What is the answer to the JEE Advanced 2025 dichromate current question?
The JEE Advanced 2025 dichromate electrolysis numerical gives a current of 100.00 amperes, because one mole of chromium(III) ions needs three moles of electrons, not six. From JEE Advanced 2025, Paper 1, Chemistry, it falls under Redox Reactions and Electrochemistry and tests Faraday’s laws of electrolysis.
An aqueous acidic solution containing dichromate is reduced to chromium(III) ions. Find the constant current needed to form one mole of these ions in 48.25 minutes, using:
This is a numerical-answer question, not an MCQ; no answer options are supplied. “Medium” and “90 seconds” are question-bank tags, not official exam classifications or a universal time limit.
How do you balance the dichromate reduction half-reaction?
Start with the chromium skeleton:
The balanced reaction requires six electrons to produce two chromium(III) ions. Use the official solution’s half-reaction method to establish both coefficients before calculating charge.
Balance the seven oxygen atoms by adding seven water molecules on the right. Since the medium is acidic, balance the resulting fourteen hydrogen atoms with hydrogen ions on the left.
Now calculate each side’s charge before adding electrons. Water is neutral, so it contributes nothing to the right-hand charge.
Add six electrons to the left to lower its charge by six units. This gives the balanced reduction half-reaction used in the official solution:
Both sides contain two chromium atoms, seven oxygen atoms and fourteen hydrogen atoms. The net charge on each side is positive six, so both atoms and charge balance.
Why does one mole of chromium(III) need three faradays, not six?
The balanced equation says that two moles of chromium(III) ions require six moles of electrons. The requested amount is one mole of product, not one mole of dichromate reactant. Keep the species attached to the mole units so the conversion is explicit:
Faraday’s constant is the charge carried by one mole of electrons. Multiply the electron amount by this charge per mole, not by the electron coefficient alone:
The electron-mole units cancel, leaving coulombs. Therefore, the charge required for the specified one mole of product is:
How do you convert the time and calculate the current?
Convert the given time to seconds, then divide the charge by that time. This gives 100 amperes because current is charge transferred per unit time, and an ampere means one coulomb per second.
Dividing by the time in minutes would give coulombs per minute, not amperes. Check the answer by multiplying current by time:
That is exactly the charge needed to produce one mole of chromium(III) ions. Use this sequence for the next numerical:
Balanced half-reaction → electron moles → charge → time in seconds → current.
Why is 200 amperes wrong for the stated question?
200 amperes is a derived incorrect numerical result, not an official option. It comes from assigning six moles of electrons directly to one mole of chromium(III) product. The arithmetic works, but the stoichiometry does not:
The coefficient six belongs to a reaction producing two moles of chromium(III) ions. It cannot be carried unchanged into a calculation asking for one mole.
At 200 amperes, the charge passed in the stated time would produce two moles of chromium(III) ions, assuming the same reaction. Correct the method by multiplying the desired product amount by the ratio of the electron coefficient to the product coefficient:
Write that ratio before inserting Faraday’s constant. It prevents the factor-of-two error at its source.
Which three practice questions check whether the method has stuck?
These three newly written practice questions are not verified JEE PYQs. They change the stoichiometric target, the unknown quantity and the reacting species, while retaining the electron-mole conversion in every solution. Use the following constant throughout:
What current reduces one mole of dichromate completely in 48.25 minutes?
The required current is 200 amperes. Here, one mole refers to the dichromate reactant, which produces two moles of chromium(III), rather than the original one mole of product.
At 50 amperes, how long does dichromate reduction take to form 0.50 mole of chromium(III)?
The time is 48.25 minutes for the original dichromate reduction. Convert product moles into electron moles first, then divide the resulting charge by the given current.
What current forms 0.20 mole of manganese(II) from acidic permanganate in 1930 seconds?
The required current is 50 amperes. Use the supplied acidic half-reaction: each mole of manganese(II) product requires five moles of electrons.
Cover the solutions and rewrite each electron-to-product ratio before calculating charge. For further PYQs, use JEEnius’s free archive of Advanced papers from 2007, searchable by year, subject or chapter, with worked solutions.
Next step: photograph a doubt on JEEnius and photograph any question you are stuck on and get a step-by-step solution, with a free-body diagram when the question needs one (20 free).
For a worked example of the same idea, see Rotational Motion JEE 2021: Torque and Angular Momentum.
Frequently asked questions
What is the answer to the JEE Advanced 2025 dichromate current question?
The required constant current is 100.00 A. Forming one mole of chromium(III) ions from acidic dichromate requires three moles of electrons, giving Q = 3 × 96500 = 289500 C. With 48.25 minutes equal to 2895 seconds, I = Q/t = 100 A.
How do you balance dichromate reduction in acidic medium?
The balanced half-reaction is Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O. Balance oxygen with water, hydrogen with H⁺, and then charge with electrons. Both sides have the same atom counts and a net charge of +6.
Why does one mole of chromium(III) need three faradays instead of six?
In acidic dichromate reduction, six moles of electrons produce two moles of chromium(III) ions. One mole of chromium(III) therefore requires 6/2 = 3 moles of electrons, or three faradays of charge. Six faradays would produce two moles of chromium(III).
Why is 200 A wrong for the JEE 2025 dichromate question?
The stated question asks for one mole of chromium(III) product, which requires three moles of electrons, not six. Using six moles of electrons gives 200 A and would produce two moles of chromium(III) in 48.25 minutes. The 200 A result is a stoichiometric error, not an official answer option.