JEE Advanced 2025 Paper 1 · Q08 · Electrolysis & Faraday's Laws
In an electrochemical cell, dichromate ion ($Cr_2O_7^{2-}$) in aqueous acidic medium is reduced to $Cr^{3+}$. Find the current (in amperes) that flows through the cell for 48.25 minutes to produce 1 mole of $Cr^{3+}$. Use 1 Faraday = 96500 C/mol.
Reveal answer + step-by-step solution
Correct answer:100
Solution
Half-reaction: $Cr_2O_7^{2-} + 14 H^+ + 6 e^- \to 2 Cr^{3+} + 7 H_2O$. So per mole of $Cr^{3+}$ produced, 3 moles of electrons are required. Charge needed = $3 \times 96500 = 289500$ C. Time = $48.25 \text{ min} \times 60 \text{ s/min} = 2895$ s. Current = $Q/t = 289500 / 2895 = 100$ A. Answer: 100.
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