How Much Charge in Faradays is Required to Reduce 1 Mole of Dichromate (Cr2O72-)?
One of the most frequently asked questions in the JEE Main and NEET Electrochemistry section involves calculating the quantity of electricity required for a specific redox transformation.
While the concept relies on Faraday's Laws of Electrolysis, the calculation is purely based on mastering Oxidation States and stoichiometry. Let's look at a classic Previous Year Question (PYQ) involving the reduction of the Dichromate ion.
Video Tutorial: The Balancing Trick
Watch Abhishek Sengar sir from CHEMCA break down the exact step-by-step method to solve this numerical without making the most common student error.
Step-by-Step Reaction Breakdown
Quantity of Electricity in Faradays (F) = Total number of Moles of Electrons transferred in the balanced half-reaction.
Remember: 1 mole of electrons carries a charge of exactly 1 Faraday (approx. 96,500 Coulombs).
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Write the Skeletal Equation:
Identify the reactant and the product. We are reducing Dichromate to Chromium (III) ions.
Cr2O72- → Cr3+ -
Balance the Central Atoms (Crucial Step):
This is where 90% of students make a mistake! There are 2 Chromium atoms on the reactant side, so you MUST place a coefficient of 2 on the product side before doing anything else.
Cr2O72- → 2 Cr3+ -
Calculate Total Oxidation Number Change:
Find the oxidation state of Cr in Cr2O72-: Let it be x.
2x + 7(-2) = -2 → 2x = +12 → x = +6.
Now, calculate the total charge on both sides for all Chromium atoms:
Reactant Side: 2 atoms × (+6) = +12
Product Side: 2 atoms × (+3) = +6
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Find the Difference (Moles of Electrons):
The difference between +12 and +6 is 6. Since oxidation number is decreasing, it is a reduction, meaning electrons are gained (added to the left side).
Cr2O72- + 6e- → 2 Cr3+
Fig: Stoichiometry of Dichromate Reduction. 6 moles of electrons are required.
Therefore, the charge required = 6 Faradays (6F).
Practice Questions for JEE & NEET
Ready to test your speed? Apply this exact logic to two other highly-repeated textbook examples.
Question 1: How much charge is required to reduce 1 mole of Permanganate ion (MnO4-) to Mn2+ in an acidic medium?
Answer: 5 Faradays (5F)
Reasoning:
- Skeletal Equation: MnO4- → Mn2+. (Mn is already balanced: 1 atom on both sides).
- Oxidation state of Mn in reactant (MnO4-): Let it be x. x + 4(-2) = -1 → x = +7.
- Oxidation state of Mn in product: +2.
- Difference: 7 - 2 = 5.
- Since 5 moles of electrons are required per mole of Permanganate, the charge is 5F.
Question 2: Calculate the charge in Faradays required to deposit 1 mole of Aluminum (Al) metal from a solution of Al2O3.
Answer: 3 Faradays (3F)
Reasoning:
Pay close attention to the wording! The question asks for the charge to deposit 1 mole of Aluminum (Al) metal, NOT 1 mole of Al2O3.
- Equation for the reduction of the ion: Al3+ + 3e- → Al(s)
- To produce 1 mole of Al(s) from Al3+, exactly 3 moles of electrons are needed.
- Therefore, the required charge is 3F.
(Note: If the question had asked for the reduction of 1 mole of Al2O3, the answer would be 6F, because 1 mole of Al2O3 contains 2 moles of Al3+).
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