One mole of electron passes through each of the solution of $A g N O_{3}, C u S O_{4}$ and $A l C l_{3}$ then $A g, C u$ and $A l$ are deposited at cathode. The molar ratio of $A g, C u$ and $A l$ deposited are:

1 : 1 : 1

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6 : 3 : 2

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6 : 3 : 1

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1 : 3 : 6

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Solution

The correct option is B 6 : 3 : 2
Reduction of metals in each solution with one mole of electron is:

$A g^{+} + e \to A g$

$\frac{1}{2} C u^{2 +} + e^{-} \to \frac{1}{2} C u$

$\frac{1}{3} A l^{3 +} + e^{-} \to \frac{1}{3} A l$

Thus one mole of electron will reduce 1 mol $A g N O_{3}$ , 0.5 mol $C u S O_{4}$ and 0.33 mol $A l C l_{3}$

Molar ratio of $A g$ , $C u$ and $A l$ are $1 : 0.5 : 0.33$ or $6 : 3 : 2$

Hence, option B is correct.

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