JEE Advanced 2025 · Paper 1 · Chemistry Q8

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ChemistryNumerical+4Question 40 of 48

In an electrochemical cell, dichromate ions in aqueous acidic medium are reduced to $\ce{Cr^3+}$ . The current (in amperes) that flows through the cell for 48.25 minutes to produce 1 mole of $\ce{Cr^3+}$ is _____.

Use: 1 Faraday $= 96500\ \mathrm{C\,mol^{-1}}$

Your answer

Answer and worked solution

Official answer: 100

In $\ce{Cr2O7^2-}$ chromium is +6, so each $\ce{Cr^3+}$ formed needs 3 electrons.

Charge. For 1 mol $\ce{Cr^3+}$ : $Q=3\times96500=289500$ C.

Current. $t=48.25\times60=2895$ s, so
$I=\frac{289500}{2895}=100\ \text{A}.$

Topics:Electrolysis and Faraday's laws Oxidation number

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