Chem help involving REDOX..?

Question

I dont really know how to do this and my friend asked me to explain it to her..any help please? thankss
So far i did what i know, but i am not even sure if that is right, if you could explain it that would be helpful too :), if not its ok

For the equation Ag + NO3 - ==> Ag + + NO
(Note: This reaction takes place in an acidic solution.)

Step 1: What substance is reduced?
NO

Step 2: What substance is oxidized?
AG

Step 3: What is the half reaction for oxidation?

Step 4: What is the half reaction for reduction?

Step 5: What is the net balanced equation?

Step 6: What is the reduced equation?

Answer 1

The first two steps are correct; electrons are flowing from silver (0 --> 1+) to the nitrogen (5+ --> 2+). Elemental substances have an oxidation state of 0 (Ag, Fe, O2, ect.) and oxidation state of oxygen is almost always 2- beside in peroxides which would be 1-. The oxidation state of the nitrogen in the nitrate was found by finding a combination of oxidation states equates to a total of 1-. Since oxygen is 2- in this case, you can solve for nitrogen by the equation X + 3(2-)= -1 thus X(nitrogen) = 5+. Same thing for nitric oxide except it has a net charge of 0. Just it would better to state nitrate was reduce instead of nitric oxide.
Step 3: oxidation reaction
Ag ---> Ag+
Since the conservation of matter is already upheld, one needs to only make sure the conservation of charge is followed and this is done by balancing the more positive side with electrons.
Ag ---> Ag+ + 1e-
Step 4: reduction reaction
NO3- ---> NO
Balance oxygens,
NO3- ---> NO + 2 H2O
Balance the hydrogens
NO3- + 4H+ ---> NO + 2 H2O
Conservation of matter is upheld now to the conservation of charge.
NO3- + 4H+ + 3e- ---> NO + 2 H2O
Step 5: net balance equation
This just involves combining the reduction and oxidation reactions and making sure the electrons cancel. This is done similarly to finding the lowest common denominator in fractions. Since there is 1e- on the product side of the oxidation reaction and 3e- on the reactant side of the reduction reaction, one must multiply the oxidation reaction by 3 to obtain the correct flow of electrons.
3(Ag ---> Ag+ + 1e-) = 3 Ag ---> 3Ag+ + 3e-
Combining the two
3Ag ---> 3Ag+ + 3e-
NO3- + 4H+ + 3e- ---> NO + 2 H2O
----------------------------------------------
3Ag + NO3- + 4H+ ---> 3Ag+ + NO + 2 H2O
Step 6: reduced equation
Since there is no repeated molecule, ion or element, the net balanced equation is the reduced equation. If your professor is really detailed oriented, you may need to change the H+ to H3O+ because radical protons don't float around in solution. Just replace H+ by H3O+ and add equivalent amounts of H2O to the other side.
3Ag + NO3- + 4H3O+ ---> 3Ag+ + NO + 2 H2O + 4 H2O =>
3Ag + NO3- + 4H3O+ ---> 3Ag+ + NO + 6 H2O
And that’s how you do a redox reaction in acidic conditions. This is just like if you did a redox reaction without considering the conditions, but this is not true for basic condition. Just FYI, for basic condition you would add OH- to each side of the balanced equation of every H+
3Ag + NO3- + 4H+ + 4OH- ---> 3Ag+ + NO + 2 H2O + 4OH-
The H+ and OH- from H2O on the reactant side and the balance equation for basic conditions would be:
3Ag + NO3- + 4H2O ---> 3Ag+ + NO + 2 H2O + 4OH- =
3Ag + NO3- + 2H2O ---> 3Ag+ + NO + 4OH-

Hopes this helps.

Answer 2

Steps one and two are correct! Remember that the oxidation state of a solid metal (Ag in this case) is 0. On the product side is Ag+, so Ag has lost 1 electron and that electron was added to NO3.

3. Since Ag lost an e-, then it was oxidized: Ag --> Ag+ + 1e-
Half-reactions look only at what is happening to one aspect of reaction, ie the reduction or oxidation process.

See for a great explanation for the rest.
http://dbhs.wvusd.k12.ca.us/webdocs/Redox/Balance-HalfReactions-Acid.html

Answer 3

A key question my friend is are you effectively making silver I oxide[ {(Ag)2}1+ (O)2-] or silver ii oxide [(Ag) 2+ (O)2-] in this reaction?

I am guessing that you are making silver I oxide because silver II oxide is rare. Therfore I think that....

The NO3 is losing oxygen so it is being reduced to form NO.

The Silver is gaining oxygen so it is being oxidised to form Ag2O or {Ag+, Ag+, O2-}

The half reaction for reduction is
NO3- ---> NO + 2O- (I guess your acid solution provides a balancing +)
The half reaction for oxidation is
2Ag ---> 2Ag+ + 2e-

Add the two half reactions together and then balance it to give you the net balanced equation...
NO3- + 4Ag ----> NO + 4Ag+ + 2O- +H+ (acid)

The reduced equation is where you don't show the spectator ions. It's 4Ag + 2O- ---> 2Ag2O

Answer 4

A 0.5 reaction is both for oxidation or relief. In any 0.5 reaction you should stability the replace in oxidation huge form with electron, stability the fee (that could want to require utilizing H+, OH- and or water) and stability the atoms. common case: Zn ----------> Zn++ atoms balanced yet not fee, upload 2 electrons Zn ---------> Zn++ + 2 e- and this oxidation 0.5 reaction is balanced. maximum such reactions ensue in water so each so often it takes position or is produced contained in the redox reactions and extra importantly, those reactions might want to ensue in acidic or undemanding thoughts, so those ions take area. evaluate the relief of Chromate ion to chromite ion in undemanding answer: CrO4^=(aq) -----------> Cr(OH)4^a million-(aq) replace in oxidation # is +6 to +3 3 e^- + CrO4^=(aq) -----------> Cr(OH)4^a million-(aq) oxidation # fee is balanced, yet desire extra hydrogen on the left, 3 e^- + CrO4^=(aq) + 4 H2O --------> Cr(OH)4^a million-(aq) so now you want extra - ions on the right 3 e^- + CrO4^=(aq) + 4 H2O --------> Cr(OH)4^a million-(aq) + 4 OH^- and now each little thing balances. water change into extra to furnish hydrogen on the left and hydroxide change into extra to stability the H and stability the overall - fee on ions and electrons. wish this allows.

Answer 5

n