For the following redox reactions
i) `2Fe^(2+) + H_2O_2 rarr xA + yB`
ii) `2MnO_4^- + 6H^+ + 5H_2O_2 rarr pA+ qB + rC`
so sum of (x+y+p+q+r)

Answer

To solve the given redox reactions and find the sum of \(x + y + p + q + r\), we will analyze each reaction step by step.

### Step 1: Analyze the first reaction
The first reaction is:
\[ 2Fe^{2+} + H_2O_2 \rightarrow xA + yB \]

1. **Identify the oxidation states**:
- In \(Fe^{2+}\), iron has an oxidation state of +2.
- In \(H_2O_2\) (hydrogen peroxide), the oxidation state of oxygen is -1.

2. **Determine the products**:
- \(Fe^{2+}\) can be oxidized to \(Fe^{3+}\) and \(H_2O_2\) can be reduced to water (\(H_2O\)).
- The balanced reaction can be written as:
\[ 2Fe^{2+} + H_2O_2 \rightarrow 2Fe^{3+} + 2H_2O \]

3. **Assign values to \(x\) and \(y\)**:
- Here, \(x = 2Fe^{3+}\) and \(y = 2H_2O\).
- Thus, \(x = 2\) and \(y = 2\).

### Step 2: Analyze the second reaction
The second reaction is:
\[ 2MnO_4^- + 6H^+ + 5H_2O_2 \rightarrow pA + qB + rC \]

1. **Identify the oxidation states**:
- In \(MnO_4^-\), manganese has an oxidation state of +7.
- In \(H_2O_2\), the oxidation state of oxygen is -1.

2. **Determine the products**:
- \(MnO_4^-\) is reduced to \(Mn^{2+}\) and \(H_2O_2\) is oxidized to \(O_2\).
- The balanced reaction can be written as:
\[ 2MnO_4^- + 6H^+ + 5H_2O_2 \rightarrow 2Mn^{2+} + 5O_2 + 8H_2O \]

3. **Assign values to \(p\), \(q\), and \(r\)**:
- Here, \(p = 2Mn^{2+}\), \(q = 5O_2\), and \(r = 8H_2O\).
- Thus, \(p = 2\), \(q = 5\), and \(r = 8\).

### Step 3: Calculate the sum
Now, we can calculate the total sum:
\[ x + y + p + q + r = 2 + 2 + 2 + 5 + 8 \]
\[ = 19 \]

### Final Answer
The sum of \(x + y + p + q + r\) is \(19\).