Keroro

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KMnO4 + NH3 -> KNO3 +MNO2 + KOH + H20



I KNOW HOW TO BALANCE REDOX RXN
BUT WHEN I SAW THIS, AND TRIED AT FIRST.. I WAS LIKE.. WHAT THE..



HOW CAN YOU DIVIDE INTO TWO HAF RXNS AT FIRST?


CAN ANY EXPERT EXPLAIN THIS STEP BY STEP?

THANKS A LOT !
 

DonExodus

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Did it on my own, googled my answer to verify :p
I got 8 KMnO4 + 3 NH3 --> 3 KNO3 + 8 MnO2 + 2 H2O + 5 KOH

Blatant plagarism from elsewhere yields the explanation Im too lazy to type :p
Person who got plagarized said:
Half reactions are useful when dealing with batteries or electrolysis,
but they aren't necessary when you only want to balance a redox
reaction.

So try this way:

-write the reactions without spectator species and ions (which do not
oxidise nor reduce); balance oxidising and reducing atoms (not needed
in this case).

MnO4- + NH3 --> NO3- + MnO2

-find the oxidation states of species and, thus, the number of
electrons they exchange (but do not balance the half reactions!)

+7 -3 +5 +4
MnO4- + NH3 --> NO3- + MnO2

MnO4- --> MnO2 +3e-
NH3 --> NO3- -8e-

-balance electron exchange:

8 MnO4- + 3 NH3 --> 3 NO3- + 8 MnO2

- balance oxigen and hydrogen. This could be a little tricky,
particularly in alcaline solution and so here's a way how to do it in
this case.

# calulate the excess of oxigen and hydrogen in the LEFT side of
the reaction. Should
it happen that the excess is on the right side, take it as a
negative value:

32 O --> 25 O +7 O
9 H --> 0 H +9 H

# The water to add to the RIGHT side of the reaction is then the
difference between
excess of hydorgen and oxigen; the hydroxide ion to add to the
RIGHT side of the
reaction is double the excess of oxigen minus the excess of
hydrogen; if one of
these quantities (or both) are negative, add it (or both) to the
left side of the
reaction instead (as a positive value).

exH - exO = 9 - 7 --> 2 H20
2 exO - exH = 14 - 9 --> 5 OH-

Thus:

8 MnO4- + 3 NH3 --> 3 NO3- + 8 MnO2 + 2 H2O + 5 OH-

- Add the appriopiate ions to reach neutrality (they are balanced, if
you have done everything correctly)

8 KMnO4 + 3 NH3 --> 3 KNO3 + 8 MnO2 + 2 H2O + 5 KOH


So, here's the balanced reaction.

Some people prefer to balance charges by adding OH- (or H+ in acid
solutions) and then oxigen adding water, instead of balancing oxigen
and hydrogen. The result is the same (that's to say oxigen, hydrogen
and charges are all balanced). Use the method you feel more comfortable
with.
Hope this helps,
Don
 
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slayerdeus

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there's another way you can do this by substituting different variables for each coefficient, but you have to be very quick and efficient with your algebra skills. on the DAT it probably isn't worth the time.
 

dat_student

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slayerdeus said:
there's another way you can do this by substituting different variables for each coefficient, but you have to be very quick and efficient with your algebra skills. on the DAT it probably isn't worth the time.
yes I think there is a faster way. Look at the oxidation state of each element and try to balance it that way ;)
 
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