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

Explanation

To solve the given redox reaction and find the values of \( p \), \( q \), \( r \), \( s \), and \( t \) in the reaction:

\[ p \text{Br}_2 + q \text{OH}^- \rightarrow r \text{Br}^- + s \text{BrO}_3^- + t \text{H}_2\text{O} \]

we will follow these steps:

### Step 1: Write the unbalanced reaction
The unbalanced reaction is:
\[ \text{Br}_2 + \text{OH}^- \rightarrow \text{Br}^- + \text{BrO}_3^- + \text{H}_2\text{O} \]

### Step 2: Assign oxidation states
- For \( \text{Br}_2 \): 0 (elemental state)
- For \( \text{Br}^- \): -1
- For \( \text{BrO}_3^- \): +5 (since each oxygen is -2 and the overall charge is -1)
- For \( \text{OH}^- \): -1 (hydroxide ion)
- For \( \text{H}_2\text{O} \): 0 (neutral)

### Step 3: Identify oxidation and reduction
- \( \text{Br}_2 \) is reduced to \( \text{Br}^- \) (gain of electrons).
- \( \text{Br}_2 \) is oxidized to \( \text{BrO}_3^- \) (loss of electrons).

### Step 4: Write half-reactions
**Reduction half-reaction:**
\[ \text{Br}_2 + 2e^- \rightarrow 2 \text{Br}^- \]

**Oxidation half-reaction:**
\[ \text{Br}_2 \rightarrow \text{BrO}_3^- + 10e^- \]

### Step 5: Balance the half-reactions
1. For the reduction half-reaction, we have:
\[ \text{Br}_2 + 2e^- \rightarrow 2 \text{Br}^- \]
(Multiply by 5 to balance with the oxidation half-reaction)

2. For the oxidation half-reaction, we have:
\[ 5 \text{Br}_2 + 12 \text{OH}^- \rightarrow 10 \text{Br}^- + 2 \text{BrO}_3^- + 6 \text{H}_2\text{O} \]

### Step 6: Combine the half-reactions
Combining the balanced half-reactions gives:
\[ 5 \text{Br}_2 + 12 \text{OH}^- \rightarrow 10 \text{Br}^- + 2 \text{BrO}_3^- + 6 \text{H}_2\text{O} \]

### Step 7: Identify coefficients \( p, q, r, s, t \)
From the balanced equation:
- \( p = 5 \)
- \( q = 12 \)
- \( r = 10 \)
- \( s = 2 \)
- \( t = 6 \)

### Final Values
Thus, the values are:
- \( p = 5 \)
- \( q = 12 \)
- \( r = 10 \)
- \( s = 2 \)
- \( t = 6 \)

---