Can you show me how to balance this equation using 1/2 reactions?
Zn + NO3- ----> NH3 + Zn(OH4)-2
Zn + OH^- ==> Zn(OH)4^- + 2e
6H2O + NO3^- + 8e ==>NH3 + 9OH^-
Here is a site that you can access for rules for assigning oxidation states up to an including how to balance redox equations (and half reactions).
http://www.chemteam.info/Redox/Redox.html
Sure! Balancing equations using half-reactions involves breaking down the overall reaction into two half-reactions, one for the reduction and one for the oxidation.
Let's start by identifying the oxidation state of each element in the reactants and products.
In Zn, the oxidation state is 0 (since it is an element).
In NO3-, the oxidation state of N is +5 (since it is in the nitrate ion). The oxidation state of O is -2, so the overall oxidation state of NO3- is -1.
In NH3, the oxidation state of N is -3 (since it is in ammonia).
In Zn(OH)4^-2, the oxidation state of Zn is +2 (since it is in Zn(OH)4^-2). The oxidation state of O is -2, so the overall oxidation state of OH^- is -1.
Now, let's write out the two half-reactions:
Oxidation half-reaction (loss of electrons):
Zn ---> Zn(OH)4^-2
Reduction half-reaction (gain of electrons):
NO3- ---> NH3
Next, we need to balance each half-reaction individually.
For the oxidation half-reaction:
We can see that there is one Zn atom on the reactant side, and one Zn atom on the product side. Therefore, the number of Zns is already balanced.
For the reduction half-reaction:
We can see that there is one N atom on the reactant side and one N atom on the product side. Therefore, the number of Ns is already balanced.
Now, let's balance the charges in each half-reaction.
In the oxidation half-reaction:
The overall charge on the product side is -2 (from Zn(OH)4^-2), while the charge on the reactant side is 0 (from Zn). To balance the charges, we need to add 2 extra electrons (e^-) on the reactant side.
In the reduction half-reaction:
The overall charge on the reactant side is -1 (from NO3^-), while the charge on the product side is 0 (from NH3). To balance the charges, we need to add 1 extra electron (e^-) on the product side.
Now, let's add the two half-reactions together:
2 e^- + Zn ---> Zn(OH)4^-2
NO3^- + e^- ---> NH3
To balance the number of electrons, we can multiply the oxidation half-reaction by 1, and multiply the reduction half-reaction by 2:
2 e^- + Zn ---> Zn(OH)4^-2
2 NO3^- + 2 e^- ---> 2 NH3
Now, let's check if the charges and elements on both sides of the equation are balanced:
On the reactant side:
Charge: -2 + 2(-1) = -4
Zn: 1
NO3-: 2
On the product side:
Charge: -2(-2) = +4
Zn(OH)4^-2: 1
NH3: 2
Since the charges and elements are now balanced on both sides, the equation is balanced using half-reactions.
The balanced equation is:
2 NO3- + 2 Zn ---> 1 Zn(OH)4^-2 + 2 NH3