JEE Advance - Chemistry (1993 - No. 19)

Chromium metal can be plated out from an acidic solution containing CrO₃ according to the following equation
CrO₃ (aq) + 6H⁺ (aq) + 6e^- $$\to$$ Cr(s) + 3H₂O
Calculate (i) how many grams of chromium will be plated out by 24,000 coulombs and (ii) how long will it take to plate out 1.5 g of chromium by using 12.5 amp current.

i) 2.155 g Cr, ii) 1336.15 s

i) 4.311 g Cr, ii) 668.08 s

i) 1.078 g Cr, ii) 2672.30 s

i) 2.155 g Cr, ii) 668.08 s

i) 4.311 g Cr, ii) 1336.15 s

Explanation

$$Cr{O_3}(aq) + 6{H^ + }(aq) + 6e \to Cr(s) + 3{H_2}O$$

52g of Cr is deposited by 6 $$\times$$ 96500 C of electric current. [$$\because$$ atomic mass of Cr = 52 g mol^$$-$$1]

(1) Given, 24000 C current is passed.

$$\therefore$$ Amount of Cr plated out = $${{52 \times 24000} \over {6 \times 96500}}$$ = 2.155 g

(2) According to Faraday's 1st law, W = Z $$\times$$ I $$\times$$ t

$$\therefore$$ 1.5 = Z $$\times$$ 12.5 $$\times$$ t

or, $$1.5 = {{52} \over {6 \times 96500}} \times 12.5 \times t$$

or, $$t = {{1.5 \times 6 \times 96500} \over {52 \times 12.5}} = 1336.15$$

JEE Advance - Chemistry (1993 - No. 19)

Explanation

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