[boroncarbon]

Hi i need help balancing this equation. I suppose you should split it into redox half equations. I can see the iron gets oxidised and the manganese gets reduced. But the oxidation states of the cyanide also changes- which confuses me abit

K₄Fe(CN)₆ + KMnO₄ + H₂SO₄ --> KHSO₄ + Fe₂(SO₄)₃ +MnSO₄ + HNO₃ + CO₂ + H₂O

[AWK]

Use method of oxidation numbers (ONs) in trick mode. Use oxidation numbers for C and N from right side of equation on the left side and calculate artificial oxidation number for Fe on the left side (I guess ON for Fe = -58, but check me). You will get a "polite" redox reaction with Mn and Fe that changed ONs on both sides. (the highest coefficient is 299)

[magician4]

a slightly different approach:

you already analysed correctly the "elements" being oxidized, reduced
now you should take a closer look at exactly which atoms are being oxidized, reduced , and how they are stoechiometrically coupled

Fe in K₄[Fe(CN)₆] is (+II) , Fe in Fe₂(SO₄)₃ is (+III)
  you'll oxidize iron by 1 per iron
... and as you need two irons to create one iron(III)sulfate, that's a total of two electrons being released per basic transformation:

(from 2 K₄[Fe(CN)₆] ) : 2 Fe(+II)  2 Fe(+III) + 2 e- (in Fe₂(SO₄)₃)

now, those irons are coupled to 2 * 6 = 12 cyanides
in cyanide, carbon is (+II) and nitrogen is (-III)
carbon goes from (+II) to (+IV) :
(in cyanide) C(+II)  C(+IV) + 2 e^- (in CO₂ )
nitrogen goes from (-III) to (+V)
(in cyanide) N(-III)  N(+V) + 8 e^- (in HNO₃)
total: CN^-  C(+IV) + N (+V) + 10 e^-
for 12 cyanides: 12 CN^-  12C(+IV) + 12 N (+V) + 120 e^-

considering the coupling once again:
(in 2 K₄[Fe(CN)₆] ) 2 [Fe(CN)₆]^4-  2 Fe(+III) + 12C(+IV) + 12 N (+V) + 122 e^- (in named substances)

... and this is the total oxidation taking place per 2 units of K₄[Fe(CN)₆] , and has to be balanced with the electron uptake, i.e. reduction of manganese per transformation Mn(+VII) (in permanganate) + 5 e^- Mn(+II) (in manganese-(II)-sulfate)

think you can manage the rest?

regards

Ingo

[AWK]

Both methods are exactly equivalent. One of them needs more time.

[Borek]

This kind of reaction is probably easiest to balance by the algebraic method. No tricks involved, just a frontal attack.

http://www.chembuddy.com/?left=balancing-stoichiometry&right=algebraic-method