[ajones51]

Could someone out there please help me try and balance the following redox equation? I have all the reagents and products, but just need to balance them with stoichiometric coefficients.

[Co(NH3)5Cl]Cl2  + HNO3  ------> Co(NO3)2 + HCl + NO + H2O

I know that these are the following half reactions:
Co3+  ----> Co2+  + 1 electron
N(-3) ----> N(+2) + 5 electrons

But, both of these elements are in the complex, and since the number of electrons being reduced must equal the number of electrons oxidized, how can balance this if, there are 5 NH3 ligands in the complex (which would be 25 electrons transferred) and only 1 electron transferred per cobalt atom?  Or do the other nitrogens somehow get involved?

If someone could please provide me with an explanation, that would be awesome.  Thanks!

[sondakem]

Good instinct!

Try involving other nitrogens.
Make sure to keep track of which half reactions are oxidations and which are reductions.

Good luck.

[AWK]

Ajones51,
note the whole molecule loose or take electrons in the redox reaction and in this case you can solve it in the easiest way taking into account the rule - sum of oxidation numbers for a neutral molecule should be null.
The easiest way is: the complex looses electrons - is oxidized, nitric acid takes electrons - is reduced.
Hence cumulate all loosed electrons on Co (since this is a single atom), and assign oxidation numbers for H, N (as in NO)  and Cl the same for both sides
Then Co(-22) => Co(+2) + 24e(-)
and N(+5) + 3e(-) = N(+2)
Common multiplier is 24, and reaction is:
[Co(NH3)5Cl]Cl2  + 10HNO3  ------> Co(NO3)2 + 3HCl + 13NO + 11H2O

[Mr.science]

Can some one please help me?
There is this redox equation that i just cant seem to solve...
please see if you can aid me.

[Cr(N2H4CO)6]4 [Cr(CN)6]3 + MnO4^(-1) --> Cr2O7^(-2) + Mn^(+2) + CO2 + NO3^(-1) (acid)

I am not even sure how to break this.  Any help would be much appreciated!

[AWK]

Use exactly the same method as above for "pure" empirical formula with cumulation all electrons on Cr. Note, H+ are missing on the left side of equation.

Look at:
 http://siorpc12.chem.pg.gda.pl/smf/index.php?topic=2.0
and download:
Trudne_redoksy.pdf
Do not be afraid that language is not English. Equation are international. Solutions are only sketched! In this case two methods are used

[Borek]

You may also go for algebraic method - it does wonders in such cases and doesn't require any artficial assumptions:

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

[AWK]

But in this case an algebraic method without computer program may need a few days of work. An artificial assumption works in ten minutes.

[Borek]

7 unknowns, an hour at worst. Don't scare people.

[AWK]

7 unknowns, an hour at worst. Don't scare people.

For you and me (amongst that discussing this subject)

[lutesium]

Hey AWK I didn't say that an Organic Chemist can't solve a redox equation! I'd like to say that an Organic Chemist would like to help thanx borek that he has moved this topic to its right place !!!


Lutesium...

[Alpha-Omega]

I will start this for you....now just balance it:

Cobalt ammonium chloride in nitric acid solution will form HCl and NO gas (nitrogen monoxide) and water.  The soluble salt, Cobalt II Nitrate remains in the aqueous solution.

Co3+ [(NH3)5Cl)Cl2 + H+ + (NO3)- --> Co2+ + (NO3)2- + H+ + Cl- + NO + H2O

The two half reactions are as follows:


2(1e- + Co3+ --> Co2+) Reduced
(where Co+3 goes to Co+2)

((NO3)- --> NO0 + 1e) Oxidized 
(where for (NO3)- the N = +5 and O = +2 goes to NO where N = +2 and O = -2)

[Alpha-Omega]

It has been my experience that everyone has thier own approach to balancing redox equations.  If I were teaching this and giving a detailed explanation, I would do as follows:

You have to balance the charges and the masses.  So since she is using hald reactions, the coefficient for her cobalt species is 2 since that balances the charge in the reduction of the cobalt Co+3 ----Co+2, then I would proceed as follows:

2[Co(NH3)5Cl]Cl2     +     HNO3     ------>     2Co(NO3)2    +     HCl    + NO    +   H2O

Then balancing the Nitrogen:

2[Co(NH3)5Cl]Cl2     +     20HNO3     ------>     2Co(NO3)2    +     HCl    + 26NO    +   H2O

Then balancing the Cl:

2[Co(NH3)5Cl]Cl2     +     20HNO3     ------>     2Co(NO3)2    +     6HCl    + 26NO    +   H2O

Then balancing the waterlast to account for the H and O:

2[Co(NH3)5Cl]Cl2     +     20HNO3     ------>     2Co(NO3)2    +     6HCl    + 26NO    +   22H2O

Divde thru by 2  and reduces to lowest terms:

[Co(NH3)5Cl]Cl2     +     10HNO3     ------>     Co(NO3)2    +     3HCl    + 13NO    +   11H2O

Old fashioned but it always works....HAPPY NEW YEAR!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

THANKS FOR THE PRACTICE.....ALWAYS A GOOD THING!!!!!!!!!!!!!!!!!!!!!!!!!!!!11