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Standard XII

Chemistry

Faraday's First Law

Three faraday...

Question

Three faraday of electricity is passed through three electrolytic cells connected in series containing $A g^{+}, C a^{2 +}$ and $A l^{+ 3}$ ions respectively. The molar ratio in which the three metal ions are liberated at the electrodes is:

1 : 2 : 3

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3 : 2 : 1

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6 : 3 : 2

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3 : 4 : 2

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Solution

The correct option is C $6 : 3 : 2$
$A g^{+} + e \to A g; C a^{2 +} + 2 e^{-} \to C l; A l^{3 +} + 3 e^{-} \to A l$
$2 F produce = 1 mole of A g; 2 F produce = 1 mole of C a; 3 F produce = 1 mole of A l$
$1 F produce = 1;$ $1 F produce = \frac{1}{2} mole;$ $1 F produce = \frac{1}{3} mole;$
Ratio of librated ions $= 1 : \frac{1}{2} : \frac{1}{3}$

$= 6 : \frac{6}{2} : \frac{6}{3}$

$= 6 : 3 : 2$

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Three faraday of electricity is passed through three electrolytic cells connected in series containing Ag+,Ca2+ and Al+3 ions respectively. The molar ratio in which the three metal ions are liberated at the electrodes is:

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Three faraday of electricity is passed through three electrolytic cells connected in series containing $A g^{+}, C a^{2 +}$ and $A l^{+ 3}$ ions respectively. The molar ratio in which the three metal ions are liberated at the electrodes is: