How many Faraday`s required to reduce 1mole dichromate to chronic salt

To determine how many Faradays are required to reduce one mole of dichromate (Cr2O7^2-) to chromium(III) salt (Cr^3+), we first need to understand the reduction process and the concept of Faraday's laws of electrolysis.

The Reduction Process

The dichromate ion, Cr2O7^2-, undergoes a reduction reaction in acidic conditions to form chromium(III) ions. The balanced half-reaction for this process can be represented as follows:

Half-Reaction

In acidic solution, the reduction of dichromate can be expressed as:

• Cr2O7^2- + 14H+ + 6e- → 2Cr^3+ + 7H2O

This equation shows that for every mole of dichromate ion, six electrons (6e-) are required to reduce it to chromium(III) ions.

Understanding Faraday's Constant

Faraday's constant (approximately 96485 C/mol) represents the charge of one mole of electrons. Therefore, to find out how many Faradays are needed, we can use the relationship between the number of moles of electrons and Faraday's constant.

Calculating Faradays

Since we need 6 moles of electrons to reduce 1 mole of dichromate, we can calculate the required Faradays as follows:

• 1 Faraday = 1 mole of electrons
• 6 moles of electrons = 6 Faradays

This means that to fully reduce 1 mole of dichromate to chromium(III) salt, you would need 6 Faradays of electric charge.

Summary

In summary, the reduction of 1 mole of dichromate to chromium(III) salt requires 6 Faradays. This is derived from the balanced half-reaction, which indicates that 6 electrons are necessary for the reduction process. Understanding this relationship is crucial in electrochemistry, especially when dealing with redox reactions and electrolysis.