Values of `p`, `q`, `r`, `s` and `t` are in the following redox reaction
`pBr_(2)+qOH^(-) rarr rBr^(-)+sBrO_(3)^(-)+tH_(2)O`

Options

A. (`p=3`, `q=6`, `r=1`, `s=5`, `t=3`)
B. (`p=3`, `q=6`, `r=5`, `s=3`, `t=1`)
C. (`p=3`, `q=6`, `r=5`, `s=1`, `t=3`)
D. (`p=3`, `q=5`, `r=1`, `s=6`, `t=3`)

Explanation

To solve the redox reaction \( pBr_2 + qOH^- \rightarrow rBr^- + sBrO_3^- + tH_2O \), we need to balance the reaction using the ion-electron method. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the Half Reactions
First, we identify the reduction and oxidation half-reactions.

- **Reduction Half-Reaction**: \( Br_2 \) is reduced to \( Br^- \).
- **Oxidation Half-Reaction**: \( Br_2 \) is oxidized to \( BrO_3^- \).

### Step 2: Balance the Reduction Half-Reaction
For the reduction half-reaction:
\[ Br_2 + 2e^- \rightarrow 2Br^- \]
- Here, we have 2 bromine atoms on the left side, so we need 2 \( Br^- \) on the right side.
- Each \( Br \) goes from an oxidation state of 0 to -1, meaning each \( Br \) gains 1 electron, thus 2 \( Br \) gain 2 electrons.

### Step 3: Balance the Oxidation Half-Reaction
For the oxidation half-reaction:
\[ Br_2 \rightarrow 2BrO_3 + 10e^- \]
- We balance the bromine first. Since we have 2 \( Br \) on the left, we need 2 \( BrO_3^- \) on the right.
- Each \( Br \) goes from 0 to +5 (in \( BrO_3^- \)), meaning each \( Br \) loses 5 electrons, thus 2 \( Br \) lose 10 electrons.

### Step 4: Balance Oxygen and Hydrogen
In the oxidation half-reaction, we have 6 oxygen atoms in \( 2BrO_3^- \). To balance the oxygen, we add 6 water molecules to the left side:
\[ Br_2 + 6H_2O \rightarrow 2BrO_3^- + 10e^- + 12H^+ \]
- Now, we have 12 hydrogen atoms on the right side, so we need to balance hydrogen by adding \( 12OH^- \) to both sides (since we are in a basic medium):
\[ Br_2 + 6H_2O + 12OH^- \rightarrow 2BrO_3^- + 10e^- + 12H_2O \]
- The \( 12H^+ \) and \( 12OH^- \) combine to form 12 water molecules, and we can cancel out 6 \( H_2O \) from both sides.

### Step 5: Combine the Half-Reactions
Now we combine the balanced half-reactions:
\[ Br_2 + 12OH^- \rightarrow 2BrO_3^- + 10e^- + 6H_2O \]
\[ 2Br_2 + 2e^- \rightarrow 2Br^- \]

### Step 6: Equalize Electrons and Final Reaction
To equalize the electrons, we multiply the reduction half-reaction by 5:
\[ 5(Br_2 + 12OH^- \rightarrow 2BrO_3^- + 10e^- + 6H_2O) \]
This gives us:
\[ 3Br_2 + 6OH^- \rightarrow 5Br^- + BrO_3^- + 3H_2O \]

### Final Balanced Equation
The final balanced equation is:
\[ 3Br_2 + 6OH^- \rightarrow 5Br^- + BrO_3^- + 3H_2O \]

### Step 7: Identify Values of p, q, r, s, and t
From the balanced equation, we can identify:
- \( p = 3 \)
- \( q = 6 \)
- \( r = 5 \)
- \( s = 1 \)
- \( t = 3 \)

### Final Answer
The values are:
- \( p = 3 \)
- \( q = 6 \)
- \( r = 5 \)
- \( s = 1 \)
- \( t = 3 \)