7(aq))^(2-) + 6"e"^(-)-> 2stackrel(color(blue)(+3))("Cr")""_ ((aq))^(3+) + 7"H"_ 2"O"_ ((l)) + 14"OH"_ ((aq))^(-)#
which is equivalent to
#stackrel(color(red)(7))color(red)(cancel(color(black)(14)))"H"_ 2"O"_ ((l)) + stackrel(color(blue)(+6))("Cr")_ 2"O"_ (7(aq))^(2-) + 6"e"^(-)-> 2stackrel(color(blue)(+3))("Cr")""_ ((aq))^(3+) + color(red)(cancel(color(black)(7"H"_ 2"O"_ ((l))))) + 14"OH"_ ((aq))^(-)#
You can simplify the half-reaction to
#7"H"_ 2"O"_ ((l)) + stackrel(color(blue)(+6))("Cr")_ 2"O"_ (7(aq))^(2-) + 6"e"^(-)-> 2stackrel(color(blue)(+3))("Cr")""_ ((aq))^(3+) + 14"OH"_ ((aq))^(-)#
Notice that the half-reaction is balanced in terms of charge because you have
#(2-) + 6 xx (1-) = 2 xx (3+) + 14 xx (1-)#
So there you have it, the two balanced half-reactions are
# {(color(white)(aaaaaaaaaaaaaaaaaa)stackrel(color(blue)(+2))("Fe") ""_ ((aq))^(2+) -> stackrel(color(blue)(+3))("Fe") ""_ ((aq))^(3+) + "e"^(-)), (7"H"_ 2"O"_ ((l)) + stackrel(color(blue)(+6))("Cr")_ 2"O"_ (7(aq))^(2-) + 6"e"^(-)-> 2stackrel(color(blue)(+3))("Cr")""_ ((aq))^(3+) + 14"OH"_ ((aq))^(-)) :}#
To get the balanced chemical equation, you need to use the fact that in every redox reactions, the number of electrons lost in the oxidation half-reaction must be equal to the number of electrons gained in the reduction half-reaction.
So you would multiply the oxidation half-reaction by
# {(color(white)(aaaaaaaaaaaaaaaaa)6stackrel(color(blue)(+2))("Fe") ""_ ((aq))^(2+) -> 6stackrel(color(blue)(+3))("Fe") ""_ ((aq))^(3+) + 6"e"^(-)), (7"H"_ 2"O"_ ((l)) + stackrel(color(blue)(+6))("Cr")_ 2"O"_ (7(aq))^(2-) + 6"e"^(-)-> 2stackrel(color(blue)(+3))("Cr")""_ ((aq))^(3+) + 14"OH"_ ((aq))^(-)) :}#
#color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa)/color(white)(a)#
#stackrel(color(blue)(+6))("Cr")_ 2"O"_ (7(aq))^(2-) + 7"H"_ 2"O"_ ((l)) + 6stackrel(color(blue)(+2))("Fe") ""_ ((aq))^(2+) + color(red)(cancel(color(black)(6"e"^(-)))) -> 2stackrel(color(blue)(+3))("Cr")""_ ((aq))^(3+) + color(red)(cancel(color(black)(6"e"^(-)))) + 6stackrel(color(blue)(+3))("Fe") ""_ ((aq))^(3+) + 14"OH"_ ((aq))^(-)#
which is exactly what you started with
#stackrel(color(blue)(+6))("Cr")_ 2"O"_ (7(aq))^(2-) + 7"H"_ 2"O"_ ((l)) + 6stackrel(color(blue)(+2))("Fe") ""_ ((aq))^(2+) -> 2stackrel(color(blue)(+3))("Cr")""_ ((aq))^(3+) + 6stackrel(color(blue)(+3))("Fe") ""_ ((aq))^(3+) + 14"OH"_ ((aq))^(-)#