Writing chemical equations help please?
I'm currently going over past papers and one of the Qs is
1) Write an ionic half-equation for the conversion of NH3 to HNO3 in the presence of H2O, given that the number of electrons involved is the same as the oxidation state change.
I just don't understand how to go about actually working out the answers, like in q1, how would I know how many e- and H+ to add
would realllllly appreciate some step to step working out for the q please
Thank you !
just attached the ans
1) Write an ionic half-equation for the conversion of NH3 to HNO3 in the presence of H2O, given that the number of electrons involved is the same as the oxidation state change.
I just don't understand how to go about actually working out the answers, like in q1, how would I know how many e- and H+ to add
would realllllly appreciate some step to step working out for the q please
Thank you !
just attached the ans
(edited 11 years ago)
You have to add things to the half-equation in order to make it balance completely.
All you are allowed to add are:
electrons, water, hydrogen ions(unless the reaction is being done under alkaline conditions - in which case, you can add hydroxide ions instead).
First balance the number of species on both sides of the equations. Then determine the change in oxidation state of nitrogen. If it's increasing, then there's loss of electrons, so you should add that much number of electrons on right hand side of the equation. If the oxidation state decreases, then there is gain of electrons, so you should write that much electrons at left hand side of the equation.
All you are allowed to add are:
electrons, water, hydrogen ions(unless the reaction is being done under alkaline conditions - in which case, you can add hydroxide ions instead).
First balance the number of species on both sides of the equations. Then determine the change in oxidation state of nitrogen. If it's increasing, then there's loss of electrons, so you should add that much number of electrons on right hand side of the equation. If the oxidation state decreases, then there is gain of electrons, so you should write that much electrons at left hand side of the equation.
(edited 11 years ago)
Original post by -James-
I'm currently going over past papers and one of the Qs is
1) Write an ionic half-equation for the conversion of NH3 to HNO3 in the presence of H2O, given that the number of electrons involved is the same as the oxidation state change.
I just don't understand how to go about actually working out the answers, like in q1, how would I know how many e- and H+ to add
would realllllly appreciate some step to step working out for the q please
Thank you !
just attached the ans
1) Write an ionic half-equation for the conversion of NH3 to HNO3 in the presence of H2O, given that the number of electrons involved is the same as the oxidation state change.
I just don't understand how to go about actually working out the answers, like in q1, how would I know how many e- and H+ to add
would realllllly appreciate some step to step working out for the q please
Thank you !
just attached the ans
It's a typical question asking you how to write an ionic equation.
1) Start off with writing what you know:
NH3 + H2O ---> HNO3
2) Work out the oxidation states of the element that appears to be changing in oxidation state (This is the nitrogen atom).
The N in NH3 has an oxidation state of -3.
The N in HNO3 has an oxidation state of +5.
3) Add electrons to the side which allows the oxidation states to be balanced on both sides.
The left hand side has -3, the right hand side has +5, so if you add 8 electrons to the right hand side, then both sides will have -3.
NH3 + H2O ---> HNO3 + 8e-
4) Add hydrogen ions so the charges on both sides are the same.
The right hand side has a total charge of -8, the left hand side has a total charge of 0, so the right hand side needs 8H+ to make the charges on the right hand side equal to 0.
NH3 + H2O ---> HNO3 + 8e- + 8H+
5) Add water molecules to balance the number of oxygen atoms.
The right hand side has 3 oxygen atoms, the left hand side only has 1, so times the number of water molecules on the left hand side by three to get three water molecules.
NH3 + 3H2O ---> HNO3 + 8e- + 8H+
Hopefully this clears everything up =]
By the way, the equation on the bottom of the mark scheme is also correct, HNO3 is an acid and it dissociates into NO3- and H+, so the H+ ion can be added to the 8H+ to make 9H+.
Original post by TiTo20
It's a typical question asking you how to write an ionic equation.
1) Start off with writing what you know:
NH3 + H2O ---> HNO3
2) Work out the oxidation states of the element that appears to be changing in oxidation state (This is the nitrogen atom).
The N in NH3 has an oxidation state of -3.
The N in HNO3 has an oxidation state of +5.
3) Add electrons to the side which allows the oxidation states to be balanced on both sides.
The left hand side has -3, the right hand side has +5, so if you add 8 electrons to the right hand side, then both sides will have -3.
NH3 + H2O ---> HNO3 + 8e-
4) Add hydrogen ions so the charges on both sides are the same.
The right hand side has a total charge of -8, the left hand side has a total charge of 0, so the right hand side needs 8H+ to make the charges on the right hand side equal to 0.
NH3 + H2O ---> HNO3 + 8e- + 8H+
5) Add water molecules to balance the number of oxygen atoms.
The right hand side has 3 oxygen atoms, the left hand side only has 1, so times the number of water molecules on the left hand side by three to get three water molecules.
NH3 + 3H2O ---> HNO3 + 8e- + 8H+
Hopefully this clears everything up =]
By the way, the equation on the bottom of the mark scheme is also correct, HNO3 is an acid and it dissociates into NO3- and H+, so the H+ ion can be added to the 8H+ to make 9H+.
1) Start off with writing what you know:
NH3 + H2O ---> HNO3
2) Work out the oxidation states of the element that appears to be changing in oxidation state (This is the nitrogen atom).
The N in NH3 has an oxidation state of -3.
The N in HNO3 has an oxidation state of +5.
3) Add electrons to the side which allows the oxidation states to be balanced on both sides.
The left hand side has -3, the right hand side has +5, so if you add 8 electrons to the right hand side, then both sides will have -3.
NH3 + H2O ---> HNO3 + 8e-
4) Add hydrogen ions so the charges on both sides are the same.
The right hand side has a total charge of -8, the left hand side has a total charge of 0, so the right hand side needs 8H+ to make the charges on the right hand side equal to 0.
NH3 + H2O ---> HNO3 + 8e- + 8H+
5) Add water molecules to balance the number of oxygen atoms.
The right hand side has 3 oxygen atoms, the left hand side only has 1, so times the number of water molecules on the left hand side by three to get three water molecules.
NH3 + 3H2O ---> HNO3 + 8e- + 8H+
Hopefully this clears everything up =]
By the way, the equation on the bottom of the mark scheme is also correct, HNO3 is an acid and it dissociates into NO3- and H+, so the H+ ion can be added to the 8H+ to make 9H+.
thankkk you so much!
So you should start of balancing the charge, then add H+ then balance the equation?
Original post by -James-
thankkk you so much!
So you should start of balancing the charge, then add H+ then balance the equation?
So you should start of balancing the charge, then add H+ then balance the equation?
Not the charges, the oxidation states =P
Start off ALWAYS by identifying the atom that changes oxidation state, then add electrons to balance the oxidation states, then balance the charges, then the oxygen atoms.
So basically, you add electrons, then hydrogen ions, then water molecules.
Hope this helped! =]
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