In the following unbalanced redox reaction,
`Cu_(3)P+Cr_(2)O_(7)^(2-) rarr Cu^(2+)+H_(3)PO_(4)+Cr^(3+)`
Equivalent weight of `H_(3)PO_(4)` is

Explanation

To find the equivalent weight of \( H_3PO_4 \) in the given redox reaction, we can follow these steps:

### Step 1: Identify the Oxidation States
In the reaction:
\[ Cu_3P + Cr_2O_7^{2-} \rightarrow Cu^{2+} + H_3PO_4 + Cr^{3+} \]

- **Copper (Cu)** changes from +1 to +2 (oxidation).
- **Phosphorus (P)** changes from -3 in \( Cu_3P \) to +5 in \( H_3PO_4 \) (oxidation).
- **Chromium (Cr)** changes from +6 in \( Cr_2O_7^{2-} \) to +3 (reduction).

### Step 2: Determine the Change in Oxidation States
- For **Copper**: The change is \( +1 \) (from +1 to +2).
- For **Phosphorus**: The change is \( +8 \) (from -3 to +5).
- For **Chromium**: The change is \( -3 \) (from +6 to +3).

### Step 3: Write the Balanced Redox Reaction
The balanced equation for the reaction is:
\[ 3 Cu_3P + 3 Cr_2O_7^{2-} + 26 H^+ \rightarrow 6 Cu^{2+} + 6 Cr^{3+} + 2 H_3PO_4 + 13 H_2O \]

### Step 4: Calculate the Molar Mass of \( H_3PO_4 \)
The molar mass of \( H_3PO_4 \) is calculated as follows:
- Hydrogen (H): 1 g/mol × 3 = 3 g/mol
- Phosphorus (P): 31 g/mol × 1 = 31 g/mol
- Oxygen (O): 16 g/mol × 4 = 64 g/mol

Total molar mass of \( H_3PO_4 \):
\[ 3 + 31 + 64 = 98 \, \text{g/mol} \]

### Step 5: Calculate the Equivalent Weight
The equivalent weight is calculated using the formula:
\[ \text{Equivalent Weight} = \frac{\text{Molar Mass}}{\text{Change in Oxidation State}} \]

For \( H_3PO_4 \):
- Molar Mass = 98 g/mol
- Change in Oxidation State = 8 (from -3 to +5)

Thus, the equivalent weight of \( H_3PO_4 \) is:
\[ \text{Equivalent Weight} = \frac{98}{8} = 12.25 \, \text{g/equiv} \]

### Final Answer
The equivalent weight of \( H_3PO_4 \) is **12.25 g/equiv**.

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