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Electrolytic Cell

Passage of th...

Question

Passage of three faradays of charge through an aqueous solution of $A g N O_{3}, C u S O_{4}, A l (N O_{3})_{3} and N a C l$ will deposit metals at the cathode in the molar ratio of :

1 : 2 : 3 : 1

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6 : 3 : 2 : 6

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6 : 3 : 2 : 2

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3 : 2 : 1 : 2

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Solution

The correct option is B 6 : 3 : 2 : 6
Electrolysis of $a q . A g N O_{3}$
$A g N O_{3} (a q) + e^{-} \to A g (s) + N O_{3}^{-} (a q)$
Thus,
1 mole of $A g N O_{3}$ requires 1 mole of $e^{-}$ or 1 Faraday of electricity to produce one mole of $A g$ .
Hence, $3 F$ is used to produce 3 mole of $A g$ .

Electrolysis of $a q . C u S O_{4}$
$C u S O_{4} (a q) + 2 e^{-} \to C u (s) + S O_{4}^{2 -} (a q)$
1 mole of $C u S O_{4}$ requires 2 Faraday of Charge to produce 1 mole of $C u$
Hence, $3 F$ is used to produce $\frac{3}{2}$ mole of $C u$ .

Electrolysis of $a q . A l (N O_{3})_{3}$
$A l N O_{3} (a q) + 3 e^{-} \to A l (s) + 3 N O_{3}^{-} (a q)$
1 mole of $A l (N O_{3})_{3}$ requires 3 mole of $e^{-}$ or 3 Faraday of electricity to produce one mole of $A l$ .
Hence, $3 F$ is used to produce 1 mole of $A l$ .

Electrolysis of $a q . N a C l$
$N a C l (a q) + e^{-} \to N a (s) + C l^{-} (a q)$
1 mole of $N a C l$ requires 1 mole of $e^{-}$ or 1 Faraday of electricity to produce one mole of $N a$ .
Hence, $3 F$ is used to produce 3 mole of $N a$ .

The molar ratio will be :
$3 : \frac{3}{2} : 1 : 3$
or
$6 : 3 : 2 : 6$

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Electrolytic Cell

Standard XII Chemistry

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