a) Use the standard reduction potentials table to balance the following redox equation:
H2O2 + I + H = H2O + I2
b) Balance the following redox equation by the oxidation number method:
Ag + HNO3 = AgNO3 + NO + H2O
I can't help you without doing all of the work. You need to learn how to do this yourself. What, specifically, do you not understand?
i don't understand how i'm supposed to use the reduction potentials table to blance it
To be completely honest, I don't understand that either since I've never been asked to use reduction potentials to balance a redox equation. However, I think it is done this way.
Look up the standard reduction potential of H2O2 ==> H2O and I see the equation as follows (see +1.77 volts in the table):
H2O2 + 2H^+ + 2e ==> 2H2O
Then look up the I2 ==> I^- couple (see +0.535 volts) and I see the following:
I2 + 2e ==> 2I^- but we want the reverse of that which is 2I^- ==>I2 + 2e, then we add the two together.
H2O2 + 2H^+ + 2e ==> 2H2O
2I^- ==> I2 + 2e
-----------------------------
H2O2 + 2H^+ + 2I^- +2e ==> 2H2O + I2 + 2e
Check that I didn't make a typo in getting all of this down. Apparently one positive for this method is that a look at the standard reduction tables gives you the half reaction ALREADY balanced, then you balance the electrons (they are already ok in this example) and add them.
For the oxidation number method, here is a very good site. Post with specific questions if you still have questions and I'll be happy to help you through them.
http://www.chemteam.info/Redox/Redox.html
wow thank you very much, that site is very helpful and you've been very helpful! Thank you!
how are we supposed to add them together when there is no I in the first equation?
You are adding the two halves. You add the H2O2 + 2H^+ + 2e ==> 2H2O which is the first half, then you place the second half below it (as I've done in my previous post) which is
2I^- ==> I2 + 2e and add those two halves together to obtain the final of
H2O2 + 2H^+ + 2I^- ==> 2H2O + I2
a) To balance the given redox equation using the standard reduction potentials table, we need to follow these steps:
1. Assign oxidation numbers to the relevant elements in the equation:
H2O2 + I + H = H2O + I2
Here, the oxidation number of O in H2O2 is -1, I is 0, H is +1, O in H2O is -2, and I in I2 is 0.
2. Identify the species being reduced and oxidized:
In this equation, I is being oxidized (from 0 to +2) and H2O2 is being reduced (from -1 to -2).
3. Look up the standard reduction potentials for the species involved:
By referring to the standard reduction potentials table, we find that the reduction potential for I2 reduction to 2I- is +0.54 V, and the reduction potential for H2O2 reduction to 2OH- is +1.77 V.
4. Write the two half-reactions for oxidation and reduction:
Oxidation half-reaction: I → I2 (multiply by 2 to balance the iodine atoms)
Reduction half-reaction: H2O2 → 2OH- (multiply by 2 to balance the hydrogen and oxygen atoms)
The balanced half-reactions are:
Oxidation: 2I → I2
Reduction: H2O2 + 2H+ + 2e- → 2H2O
5. Balance the number of electrons transferred:
Multiply the reduction half-reaction by 2 to balance the number of electrons. Now the number of electrons is equal in both half-reactions.
The balanced half-reactions are:
Oxidation: 2I → I2
Reduction: 2H2O2 + 4H+ + 4e- → 4H2O
6. Finally, combine the two half-reactions:
Multiply the oxidation half-reaction by 2 and combine them with the reduction half-reaction.
The balanced equation is:
2H2O2 + 4H+ + 4I → 4H2O + 2I2
b) To balance the given redox equation using the oxidation number method, we need to follow these steps:
1. Assign oxidation numbers to the relevant elements in the equation:
Ag + HNO3 = AgNO3 + NO + H2O
Here, the oxidation number of Ag is 0, N in HNO3 is +5, O in HNO3 is -2, Ag in AgNO3 is +1, N in NO is +2, O in NO is -2, and H in H2O is +1.
2. Identify the species being reduced and oxidized:
In this equation, Ag is being oxidized (from 0 to +1), and N in HNO3 is being reduced (from +5 to +2).
3. Split the equation into two half-reactions:
Oxidation half-reaction: Ag → Ag+
Reduction half-reaction: N in HNO3 → N in NO
4. Balance the number of atoms other than hydrogen and oxygen in each half-reaction:
Since there is only one atom of Ag in both reactant and product, the oxidation half-reaction is already balanced.
For the reduction half-reaction, we need to balance the number of nitrogen atoms by multiplying the whole half-reaction by 2.
The balanced half-reactions are:
Oxidation: Ag → Ag+
Reduction: 2HNO3 + 4H+ + 4e- → 2NO + 2H2O
5. Balance the number of electrons transferred:
To balance the number of electrons, we multiply the oxidation half-reaction by 4, so 4 electrons are transferred in both reactions.
The balanced half-reactions are:
Oxidation: 4Ag → 4Ag+
Reduction: 8HNO3 + 16H+ + 16e- → 8NO + 8H2O
6. Finally, combine the two half-reactions:
Multiply the oxidation half-reaction by 2 and combine them with the reduction half-reaction.
The balanced equation is:
8Ag + 8HNO3 + 16H+ → 8AgNO3 + 8NO + 8H2O