P(s)+ H+ + NO3- ---> H2PO4 + NO + H2O
I am trying to write the two half rxns and I can get one, but not the other. I have:
red: NO3- +4H+ +3e- ---> NO + 2H2O
Ox: P(s) + ? ----> H2PO4- + x e-
thank you!
Are you stuck on x e^-? If so that is 5e on the right BECAUSE P changes from zero to +5 so 5e must have been added to do that. Then add H^+ and water for the rest of it balance.
To write the second half-reaction for the given reaction, let's identify the changes in oxidation states for each element involved.
In the given equation, we have P(s) on the left and H2PO4 on the right. The oxidation state of P changes from 0 to +5, while the oxidation state of H changes from +1 to +1 (no change). Therefore, the oxidation state of P increases, indicating oxidation.
Now let's balance the number of electrons in the half-reaction. Since the oxidation state of P increases by 5, we need to add 5 electrons (e-) to the left side of the equation to balance the charge.
The half-reaction for the oxidation of P(s) can be written as follows:
Ox: P(s) + 5e- ---> 5H2PO4-
Here, we added 5 electrons to the left side to balance the increase in oxidation state of P, and the resulting species is 5H2PO4-.
So, the balanced half-reactions for the given redox reaction are:
Reduction half-reaction: NO3- + 4H+ + 3e- ---> NO + 2H2O
Oxidation half-reaction: P(s) + 5e- ---> 5H2PO4-