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

9gm→1F;4.5gm→?=(4.5×1)/9=0.5F1F→11.2l;0.5F→?;VH2=0.5×11.2=5.6L

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

9gm→1F;4.5gm→?=(4.5×1)/9=0.5F1F→11.2l;0.5F→?;VH2=0.5×11.2=5.6L

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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

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5.6 L

11.2 L

44.8 L

22.4 L

answer is D.

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Detailed Solution

$\begin{array}{l} 9 gm \to 1 F; 4 .5 gm \to ? = (4 .5 \times 1) / 9 = 0 .5 F \\ 1 F \to 11 .2 l; 0 .5 F \to ?; V_{H_{2}} = 0 .5 \times 11 .2 = 5 .6 L \end{array}$

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