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Balancing Cu+hno3-cu(no3)2+h2o+no2

Balancing Cu+hno3-cu(no3)2+h2o+no2


1 Answers

Harishwar IIT Roorkee
askIITians Faculty 50 Points
6 years ago
It's not easy unless you understand the concept of oxidation states (these are the apparent valencies of the atoms within the species)

On the left hand side of the equation the copper is in the oxidation state zero (the element) and on the right hand side it is in the oxidation state +2. This means that it has to lose two electrons:

Cu --> Cu2+ + 2e

Meanwhile, the nitrogen atom on the left hand side is in the oxidation state +5. On the right hand side (in the species NO) it is in the oxidation state +2. Hence it must have gained three electrons. To remove the oxygen atoms you need hydrogen ions to make water.

NO3- + 4H+ + 3e --> NO + 2H2O

Now you have to combine the two equations by equalising the electrons. Multiply the first equation by 3 and the second by 2

3Cu --> 3Cu2+ + 6e
2NO3- + 8H+ + 6e --> 2NO + 4H2O
3Cu + 2NO3- + 8H+ --> 3Cu2+ + 2NO + 4H2O

Now we have to enter the spectator ions. You can combine two of the hydrogen atoms with the two nitrate ions to make nitric acid, but you need another six nitrate ions to combine with the six remaining hydrogen ions. The six nitrate ions are used to join up with the three copper(II) ions:

3Cu + 8HNO3 --> 3Cu(NO3)2 + 2NO + 4H2O

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