Question #aec6a

2 Answers
Sep 11, 2017

Answer:

Well, oxalato ion is OXIDIZED to give carbon dioxide gas.....

Explanation:

#"Oxidation:"# #C(+III) rarrC(+IV)#

#C_2O_4^(2-) rarr 2CO_2(g)+2e^(-)# #(i)#

Mass and charge are balanced so this is a valid oxidation reaction. And for every oxidation there is a corresponding reduction.

#"Reduction:"# #Fe(+III) rarrFe(+II)#

#Fe^(3+)+ e^(-)rarrFe^(2+)# #(ii)#

We add #(i)# and #(ii)# in such a way that the electrons are REMOVED from the equation, and in this way we formalize the electron transfer conceived to occur in a redox reaction, i.e. #(i) +2xx(ii)# gives......

#underbrace(2Fe^(3+)+C_2O_4^(2-) + cancel(2e^(-))rarr2Fe^(2+)+2CO_2(g)+cancel(2e^(-)))_"to give finally"#

#2Fe^(3+)+C_2O_4^(2-) rarr2Fe^(2+)+2CO_2(g)#

Ferric ion is reduced to ferrous ion, and oxalate ion is oxidized to carbon dioxide.

Do you follow?

Sep 12, 2017

Answer:

Here's what I get.

Explanation:

You can find the general technique for balancing redox equations in acid solution here.

Step 1: Write the two half-reactions

#"C"_2"O"_4^"2-" → "CO"_2#
#"Fe"^"3+" → "Fe"^"2+"#

Step 2: Balance all atoms other than #"H"# and #"O"#

#"C"_2"O"_4^"2-" → "2CO"_2#
#"Fe"^"3+" → "Fe"^"2+"#

Step 3: Balance #"O"#

Done.

Step 4: Balance #"H"#

Done.

Step 5: Balance charge

#"C"_2"O"_4^"2-" → "2CO"_2 +2"e"^"-"#
#"Fe"^"3+" +"e"^"-" → "Fe"^"2+"#

We see that #bbcolor(red)("C"_2"O"_4^"2-"color(white)(l) "is oxidized")# because it loses electrons, and #bbcolor(red)("Fe"^"3+"color(white)(l)"is reduced")# because it gains an electron.

There are #bbcolor(red)("no disproportionations")#.

Step 6: Equalize electrons transferred

#1 × ["C"_2"O"_4^"2-" → "2CO"_2 +2"e"^"-"]#
#2 × ["Fe"^"3+" +"e"^"-" → "Fe"^"2+"]#

Step 7: Add the two half-reactions

#"C"_2"O"_4^"2-" → "2CO"_2 +color(red)(cancel(color(black)(2"e"^"-")))#
#ul("2Fe"^"3+" + color(red)(cancel(color(black)("2e"^"-"))) → "2Fe"^"2+"color(white)(mmmmm))#
#"C"_2"O"_4^"2-" + "2Fe"^"3+" → "2CO"_2 + "2Fe"^"2+"#

Step 8: Check mass balance

#ul(mathbf("Atom"color(white)(m)"On the left"color(white)(m)"On the right"))#
#color(white)(ml)"C"color(white)(mmmmm)2color(white)(mmmmmmm)2#
#color(white)(ml)"O"color(white)(mmmmm)4color(white)(mmmmmmm)4#
#color(white)(ml)"Fe"color(white)(mmmmll)2color(white)(mmmmmmm)2#

Step 9: Check charge balance

#ulbb("On the left"color(white)(m)"On the right")#
#color(white)(mmll)"4+"color(white)(mmmmmm)"4+"#

The balanced equation is

#color(red)("C"_2"O"_4^"2-" + "2Fe"^"3+" → "2CO"_2 + "2Fe"^"2+")#