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Topic: Help balancing a Redox reaction in acidic conditions  (Read 715 times)

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Offline margit

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Help balancing a Redox reaction in acidic conditions
« on: May 05, 2024, 04:54:18 PM »
so I've been banging my head against this problem for a while now and I'm pretty sure I'm doing something wrong. this is the unbalanced reaction:
(may be important the mention: this is happening in acidic conditions.)

CuS(s) + HNO3   :rarrow: Cu(NO3)2 + H2O + NO(g) + S(s)

the first thing that I got stuck on was finding the oxidation states for the atoms in Cu(NO3)2. after doing some googling I found out that NO3 should have a minus sign on it((NO3)-) which made some sense after drawing the lewis stracture for it.
after doing that I noticed that S and N changed their oxidation number(S was -2 and became 0, N was +5 and became +2).

However when writing the half reactions for oxidation and reduction, I got stuck. this was my process:

oxidation half reaction(unbalanced):

CuS(s)+ HNO3  :rarrow: S(s)+ Cu((NO3)-)2

Cu and S quantities were already balanced, and the same goes for oxygen quantities. which only leaves balancing Hydrogen and nitrogen quantities like so:

CuS(s)+ 2HNO3  :rarrow: S(s)+ Cu((NO3)-)2 + 2H+

the weird thing is that after doing that the charges are balanced too, which makes no sense because that should mean there was no transfer of electrons and hence no oxidation.

as for reduction, I tried balancing this:
HNO3  :rarrow: NO(g)

however, the weird thing is that H2O is not present in either, and I have no idea how to include it.

summery of all of my questions:
1. Is the lack of a minus sign on (NO3)2 really a mistake or am I missing something? If it is intentional, how do I find the oxidation numbers for the atoms in Cu(NO3)2?
2. is my oxidation half-reaction correct? if not, what am I missing?
3. how do I involve H2O in my half reactions?

I honestly tried to solve this for a long time and I'm not asking people to do it for me, I'm just really, really, bad at chemistry and I'm kinda stuck.
« Last Edit: May 05, 2024, 05:10:20 PM by margit »

Offline Hunter2

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Re: Help balancing a Redox reaction in acidic conditions
« Reply #1 on: May 05, 2024, 05:33:52 PM »
Different approach

Use redox pairs and develop reduction and oxidation separately,  then adjust the charges by adding electrons. Build the least common multiple of the electrons. Add both equations
Pairs are
S2- / S and NO3- / NO

In acidic environment remove oxygen bei adding H+ and built water.
Copper is spectator ion here.

Can you go further.


Offline Borek

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Re: Help balancing a Redox reaction in acidic conditions
« Reply #2 on: May 06, 2024, 02:19:21 AM »
Cu((NO3)-)2

No such animal.

The compound in question is copper nitrate. Yes, nitrate in compounds is NO3-. But you are ignoring the fact that copper is Cu2+ (so, to follow your notation, (Cu2+)((NO3)-)2).

Quote
CuS(s)+ 2HNO3  :rarrow: S(s)+ Cu((NO3)-)2 + 2H+

Adding charges only to some ions in the formulas is a sure way of doing it all wrong. Much better to write the half reaction as net ionic, that will yield:

CuS(s) :rarrow: Cu2+ + S(s)

Which definitely needs electrons to be balanced.

Quote
HNO3 :rarrow: NO(g)

Reasonably good starting point, but apparently can't be balanced with just electrons. That's the moment to use H+ and H2O to make balancing possible at all.

Actually easier starting point is

NO3- :rarrow: NO(g)

as you don't have to combine hydrogen from different sources. Just throw in some H+ and H2O till atoms are balanced.
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