So we are doing a lab to show how, in principle, pure metal is pulled out of ore from the ground. This is where you take Cu(s) and react it with HNO3(aq) and then a bunch more reactions to get Cu(s) back. All I am looking for is to make sure that I am doing the 'half-reaction' method correctly to get a balanced net ionic equation. The information given is this: Oxidation: Cu(s) -> Cu2+(aq) + 2e- Reduction: NO3-(aq) + e- -> NO2 (g) It also notes that this is in an acidic environment so there are excess H+ in solution. so to balance the mass in the reduction side of things I added H2O (since you are starting with HNO3 there will be 2H+ for the water molecule anyway, but they can come from anywhere) so now it looks like: Oxidation: Cu(s) -> Cu2+(aq) + 2e- Reduction: NO3-(aq) + e- -> NO2 (g)+ H2O(l) now the reduction needs to be multiplied by 2 so the charges cancel out: Oxidation: Cu(s) -> Cu2+(aq) + 2e- Reduction: 2NO3-(aq) + 2e- -> 2NO2 (g)+ 2H2O(l) Now the 2 electrons cancel and you add the equations to get: Cu(s) + 2NO3-(aq) -> 2NO2 (g) + 2H2O(l) So is this my net ionic equation? The product is Cu(NO3)2 but it is water soluble so it would be aqueous and therefore not included?
So we are doing a lab to show how, in principle, pure metal is pulled out of ore from the ground. This is where you take Cu(s) and react it with HNO3(aq) and then a bunch more reactions to get Cu(s) back.
All I am looking for is to make sure that I am doing the 'half-reaction' method correctly to get a balanced net ionic equation. The information given is this:
Oxidation: Cu(s) -> Cu2+(aq) + 2e-
Reduction: NO3-(aq) + e- -> NO2 (g)
It also notes that this is in an acidic environment so there are excess H+ in solution. so to balance the mass in the reduction side of things I added H2O (since you are starting with HNO3 there will be 2H+ for the water molecule anyway, but they can come from anywhere) so now it looks like:
Oxidation: Cu(s) -> Cu2+(aq) + 2e-
Reduction: NO3-(aq) + e- -> NO2 (g)+ H2O(l)
now the reduction needs to be multiplied by 2 so the charges cancel out:
Oxidation: Cu(s) -> Cu2+(aq) + 2e-
Reduction: 2NO3-(aq) + 2e- -> 2NO2 (g)+ 2H2O(l)
Now the 2 electrons cancel and you add the equations to get:
Cu(s) + 2NO3-(aq) -> 2NO2 (g) + 2H2O(l)
So is this my net ionic equation? The product is Cu(NO3)2 but it is water soluble so it would be aqueous and therefore not included?
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