When water is electrolysed, hydrogen and oxygen gases are produced. If $1.008 g$ of $H_{2}$ is liberated at cathode, what mass of $O_{2}$ is formed at the anode?

32 g

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16 g

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8 g

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4 g

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Solution

The correct option is C $8 g$
$\frac{W_{1}}{W_{2}} = \frac{E_{1}}{E_{2}}$
$\frac{1.008}{W_{2}} = \frac{1.008}{8}$
$∴ W_{2} = 8 g$
where, $E_{1}$ and $E_{2}$ are equivalent masses of hydrogen and oxygen respectively.

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When water is electrolysed, hydrogen and oxygen gases are produced. If 1.008 g of H2 is liberated at cathode, what mass of O2 is formed at the anode?

The correct option is C 8 gW1W2=E1E21.008W2=1.0088∴W2=8 gwhere, E1 and E2 are equivalent masses of hydrogen and oxygen respectively.

When water is electrolysed, hydrogen and oxygen gases are produced. If $1.008 g$ of $H_{2}$ is liberated at cathode, what mass of $O_{2}$ is formed at the anode?

The correct option is C $8 g$
$\frac{W_{1}}{W_{2}} = \frac{E_{1}}{E_{2}}$
$\frac{1.008}{W_{2}} = \frac{1.008}{8}$
$∴ W_{2} = 8 g$
where, $E_{1}$ and $E_{2}$ are equivalent masses of hydrogen and oxygen respectively.