Balance the following equation in acidic conditions. phases are optional. cu+no3- > cu^2+ +no

Balance the following equation in acidic conditions. phases are optional. cu+no3- > cu^2+ +no

Answers

This is an oxidation reduction reaction
oxidation happens when the species gives out electrons.Oxidation state of the element Cu increases from 0 to +2
Cu  ---> Cu²⁺ +2e --1)
Reduction happens when the species gains electrons.
Oxidation state of N reduces from +5 to +2
4H⁺ + NO₃⁻ + 3e ---> NO + 2H₂O --2)
To balance the reactions, number of electrons need to be balanced. 
Multiply 1st reaction by 3
Multiply 2nd reaction by 2
3Cu  ---> 3Cu²⁺ + 6e
8H⁺ + 2NO₃⁻ + 6e ---> 2NO + 4H₂O
add the 2 equations 
3Cu + 8H⁺ + 2NO₃⁻ --> 3Cu²⁺ + 2NO + 4H₂O
add 6NO₃⁻ ions to each side 
3Cu + 8HNO₃ --> 3Cu(NO₃)₂ + 2NO + 4H₂O
the balanced redox reaction equation is as follows;

3Cu(s) + 8HNO₃(aq) --> 3Cu(NO₃)₂ (aq)+ 2NO(g) + 4H₂O(l)

To balance the the chemical reaction, the number of moles per element is balance is both side of the reaction and also the charge in both sides of the reation. to balnce the reaction:

S2O3 2- + Cu 2+ ---> S4O6 2- + Cu+

2S2O3 2- + Cu 2+ ---> S4O6 2- + Cu+ + e

Cr^2+ + H2MoO4 + H+ → Cr^3+ + Mo + H2O 
Cr goes from +2 to +3 or loses one e- 
Mo is +6 in H2MoO4 so it goes to 0 gaining 6 e- 
so we must multiply Cr by 6 to balance e- 

6Cr^2+ + H2MoO4 + H+ → 6Cr^3+ + Mo + H2O 
now we balance the spectators 
take O first - 4 on left then, change the right side 

6Cr^2+ + H2MoO4 + H+ → 6Cr^3+ + Mo + 4H2O 
now we have 8 H's on the right 
we don't want to touch the H2MoO4 
so we make H+ up to 6 for a total of 8 on the left 

6Cr^2+ + H2MoO4 + 6H+ → 6Cr^3+ + Mo + 4H2O 
Then, It's balanced

I think the reaction involved here is a redox reaction in acidic conditions. To balance this type of reaction, we do as follows:

Balance the O atoms in each side by adding H2O on either side.
Balance H atoms in each side by adding H+.
Balance the charges by adding electrons.

Hope this helps.

If your equation is Cu+NO^- 3-->Cu^2+NO, then the answer is 
2 Cu + 1 NO3{-}  → 1 Cu^{2+} + 3 NO 

To check if it is balance, this is the solution:
2- Cu- 2
3- N -3
3- O -3 

To balance the equation, Hydrogen ion should be added to the left of the equation and water to the right of the equation as follows:

H^{+} +   Cr^{2+}  + H_{2} Mo O_{4} --- textgreater    Cr^{3+} + Mo +  H_{2} O

Balancing the equation

6 H^{+} + 6  Cr^{2+} +  H_{2} Mo  O_{4} ---- textgreater   6  Cr^{3+} + Mo +  4 H_{2} O

It can be noted that: ions are balanced at 18+ on left hand side and 18+ on right and side just like all the other chemical elements.

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