Four faradays of electricity were passed through $A g N O_{3} (l), C d S O_{4} (l), A l C l_{3} (l) and P b C l_{4} (l)$ kept in four vessels using inert electrodes. The ratio of moles of $A g, C d, A l$ and $P b$ deposited will be:

12 : 4 : 6 : 3

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1 : 2 : 3 : 4

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12 : 6 : 4 : 3

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4 : 3 : 2 : 1

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Solution

The correct option is C $12 : 6 : 4 : 3$
$A g^{+} (a q) + e^{-} \to A g (s)$
$1 m o l = 1 F 1 m o l$
$3 F 3 m o l$

$C d^{2 +} (a q) + 2 e^{-} \to C d (s)$
$2 m o l = 2 F 1 m o l$
$3 F 3 / 2 m o l$

$A l^{3 +} (a q) + 3 e^{-} \to A l (s)$
$3 m o l = 3 F 1 m o l$

$P b^{4 +} (a q) + 4 e^{-} \to P b (s)$
$4 m o l = 4 F 1 m o l$
$3 F 3 / 4 m o l$

Therefore,
Ratio of moles $= 3 : \frac{3}{2} : 1 : \frac{3}{4} = 12 : 6 : 4 : 3$

Hence, option C is correct.

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B.12 : 6 : 4 : 3

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Four faradays of electricity were passed through AgNO3(l),CdSO4(l),AlCl3(l)andPbCl4(l) kept in four vessels using inert electrodes. The ratio of moles of Ag, Cd, Al and Pb deposited will be:

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Four faradays of electricity were passed through $A g N O_{3} (l), C d S O_{4} (l), A l C l_{3} (l) and P b C l_{4} (l)$ kept in four vessels using inert electrodes. The ratio of moles of $A g, C d, A l$ and $P b$ deposited will be: