Question

Cu(s) + NO3- (aq) ---> Cu2+ (aq) + NO
Balance the following oxidation-reduction reaction that occurs in acidic solution using the half-reaction method.

Answer 1

@Sheng can u plz help me with this

Answer 2

1/2 reactions if i remember correctly involves electrons
so...
1. Cu --> Cu2+ + 2e-
2. 2NO3- --> 2NO3 + e-

Answer 3

why did u put 2 in front of NO3-

Answer 4

because NO3- is only 1 negative, to balance with Cu2+ u need 2 of it

Answer 5

i am supposed to do it that way in the follwoing link

http://answers.yahoo.com/question/index?qid=20090927120252AAp3iP6

but i still do not understand what they did

Answer 6

@Sheng

Answer 7

The half-equation to balance goes:

\(NO_3^− +H^++e^- \rightarrow NO+H_2O\)

- Nitrogen is already balanced.
- What you should balance next is oxygen, by adjusting the stoichiometric number for \(H_2O\)
- Then you have to balance hydrogen, by adjusting the stoichiometric number for \(H^+\)
- Finally, balance the charges, by adjusting the number of electrons \(e^-\)

Answer 8

when u balance the oxygen does not it become like the following

\[NO _{3}^{-} + 6H ^{+} + 5e ^{-} \rightarrow NO + 3 H _{2} O\]

Answer 9

@Vincent-Lyon.Fr

Answer 10

Cu(s) + NO3^- (aq) ---> Cu^2+ (aq) + NO

in basic medium:

oxidation: Cu -> Cu^2+ + 2e (1)
reduction: 2H2O+ NO3^- + 3e -> NO + 4OH- (2)
--------------------------------------
multiply 1 by 3
multiply 2 by 2
3Cu -> 3Cu^2+ + 6e
4H2O+ 2NO3^- + 6e -> 2NO + 8OH-
-----------------------------------
cancel stuff out:

3Cu+4H2O+ 2NO3^- -> 3Cu^2+ 2NO + 8OH-

Answer 11

in acidic medium:

oxidation: Cu -> Cu^+2 + 2e (1)
reduction: 4H3O+ + NO3^- + 3e -> NO + 6H2O (2)

Balance electron transfer:
multiply (1) x 3
multiply (2) x 2

cancel stuff out:

3Cu + 8H3O+ + 2NO3^- -> 2NO + 12H2O + 3Cu^+2

Answer 12

@Nali
To answer your question about how to balance the oxygen element:
careful, there is already one oxygen present on the right side of the reaction. So you need only 2 moles of H2O to have it balanced.

The way you balanced H and charges is correct, but impacted by your mistake about oxygen.

Try again, but I am sure you are now able to manage this kind of half-equations :)

Answer 13

so does it become the following

\[NO _{3}^{-} + 4H ^{+} + 3e ^{-} \rightarrow NO +2H _{2} O\]

Answer 14

@Vincent-Lyon.Fr

Answer 15

Perfectly correct! :))

Answer 16

ok thanku

Answer 17

@Nali using H+ is incorrect as it doesn't exist in nature, the proper form is the hydronium ion H3O+