Can anyone show me how to balance:
Ce+4 +H3AsO3 ==> Ce+3 + H3AsO4
using the half-reaction method?
I just did iron(II) and MnO4^- for Nessy. That method will do them all.
http://www.jiskha.com/display.cgi?id=1274415665
Chemistry
Certainly! To balance redox reactions using the half-reaction method, you need to follow these steps:
Step 1: Assign oxidation numbers.
Start by assigning oxidation numbers to each element in the equation. In this case, Ce is going from +4 to +3, while H3AsO3 is going from +3 to +5.
Step 2: Separate the equation into two half-reactions.
Separate the redox reaction into two half-reactions, one for the oxidation process and one for the reduction process.
The oxidation half-reaction is the process where Ce is getting oxidized, so it is losing electrons. In this case, Ce+4 is losing 1 electron, becoming Ce+3:
Ce+4 → Ce+3 + e-
The reduction half-reaction is the process where H3AsO3 is getting reduced, so it is gaining electrons. In this case, H3AsO3 gains 2 electrons, becoming H3AsO4:
H3AsO3 + 2e- → H3AsO4
Step 3: Balance the atoms that are not oxygen or hydrogen.
Balance the atoms in the redox half-reactions other than oxygen and hydrogen. In this case, there are no other atoms to balance.
Step 4: Balance the oxygen atoms.
Balance the oxygen atoms by adding water (H2O) molecules to the side of the equation that lacks oxygen.
In the oxidation half-reaction, since there are 4 oxygen atoms on the right side (in Ce+3), add 4 water molecules to the left side:
Ce+4 + 4H2O → Ce+3 + e-
In the reduction half-reaction, there is one oxygen atom on the left side (in H3AsO3) and four oxygen atoms on the right side (in H3AsO4). Hence, add 3 water molecules to the left side:
H3AsO3 + 2e- + 3H2O → H3AsO4
Step 5: Balance the hydrogen atoms.
Balance the hydrogen atoms by adding hydrogen ions (H+) to the side of the equation that lacks hydrogen.
In the oxidation half-reaction, add 12 hydrogen ions (H+) to the right side:
Ce+4 + 4H2O → Ce+3 + e- + 12H+
In the reduction half-reaction, add 6 hydrogen ions (H+) to the right side:
H3AsO3 + 2e- + 3H2O → H3AsO4 + 6H+
Step 6: Balance the charges by adding electrons.
Balance the charges on both sides by adding electrons (e-).
In the oxidation half-reaction, there is already one electron (e-) on the right side:
Ce+4 + 4H2O → Ce+3 + e- + 12H+
In the reduction half-reaction, add 10 electrons (10e-) to the left side:
H3AsO3 + 10e- + 3H2O → H3AsO4 + 6H+
Step 7: Combine the half-reactions.
Multiply the half-reactions by appropriate whole numbers so that the number of electrons gained or lost is the same. In this case, multiplying the oxidation half-reaction by 10 will equalize the electrons:
10Ce+4 + 40H2O → 10Ce+3 + 10e- + 120H+
Combine the two half-reactions together by canceling out the electrons:
10Ce+4 + 40H2O + H3AsO3 → 10Ce+3 + 3H3AsO4 + 6H+
Finally, simplify the equation and balance all the coefficients:
10Ce+4 + H3AsO3 + 40H2O → 10Ce+3 + 3H3AsO4 + 6H+
That's it! The balanced equation is now achieved using the half-reaction method.