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
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.
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.
Sure! I can guide you on how to balance the given ionic equation for acidic conditions. However, I'm unable to visit the link you provided as I don't have web browsing capabilities.
To balance an ionic equation for acidic conditions, you can follow these steps:
1. Identify the unbalanced equation: In this case, the unbalanced equation should be provided in the link you shared.
2. Separate the equation into two half-reactions: Split the equation into two half-reactions, one for the oxidation half-reaction and another for the reduction half-reaction. It helps you keep track of the electrons being transferred.
3. Balance atoms other than hydrogen and oxygen: Begin by balancing the atoms other than hydrogen and oxygen in each half-reaction. This is done by adding appropriate coefficients to balance the elements on both sides of the equation.
4. Balance oxygen by adding water (H2O): Add water molecules to the side of the equation that is deficient in oxygen atoms until the oxygen is balanced.
5. Balance hydrogen by adding H+ ions: Add hydrogen ions (H+) to the side of the equation that is deficient in hydrogen atoms until the hydrogen is balanced.
6. Balance the charges by adding electrons (e-): Add electrons to the side of the equation that has a higher positive charge to balance the charges of the half-reactions.
7. Multiply the half-reactions: Multiply each half-reaction by the necessary coefficients to make the number of electrons transferred in each half-reaction equal.
8. Combine the half-reactions: Add the balanced half-reactions together and cancel out the electrons that appear on both sides of the equation.
9. Check the balanced equation: Verify that the number of atoms is equal on both sides, and the charges are balanced.
Remember, for me to provide a more specific answer, you need to provide the unbalanced equation from the link.