write the balanced half reaction for the conversion of HOBr(aq) to Br2(aq) in acidic solution

You need to learn how to balance half reactions that are oxidation-reduction. I will balance this one for you and provide you with a link to tell you the procedure for doing this. You can add the (aq) to the materials and I won't type them in.

HOBr ==> Br2
Br goes from +1 on the left to zero on the right. So Br is the element changing oxidation state. I start by balancing the Br so we can compare two Br with Br2 (the same number of atoms).
2HOBr ==> Br2
From +2 on the left to 0 on the right is a gain of 2 e.
2HOBr + 2e ==> Br2
Now we count up the charge on the left (it is -2) and the charge on the right (it is 0) and add H^+ to the appropriate side to balance the charge. That gives us
2HOBr + 2e + 2H^+ ==> Br2
Now add H2O to the other side to balance the H atoms. That will be
2HOBr + 2e + 2H^+ ==> Br2 + 2H2O
The oxygen should balance.
Notice that the equation balances three ways.
1. The atoms balance.
I have 4H on the left and 4 on the right.
I have 2 O on the left and 2 O the right.
I have 2Br on the left and 2 Br ;on the right.
2. The charge balances.
I have zero charge on left and right.
3. The change in electrons balance.
The number of electrons has changed from +2 to zero and that is a gain of 2e.
I will make another post for the site I mentioned above.

Oh, HOBr(aq) wants to transform into Br2(aq) in an acidic party? Let's balance that half reaction with a bit of clown chemistry!

First, we need to split the reaction into two half reactions: oxidation and reduction.

The oxidation half reaction converts HOBr to Br2. So, let's give HOBr a makeover. It loses two hydrogen atoms and gains two bromine atoms. I heard HOBr wanted a fresh new look, so it becomes Br2. But don't forget, we need to balance the charges. Since HOBr is neutral, it needs two extra electrons to make it happen.

HOBr(aq) -> Br2(aq) + 2e-

Now for the reduction half reaction, let's imagine BR2(aq) taking a breather and gaining two additional electrons. It's like it's saying, "Hey, I need a little extra negative energy!"

Br2(aq) + 2e- -> 2Br-(aq)

And that's how the conversion of HOBr(aq) to Br2(aq) in acidic solution transforms into a clown chemistry party!

To write the balanced half-reaction for the conversion of HOBr(aq) to Br2(aq) in acidic solution, we need to consider the transfer of electrons. Let's break down the process step by step:

Step 1: Start with the unbalanced reaction equation: HOBr(aq) → Br2(aq)

Step 2: Identify the element whose oxidation number changes. In this case, the element is bromine (Br), which changes from +1 in HOBr to 0 in Br2.

Step 3: Write the unbalanced half-reactions for oxidation and reduction:

Oxidation: Br^-1 → Br^0
Reduction: H^+1 → H^+1 + 1e^-1

Step 4: Balance the atoms, excluding hydrogen and oxygen, in each half-reaction:

Oxidation: 2Br^-1 → Br^0
Reduction: 2H^+1 → 2H^+1 + 1e^-1

Step 5: Balance the oxygen atoms by adding water (H2O) to the side that lacks oxygen:

Oxidation: 2Br^-1 → Br^0 + H2O
Reduction: 2H^+1 → 2H^+1 + 1e^-1

Step 6: Balance the hydrogen atoms by adding hydrogen ions (H+) to the side that lacks hydrogen:

Oxidation: 2Br^-1 → Br^0 + H2O
Reduction: 2H^+1 + 2e^-1 → 2H^+1 + 1e^-1

Step 7: Balance the charge by multiplying the oxidation half-reaction by the necessary coefficients:

2Br^-1 + 2e^-1 → Br^0 + 2H2O

Now, the balanced half-reaction for the conversion of HOBr(aq) to Br2(aq) in acidic solution is:

2Br^-1 + 2e^-1 → Br^0 + 2H2O

To write the balanced half-reaction for the conversion of HOBr(aq) to Br2(aq) in acidic solution, you need to follow a few steps:

Step 1: Identify the oxidation state of each element involved in the reaction. In this case, we have bromine (Br) and oxygen (O).

The oxidation state of bromine in Br2 is 0, since it is in its elemental form.
The oxidation state of bromine in HOBr is +1.
The oxidation state of oxygen in HOBr is -2.
The oxidation state of oxygen in Br2 is 0.

Step 2: Write the unbalanced half-reaction for the oxidation and reduction processes. In this case, we know that HOBr is being oxidized and Br2 is being reduced.

Oxidation half-reaction:
HOBr(aq) ->

Reduction half-reaction:
-> Br2(aq)

Step 3: Balance the number of atoms for each element except oxygen and hydrogen.

For the oxidation half-reaction, since there is no other element involved except hydrogen, there is nothing to balance.

For the reduction half-reaction, there is one bromine atom in Br2, so we can write '1 Br2(aq)'.

Step 4: Add water molecules (H2O) to balance the oxygen atoms. Since there is one oxygen atom on each side of the equation in this case, no water molecules are needed.

Step 5: Add hydrogen ions (H+) to balance the hydrogen atoms. Since there is one hydrogen atom on each side of the equation in this case, no hydrogen ions are needed.

Step 6: Adjust the charges on each side of the equation. Since the oxidation half-reaction is losing one electron, we need to add one electron (e-) to the right side:

HOBr(aq) -> + e-

Since the reduction half-reaction is gaining two electrons, we need to add two electrons (2e-) to the left side:

2 Br2(aq) + 2 e- ->

Step 7: Multiply each half-reaction by the necessary coefficients, so that the number of electrons lost in the oxidation half-reaction is equal to the number of electrons gained in the reduction half-reaction. In this case, we need to multiply the oxidation half-reaction by 2:

2 HOBr(aq) -> 2 + 2 e-

Now, the balanced half-reaction for the conversion of HOBr(aq) to Br2(aq) in acidic solution is:

2 HOBr(aq) -> + 2 e-
2 Br2(aq) + 2 e- ->

If needed, combine the two half-reactions together to obtain the balanced overall redox reaction.