JEE Advanced 2025 — Paper 1 — CHEM-III solution

JEE Advanced2025Paper 1CHEM-III

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

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

Your answer

Solution

Balanced half-reaction:

$Cr_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 Cr^{3 +} + 7 H_{2} O$

So 6 mol of electrons produce 2 mol of Cr $^{3 +}$ ; per mole of Cr $^{3 +}$ , 3 mol of electrons (3 F) are needed.

Charge required: $Q = 3 \times 96500 = 289500$ C.

Time: $t = 48.25 min \times 60 = 2895$ s.

Current: $I = \frac{Q}{t} = \frac{289500}{2895} = 100$ A.

$∴ I = 100.00$ A.

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