The number of gram molecules of hydrogen liberated by the passage of $1$ Faraday of electricity through hydrogen ion solution is _____.

0.5

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1.0

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0.2

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4.0

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Solution

The correct option is B $0.5$
The reaction taking place is given below:
$2 H^{+} + 2 e^{-} \to H_{2}$
Number of moles of electrons $= n (e^{-}) = \frac{Q}{F}$ $= \frac{1 F}{F}$ $= 1$ mol
$2$ mol of electrons produce $1$ mol of $H_{2}$ according to the reaction given above.
Therefore, $1$ mol of electrons will produce $\frac{1}{2}$ mol $H_{2}$ $= 0.5$ mol of $H_{2}$

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The number of gram molecules of hydrogen liberated by the passage of 1 Faraday of electricity through hydrogen ion solution is _____.

The correct option is B 0.5The reaction taking place is given below:2H++2e−→H2Number of moles of electrons =n(e−)=QF =1FF =1 mol2 mol of electrons produce 1 mol of H2 according to the reaction given above. Therefore, 1 mol of electrons will produce 12 mol H2 =0.5 mol of H2

The number of gram molecules of hydrogen liberated by the passage of $1$ Faraday of electricity through hydrogen ion solution is _____.