JEE Advance - Chemistry (1997 - No. 11)
How many grams of silver could be plated out on a serving tray by electrolysis of a solution containing silver in +1 oxidation state for a period of 8.0 hours at a current of 8.46 amperes? What is the area of the tray if the thickness of the silver plating is 0.00254 cm? Density of silver is 10.5 g/cm3
272.6 g, 1.02 x 10^3 cm^2
100 g, 500 cm^2
500 g, 2.0 x 10^3 cm^2
200 g, 800 cm^2
300 g, 1.5 x 10^3 cm^2
Explanation
Charges passed through the cell = 8.46 $$\times$$ 3 $$\times$$ 3600 = 243648 coulombs
Electrode reaction : $$A{g^ + }(aq) + e \to Ag(s)$$ ..... [1]
243648 coulombs $$ \equiv {{243648} \over {96500}} \equiv 2.524$$ mol of electrons
According to equation (1) 2.524 mol of electrons = 180 $$\times$$ 2.524 = 272.6 g of Ag
$$\therefore$$ Volume of Ag plated out $$ = {W \over {density}} = {{272.68} \over {10.5}} = 25.969$$ cm3
Given, thickness of the Ag plating = 0.0254 cm
$$\therefore$$ Area of tray plated $$ = {{25.969} \over {0.0254}} = 1022.4 = 1.02 \times {10^3}$$ cm2
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