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Standard XII

Chemistry

Faraday's Second Law

A quantity of...

Question

A quantity of electrical charge that brings about the deposition of $4.5 g A l$ from $A l^{3 +}$ at the cathode will also produce the following volume at $(S T P)$ of $H_{2} (g)$ from $H^{+}$ at the cathode.

44.8 L

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22.4 L

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11.2 L

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5.6 L

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Solution

The correct option is D $5.6 L$
$2 H^{+} + 2 e^{-} \to H_{2}$
$A l^{3 +} + 3 e^{-} \to A l$
$\frac{w (H_{2})}{w (A l)} = \frac{E (H_{2})}{E (A l)} = \frac{1}{9}$
$w (H_{2}) = 4.5 \times \frac{1}{g} = 0.5 g = \frac{1}{4} m o l H_{2}$
$= \frac{22.4}{4} = 5.6 L H_{2}$ at STP

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A quantity of electrical charge that brings about the deposition of 4.5 g Al from Al3+ at the cathode will also produce the following volume at (STP) of H2(g) from H+ at the cathode.

The correct option is D 5.6 L2H++2e−→H2Al3++3e−→Alw(H2)w(Al)=E(H2)E(Al)=19w(H2)=4.5×1g=0.5 g=14mol H2=22.44=5.6 L H2 at STP

A quantity of electrical charge that brings about the deposition of $4.5 g A l$ from $A l^{3 +}$ at the cathode will also produce the following volume at $(S T P)$ of $H_{2} (g)$ from $H^{+}$ at the cathode.

The correct option is D $5.6 L$
$2 H^{+} + 2 e^{-} \to H_{2}$
$A l^{3 +} + 3 e^{-} \to A l$
$\frac{w (H_{2})}{w (A l)} = \frac{E (H_{2})}{E (A l)} = \frac{1}{9}$
$w (H_{2}) = 4.5 \times \frac{1}{g} = 0.5 g = \frac{1}{4} m o l H_{2}$
$= \frac{22.4}{4} = 5.6 L H_{2}$ at STP