Balancing Oxidation-Reduction Reactions with Ion-Electron Method: A Guide

[nate808]

(Cr(N2H4CO)6)4(Cr(CN)6)3 + KMnO4 + H2SO4--yields-- K2Cr2O7 + MnSO4 +CO2 + KNO3 + K2SO4+ HOH under acidic conditions

this is an oxidation reduction reaction and i am using the ion-electron method to balance the equation, however, i am having trouble breaking it into its smaller equations--can some1 please lead me into the right direcction--thnx

 

[eNathan]

Sure, first can you put your equation into latex using gamma's and what not. I posted a very long equation in ASCII and learned my lesson :O

 

[Gokul43201]

Nate, textbook style questions go in the Homework Help section of the Science Education Zone. Anyway, I'd like to know where you came across this equation/who gave it to you.

The only way in general, to balance such equations, is to write out the individual ionic reactions; assign multiplicative factors to each equation and solve simultaneously (using matrices/determinants if you wish) for these factors. In this case though, you will have too many such factors and solving becomes a nightmare.

The coefficients end up being in the hundreds and even in the thousands...

 

[nate808]

this question was given 2 me by my chemistry teacher--she said that it was the hardest problem that she had ever seen on the topic, and that it took her hours to do. My guess is that she found it in a book or something. Also, i understand the process but am having a hard time breaking it up into the individual equations

 

[saltydog]

(Cr(N2H4CO)6)4(Cr(CN)6)3 + KMnO4 + H2SO4--yields-- K2Cr2O7 + MnSO4 +CO2 + KNO3 + K2SO4+ HOH under acidic conditions
this is an oxidation reduction reaction and i am using the ion-electron method to balance the equation, however, i am having trouble breaking it into its smaller equations--can some1 please lead me into the right direcction--thnx

$$\left(Cr(N_2H_4CO)_6\right)_4\left(Cr(CN)_6\right)_3+KMnO_4+H_2SO_4\rightarrow K_2Cr_2O_7+MnSO_4+CO_2+KNO_3+K_2SO_4+HOH$$

Sorry, can't bear to see this in ASCII.

 

[Gokul43201]

First identify all the ions species that change oxidation state(OS). Let me start off with some of the easier ones. Take Mn first. In KMnO4, it has an OS of +7 and in MnSO4 it is +2. So we write :

$$Mn^{7+} + 5e^- \rightarrow Mn^{2+}~~~~---~~(1)$$

That was easy. Let's now do something a little harder, like Cr. N2H4CO is neutral and is usually co-ordinated with $Cr^{3+}$ (this is the only thing I'm a little uncertain about). That makes the first half of that complex have +12 charge. For the second half to have -12 charge, Cr(CN)6 must have a charge of +4 (since there's 3 of them). Since CN is 1-, Cr must be 2+. In dichromate, Cr has an OS of 6+ (easy to work out). This gives :

$$Cr^{3+} + Cr^{2+} \rightarrow 2Cr^{6+} + e^-~~~~--~~(2)$$

Now there's only one more equation to write (fewer than I'd initially thought) - for N. See if you can do that...

...the rest is just messy math.

 

[nate808]

so you don't need 2 take into account how many Cr or Mn there are in the initial part of the reaction?

 

[Gokul43201]

You don't know yet, how many there are - you only know the ratio in which they exist in different compounds in the reaction. The point of balancing the equation is to figure out how many there are. And that involves (i) matching the number of species in the LHS and RHS, and (ii) matching the number of electrons gained to the number lost.

One way to do this is through the use of unknown multipliers. Multiply eq(1) by x, eq(2) by y, eq(3) by z, etc. and write out the two conditions described above in terms of the corresponding coefficients (which are expressions in x, y, z, etc.).

 

[nate808]

$$2N^{-1}\rightarrow 2N^{5+} + 12e^- ~~~~---~~(3)$$

Is this correct for the N

 

[Borek]

Funny thing is, it can be balanced without halfreactions.

Largest coefficient in the equation is 1879.

 

[nate808]

how did u do that?

 

[Borek]

Algebraic approach.

Generally speaking you have to solve simultaneous equations. In this case - 8 equations and 9 unknowns. It can be solved as there is an additional condition that all coefficients must be integer.

 

[Gokul43201]

Algebraic approach.
Generally speaking you have to solve simultaneous equations. In this case - 8 equations and 9 unknowns. It can be solved as there is an additional condition that all coefficients must be integer.

Borek

Can't do this without a computer though. Using half-reactions, you could possibly reduce the number of unknowns to a more manageable number...

Nate : What Borek is saying is to append a coefficient to each compound in the reaction and write out simultaneous equations to make the number of individual species match.

 

[Borek]

Let's try simple example:

a H2 + b O2 -> c H2O

Number of atoms of every element must be identical on both sides of equation, so:

for H: 2a + 0b = 2c
for O: 0a + 2b = c

or in simplified form:

a = c
2b = c

Two equations and three unknowns, so basically it is not possible to solve this set, but let's assume c = 1. Then:

c = 1
a = 1
b = 1/2

All three coeficients must be integers, so let's multiply them by 2:

c = 2
a = 2
b = 1

and lo - our equation is balanced:

2 H2 + 1 O2 -> 2 H2O

Gokul is right, original equation needs a computer to be solved this way.

 

[Lapin Dormant]

Sorry, Really NICE read, Thanks, by the way, is it exothermic, or indo' or is that just a Silly question?

LD
T-hink's he might just want to hear that answer, should he wait here?

 

[Gokul43201]

We don't post answers to homework problems...so don't wait.

From inspection (more bonds broken than formed), I would guess the reaction is endothermic, but wouldn't bet on it.

 

[nate808]

but don't you need to take into account oxidation numbers as well and not just the atoms?

 

[Borek]

There are rare cases when oxidation numbers are additional information needed for balancing.

In most cases it is enough to balance atoms and charge - especially in case of net ionic equations. You treat charge the same way you treat atoms, so for example:

a Cu2+ + b Al(0) -> c Cu(0) + d Al(3+)

gives simultaneous equations:

Cu: 1a + 0b = 1c + 0d
Al: 0a + 1b = 0c + 1d
charge: 2a + 0b = 0c + 3d

then it is simple algebra at work ;)

 

[ardisalis]

Use the matrix method.