Sciencemadness Discussion Board

Balancing a super hard Redox.

Julie18 - 16-6-2009 at 20:56

This is the hardest redox reaction to balance in the history of time. Try and figure out the coefficients, it's impossible.

Cr2O7(-2) + HNO2 --> Cr(+3) + NO3(-1)

If you could figure it out, I'd really appreciate it because it's driving me crazy.
You can add H+ to the left side and H2O to the right.

Good luck to anyone brave enough to try it. :P

[Edited on 17-6-2009 by Julie18]

merrlin - 16-6-2009 at 21:03

Is it possible that CrO7(-2) should be the dichromate Cr2O7?

UnintentionalChaos - 16-6-2009 at 21:33

Quote: Originally posted by merrlin  
Is it possible that CrO7(-2) should be the dichromate Cr2O7?


Almost guaranteed.

Have you learned the half-reactions method Julie18? Here is a similar example (I'm not doing your homework for you).

MnO2 + NaOCl --> MnO4(-1) + NaCl

half reaction one: MnO2 --> MnO4(-1)

1) balance the main element (manganese). done already.
2) balance oxygen by adding water:

2H2O + MnO2 --> MnO4(-1)

3) balance hydrogen with H+ ions

2H2O + MnO2 --> MnO4(-1) + 4H(+1)

4) balance charge with electrons (e-):

2H2O + MnO2 --> MnO4(-1) + 4H(+1) + 3e-

half reaction two:

NaOCl --> NaCl

1) already done (sodium and chlorine)
2) NaOCl --> NaCl +H2O
3) 2H(+1) + NaOCl --> NaCl +H2O
4) 2e- + 2H(+1) + NaOCl --> NaCl +H2O

find a way to combine the two half reactions so that the electrons cancel out:

2x the first half reaction
3x the second half reaction

combine them and cancel the electrons and any extra waters and hydrogens:


H2O + 2MnO2 + 3NaOCl --> 2MnO4(-1) + 2H(+1) + 3NaCl

does that seem so hard?

Julie18 - 16-6-2009 at 22:36

Thank you so much. My chem teacher never taught our class this method and I wish he had.
(Sorry about the Cr2O7(-2), typo)

A couple things, though:
In my rxn, the second half rxn would be HNO2 --> NO3(-1)
So, I'm confused because
1) There's oxygens on both sides, so I'm not sure if I should still add H2O to compensate for them.
2) What do I do about the hydrogen on the left?


[Edited on 17-6-2009 by Julie18]

12AX7 - 16-6-2009 at 23:37

Hydrogen becomes water. Or since you started with the acid, you might show it as ending with another acid, HNO3. Or you might show the starting material as NO2-, which may or may not be appropriate.

Tim

woelen - 16-6-2009 at 23:54

The second half reaction can be written as

HNO2 + H2O --> NO3(-) + 3H(+) + 2e

Now you have a half reaction and you see that this produces H(+) ion and two electrons. You see what I have done? At the right, I needed another oxygen, so I simply added that by introduciing a water molecule on the left. When that is done, then you still have unbalance for hydrogens and charge. Next step is to balance with H(+) ions such that at both sides the number of hydrogen atoms is the same. The final step is to add electrons such that charge is the same at both sides of the arrow.


For dichromate the half reaction is

Cr2O7(2-) + 14H(+) + 6e --> 2Cr(3+) + 7H2O

With this information and the method of UnintentionalChaos you should be able to solve the problem. Please write down all steps, so that we can check your lines of thought.

sparkgap - 17-6-2009 at 07:16

To further clarify woelen's explanation, remember that you are (at least for most General Chemistry situations) in an aqueous medium, so water will be (most of the time) your source of extra hydrogens and oxygens for your redox reaction. The thing, though, when you do this is that you have to note if the reaction is happening in acidic or basic medium. If, as in your example, the reaction happens in acidic medium, you add hydronium ion and water to the appropriate sides of the equation; if on the other hand, the action happens in basic medium, you use hydroxide instead of hydronium ion.

sparky (~_~)

Julie18 - 17-6-2009 at 10:27

Sorry, I didn't include this part, but all the chemicals had (aq) beside them (except water, of course). So that means in an aqueous medium and water's the extra source of H and O, right?

And I've actually never heard of adding hydroniums or hydroxides, someone taught me to use electrons. But it would end up working out either way, right?

woelen - 17-6-2009 at 10:57

I also used addition of hydrogen ions to get to my half reaction of HNO2 to NO3(-) in acidic aqueous solution.
First add water on one side of the arrow to make the number of oxygens the same on both sides. Then add hydrogen ions to make the number of hydrogen atoms the same on both sides. Finally add electrons to make the charge the same on both sides.

In alkaline solution you can do a similar thing, but now you need to add water molecules to make the number of oxygen atoms the same on both sides. Next, you need to replace part of the water molecules by hydroxide ions to make the number of hydrogen atoms the same on both sides. Finally, again you add electrons.