Question #e437b

1 Answer
Apr 30, 2017

Warning! Long Answer.

#"HOCH"_2"CH"_2"OH" + "10MnO"_4^"-" + "10OH"^"-" → "2CO"_2 + "10MnO"_4^"2-" + 8"H"_2"O"#

Explanation:

Let's rewrite the formula for ethylene glycol, #"HOCH"_2"CH"_2"OH"#, as #"C"_2"H"_6"O"_2#.

Then the skeleton equation is

#"C"_2"H"_6"O"_2 + "MnO"_4^"-" → "CO"_2 + "MnO"_4^"2-"#

One way to balance this equation is by the half-reaction method.

Step 1. Separate the equation into two half-reactions.

#"C"_2"H"_6"O"_2 → "CO"_2#
#"MnO"_4^"-"→"MnO"_4^"2-"#

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

#"C"_2"H"_6"O"_2 → "2CO"_2#
#"MnO"_4^"-"→"MnO"_4^"2-"#

Step 3. Balance #"O"# by adding #"H"_2"O"# molecules to the deficient side.

#"C"_2"H"_6"O"_2 + 2"H"_2"O" → "2CO"_2#
#"MnO"_4^"-"→"MnO"_4^"2-"#

Step 4. Balance #"H"# by adding #"H"^"+"# ions to the deficient side.

#"C"_2"H"_6"O"_2 + 2"H"_2"O" → "2CO"_2 + "10H"^"+"#
#"MnO"_4^"-"→"MnO"_4^"2-"#

Step 5. Balance charge by adding electrons to the deficient side.

#"C"_2"H"_6"O"_2 + 2"H"_2"O" → "2CO"_2 + "10H"^"+" + "10e"^"-"#
#"MnO"_4^"-" + "e"^"-"→"MnO"_4^"2-"#

Step 6. Multiply each half-reaction by a number to equalize the electrons transferred.

#1×["C"_2"H"_6"O"_2 + 2"H"_2"O" → "2CO"_2 + "10H"^"+" + "10e"^"-"]#
#10 × ["MnO"_4^"-" + "e"^"-"→"MnO"_4^"2-"]#

Step 7. Add the two half-reactions.

#"C"_2"H"_6"O"_2 + "10MnO"_4^"-"+ 2"H"_2"O" → "2CO"_2 + "10MnO"_4^"2-" + "10H"^"+"#

Step 8. Convert to basic solution my adding the appropriate multiples of #"H"^"+" + "OH"^"-" → "H"_2"O"#.

#"C"_2"H"_6"O"_2 + "10MnO"_4^"-"+ color(red)(cancel(color(black)("2H"_2"O"))) → "2CO"_2 + "10MnO"_4^"2-" + color(red)(cancel(color(black)("10H"^"+")))#
#color(red)(cancel(color(black)("10H"^"+"))) + 10"OH"^"-" → stackrelcolor(blue)(8)(color(red)(cancel(color(black)(10))))"H"_2"O"#
#stackrel(—————————————————————)("C"_2"H"_6"O"_2 + "10MnO"_4^"-" + "10OH"^"-" → "2CO"_2 + "10MnO"_4^"2-" + 8"H"_2"O")#

Step 9. Check that all atoms are balanced.

#bb("Atom"color(white)(m)"On the left"color(white)(m)"On the right")#
#color(white)(m)"C"color(white)(mmmmm)2color(white)(mmmmmmm)2#
#color(white)(m)"H"color(white)(mmmml)16color(white)(mmmmmmll)16#
#color(white)(m)"O"color(white)(mmmml)52color(white)(mmmmmmll)52#
#color(white)(m)"Mn"color(white)(mmmll)10color(white)(mmmmmmll)10#

Step 10. Check that charge is balanced

#bb("On the left"color(white)(m)"On the right")#
#color(white)(mml)"20-"color(white)(mmmmmm)"20-"#

Everything checks! The equation is balanced.