Use the 1/2 reaction method to balance the equation below in acidic solution.
Cl2 (g) + I- (aq) = Cl- (aq) + IO3- (aq)
i don't understand how to do this one. help would be appreciated :)
Here is the I^- ==> IO3^-
1. oxidation number of I is -1 on the left; +5 on the right.
2. Add electrons to the appropriate side. That is
I^- ==> IO3^-+ 6e
3. Add up the charges and add H^+ to balanced the charge. I see -1 on the left side and -7 on the right.
I- ==> IO3^- + 6e + 6H^+
(Check this to see that -1 charge on left is same as -1 charge on right.
4. Now add H2O to the appropriate side (USUALLY but not always) the opposite side from the H^+ addition)
I^- + 3H2O ==>IO3^- + 6e + 6H^+
The Cl half is much easier. If you have any questions about the I^-/IO3^- please let me know. Thos steps will get it done every time. By the way, for basic solutions you add up the charge on step 3 and add OH^- to balance the charge. Everything else is the same.
Hi DrBob222, I tried to finish it and as a final answer I got:
1/2 H2O + Cl2 + I- = Cl- +IO3- + 3H+ ?
Could you please tell me if this is correct? :)
Sorry I didn't get back sooner. It is not.
My half cell is correct.
The Cl should be
2e + Cl2 ==> 2Cl^-
Multiply my eqn by 1 and the Cl by 3 to end up with
I^- + 3H2O + 3Cl2 ==> IO3^- + 6H^+ + 6Cl^-
Check it.
1 I on each side
3O on each side
6H on each side
-1 charge on left and right.
atoms balance
charges balance
electrons (6 for each half cell) balance.
Everything OK.
I don't understand why you changed the 6H^+ I had to 3. and why 1/2H2O instead of the 3H2O? I suspect you DID NOT balance the electrons. There were 6 on my half cell and 2 on yours so you needed to multiply your half cell by 3 before adding it to mine. You should know your equation isn't right by checking it.
1 H on left and 3 on right.
1/2 O on left and 4 on right.
Charge on left is -1; on right is +1.
So atoms don't balance and charge doesn't balance. Repost as a NEW post if you have questions because your question will so far down the line that they sometimes get lost. This isn't difficult but it takes a little discipline to work on it.
To balance the given equation using the 1/2 reaction method in an acidic solution, follow these steps:
Step 1: Write the unbalanced equation:
Cl2 (g) + I- (aq) = Cl- (aq) + IO3- (aq)
Step 2: Split the reaction into two half-reactions. One for the oxidation process (loss of electrons) and one for the reduction process (gain of electrons).
Oxidation half-reaction (loss of electrons): Cl2 (g) → Cl- (aq)
Reduction half-reaction (gain of electrons): I- (aq) → IO3- (aq)
Step 3: Balance the atoms other than oxygen and hydrogen in each half-reaction:
Oxidation half-reaction: Cl2 (g) → 2Cl- (aq)
Reduction half-reaction: 3I- (aq) → IO3- (aq)
Step 4: Balance the oxygen atoms by adding water (H2O) molecules to the side that needs more oxygen:
Oxidation half-reaction: Cl2 (g) → 2Cl- (aq) + 2H2O (l) [added 2 water molecules to the right side]
Reduction half-reaction: 3I- (aq) → IO3- (aq)
Step 5: Balance the hydrogen atoms by adding hydrogen ions (H+) to the side that needs more hydrogen:
Oxidation half-reaction: Cl2 (g) + 2H2O (l) → 2Cl- (aq) + 4H+ (aq) [added 4H+ ions to the left side]
Reduction half-reaction: 3I- (aq) → IO3- (aq)
Step 6: Balance the charges by adding electrons (e-) to the side that needs more charge:
Oxidation half-reaction: Cl2 (g) + 2H2O (l) → 2Cl- (aq) + 4H+ (aq) + 4e- [added 4 electrons to the left side]
Reduction half-reaction: 3I- (aq) + 6H+ (aq) → IO3- (aq)
Step 7: Multiply the half-reactions by appropriate factors so that the number of electrons gained equals the number of electrons lost (equalize the electron transferred):
Oxidation half-reaction: 3Cl2 (g) + 6H2O (l) → 6Cl- (aq) + 12H+ (aq) + 12e-
Reduction half-reaction: 2IO3- (aq) + 12H+ (aq) + 12e- → 6I- (aq) + 6H2O (l)
Step 8: Add the balanced half-reactions together and simplify by canceling out the electrons (e-):
3Cl2 (g) + 6H2O (l) + 2IO3- (aq) + 12H+ (aq) → 6Cl- (aq) + 12H+ (aq) + 6I- (aq) + 6H2O (l)
Simplifying the equation gives the balanced equation:
3Cl2 (g) + 2IO3- (aq) + 6H+ (aq) → 6Cl- (aq) + 6I- (aq) + 6H2O (l)
Thus, the balanced equation in acidic solution using the 1/2 reaction method is:
3Cl2 (g) + 2IO3- (aq) + 6H+ (aq) → 6Cl- (aq) + 6I- (aq) + 6H2O (l)