I didn't make the link work but I remember the problem. Balance the following. BY the way, HgCl4^2- is not a solid. It is an complex ion and should have been written as HgCl4^2-(aq)
Hg(ℓ) + NO3^ (–1) (aq) + Cl^(–1) (aq) → HgCl4^(–2) (aq) + NO2(g)
Here is a very good link that shows you how to do any of these so read through and I'll show you how it works on these two.
https://www.chemteam.info/Redox/Balance-HalfReactions-Acid.html
1. Separate into the two half reactions. I'm sure you have that in my earlier post. Balance these one at a time with the following rules. Here is the first one. NO3^- ==> NO2
2. Determine the oxidation state of each element. I assume you know how to do that. If not google CHEMTEAM and they have a very good site that shows you how to do this. For me I know N is the element that changes so we won't worry about oxygen. N in NO3^- is 5+ and in NO2 it is 4+ so you add the proper number of electrons to make the oxidation change balance.
NO3^- + 1e ==> NO2; that is 5+ + 1- = 4+
3. Now count up the charge on the left. I see 2- on the left and zero on the right. Add H^+ (if acid solution; add OH^- if basic solution) to the appropriate side to make the charge balance like this.
NO3^- + 1e + 2H^+ ==> NO2. That makes the charge on the left zero and the charge on the right zero.
4. Now add H2O to the appropriate site to balance the H atoms like this.
NO3^- + 1e + 2H^+ ==> NO2 + H2O
5. ALWAYS check that all balances. I see 1 N on both sides. I see 3 O on the left and 3 on the right. I see two H on the left and right. AND we know the oxidation state balances because 5+ + 1- = 4+. I see a charge on the left of zero and a charge of zero. Everything OK.
6. Next half cell. Hg + 4Cl^- --> HgCl4^2-.
7. Oxidation state Hg on left is 0. On right Hg is 2+. So we must add 2 electrons to the right to balance the change in oxidation state. That is
Hg + 4Cl^- ==> HgCl4^2- + 2e. That make 0 charge on left and 2+ + 2- = 0 on the right. So change in oxidation state balances.
8. Now we count charge. On the left we have 4- and on the right we have 4- so we don't need to add any H^+. Check everything as before which I won't do here. Those are the two BALANCED half equations. Now we put those together. Here they are.
9. NO3^- + 1e + 2H^+ ==> NO2 + H2O
Hg + 4Cl^- ==> HgCl4^2- + 2e
10. Redox equations must balance with the change in the electrons; i.e., you must have the same number gained by one half cell and lost by the other one. You see the NO3^- has 1e added and the Hg one has 2e lost. So we multiply the NO3^- equation by 2 to make 2e gained and multiply the Hg equation by 1 to get 2e lost.
11.2*(NO3^- + 1e + 2H^+ ==> NO2 + H2O)
1*(Hg + 4Cl^- ==> HgCl4^2- + 2e)
12. 2NO3^- +2e + 4H^+ ==> 2NO2 + 2H2O
Hg + 4Cl^- ==> HgCl4^2- + 2e
-----------------------------------------------------------
13. Add the two balanced equations in step 12 to get this.
2NO3^- + 2e + 4H^+ + Hg + 4Cl^- ==> 2NO2 + 2H2O + HgCl4^2- + 2e
14. 2e are on both side so we cancel those to end with this.
2NO3^- + 4H^+ + Hg + 4Cl^- ==> 2NO2 + 2H2O + HgCl4^2-
15. Now you check everything again. It must be balanced with
a. atoms
b. charge
c. Most don't do this but I usually do check the oxidation state change.
I see 2 N on left and right. I see 6 O on left and right. I see 4 H on left and right. I see 2- charge on left and 2- on right. I see 2N totaling 10+ on left going to 2N on right @ 8+ or a change of 2. For Hg it goes from zero on left to 2+ on right which is a change of 2. Everything balances.
Hope this helps. Good luck. I can give you a link to the chem team post if you can't find it. They have a complete set of discussions about almost anything. It's a very good source.
can you show me how to do please for this link below?
www.jiskha.com/questions/1801585/this-is-a-repost-Balance-the-following-ionic-equation-for-acidic-conditions-Identify
2 answers
Remember that oxidation is the loss of electrons. Reduction is the gain of electrons. You can see from the two half cells which one loses and which gains.
2 answers