If `0.224 L` of `H_(2)(g)` is formed at the cathode of one cell at `STP,` how much of `Mg` is formed at the cathode of the other electrolytic cell arranged in series ?

Options

A. `0.24g`
B. `2.4g`
C. `0.48g`
D. `4.8g`

Answer

a

Explanation

Moles of `H_(2)(g)=(0.224L)/(22.4L)=10^(-2)=10^(-2)xx10^(3)`
`10m` moles
First method
`2H_(2)O+2e^(-) rarr H_(2)+overset(c-)(O)H`
`2F-=1 mol H_(2)`
`2mF=1m mol H_(2)`
or
`10 mmol of H_(2)=20mF` electricity
`-=10m mol of Mg` deposited
`-=10xx10^(-3)xx24g` of `Mg`
`-=0.24 g `of `Mg`
`(Mg^(2+)+2e^(-)rarr Mg)`
Second method
`1mF=1mEq `of `H_(2)`
`(n` factor of `H_(2)=2)`
`10mmol `of `H_(2)=20mEq` of `H_(2)`
`=20mF` electricity
`1mF =1mEq `of `Mg(n` factor for `Mg=2)`
`20mF=20mEq` or `Mg`
`=20xx10^(-3)xx12g ` of `Mg( Ew` of `Mg=(24)/(12)=12)`
`=0.24g ` of `Mg`