One Faraday of electricity is passed separately through one litre of one molar aqueous solutions of (i) AgNO₃ (ii) SnCl₄ and (iii) CuSO₄. The number of moles of Ag, Sn and Cu deposited at cathode are respectively

Easy

Solution

$1F = 1eq = \frac{{At.wt}}{n} = \frac{{mole}}{n}$
$XF = Xeq = X \times \frac{{At.wt}}{n} = X \times \frac{{mole}}{n}$

electrolyte moles of metal deposited

AgNO₃ 1

SnCl₄ 1/4 = 0.25

CuSO₄ 1/2 = 0.5

Ratio = 1 : 0.25 :0.5

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