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 (STP) of $H_{2} (g)$ from $H^{\oplus}$ at the cathode:

5.6

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11.2

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22.4

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Solution

The correct option is A 5.6
Reactions involved are:
$2 H^{\oplus} + 2 e^{-} \to H_{2}$

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

$\frac{W (H_{2})}{W (A l)} = \frac{E q (H_{2})}{E q (A l)} = \frac{1}{9}$

$W (H_{2}) = 4.5 \times \frac{1}{9} = 0.5 g o f H_{2}$

$= \frac{1}{4} m o l H_{2}$

$= 5.6 L H_{2} a t S T P$

Hence, option A is correct.

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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 (STP) of H2(g) from H⊕ at the cathode:

The correct option is A 5.6Reactions involved are:2H⊕+2e−→H2Al3++3e−→AlW(H2)W(Al)=Eq(H2)Eq(Al)=19W(H2)=4.5×19=0.5gofH2=14molH2=5.6LH2at STPHence, option A is correct.

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 (STP) of $H_{2} (g)$ from $H^{\oplus}$ at the cathode: