Number of moles of K₂Cr₂O_{7 in acidic medium required to oxidise one mole of} Cu₃P to CuSO_{4 and} H₃PO_{4 is …}

To determine the number of moles of potassium dichromate (K2Cr2O7) needed to oxidize one mole of copper phosphide (Cu3P) to copper sulfate (CuSO4) and phosphoric acid (H3PO4) in an acidic medium, we first need to look at the redox reaction involved.

Understanding the Reaction

Copper phosphide reacts with potassium dichromate in an acidic solution. In this reaction, copper (Cu) is oxidized and phosphorus (P) is oxidized to phosphoric acid (H3PO4). The key here is to balance the oxidation and reduction half-reactions to find out how many moles of K2Cr2O7 are required.

Oxidation States

Let's break it down step by step:

• The oxidation state of copper in Cu3P is +1.
• In CuSO4, copper has an oxidation state of +2.
• Phosphorus in Cu3P has an oxidation state of -3 and in H3PO4, it has an oxidation state of +5.

Half-Reactions

The oxidation half-reactions can be summarized as follows:

• For copper: Cu₃P → 3Cu²⁺ + 3e^-
• For phosphorus: P + 8H⁺ + 8e^- → H₃PO₄

Balancing the Electrons

Now, we balance the electrons transferred in both half-reactions:

• The oxidation of 1 mole of Cu3P gives 3 moles of Cu²⁺ and consumes 8 moles of electrons from the reduction of phosphorus.
• To balance these, we need to make sure the total number of electrons lost equals the total number gained.

Calculating Moles of K2Cr2O7

In acidic medium, potassium dichromate acts as an oxidizing agent and its reduction can be represented as:

• Cr2O7^2- + 14H⁺ + 6e^- → 2Cr³⁺ + 7H₂O

This means that 1 mole of K2Cr2O7 can accept 6 moles of electrons. Since we determined that the total electrons needed to oxidize 1 mole of Cu3P to CuSO4 and H3PO4 is 8, we can set up a ratio:

Final Calculation

To find out how many moles of K2Cr2O7 are required:

• Let x be the number of moles of K2Cr2O7 needed.
• Using the stoichiometry: 6x = 8 (where 6 is the number of electrons accepted by 1 mole of K2Cr2O7).

Solving for x gives:

• x = 8/6 = 4/3 or approximately 1.33 moles.

Thus, to oxidize one mole of Cu3P to CuSO4 and H3PO4, you need about 1.33 moles of K2Cr2O7 in acidic medium.