Question

If $0.224\mathrm{L}$ ${\mathrm{H}}_2$ gas is formed at the cathode, the volume of ${\mathrm{O}}_2$ gas formed at the anode under identical conditions is?

Answer 1

Step 1: Given data

The volume of ${\mathrm{H}}_2$ gas at the cathode$=0.224\mathrm{L}$

Step 2: Calculating the volume of oxygen gas

A reduction process occurs at the cathode. $\underset{\mathrm{Water}}{2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}+2{\mathrm{e}}^{-}\to \underset{\mathrm{Hydrogen}\mathrm{gas}}{{\mathrm{H}}_2\left(\mathrm{g}\right)}+2{\mathrm{OH}}^{-}$

At the anode, oxidation occurs $\underset{\mathrm{water}}{2{\mathrm{H}}_2\mathrm{O}(\mathrm{l}})\to {\mathrm{O}}_2(\mathrm{g})+4{\mathrm{H}}^{+}+4{\mathrm{e}}^{-}$

At the cathode, ${\mathrm{H}}_2$ is released; the number of electrons involved is two, and the "n" factor is two.

The number of electrons participating in the liberation of ${\mathrm{O}}_2$ at anode is four, and the n factor is four.

Also, we know Equivalent weight$=$Moles $\times$n factor

The equivalent weight of ${\mathrm{O}}_2$$=$Equivalent weight of ${\mathrm{H}}_2$

Moles of ${\mathrm{O}}_2\times$n $=$Moles of ${\mathrm{H}}_2\times$n

Moles of ${\mathrm{H}}_2=\frac{0.224}{22.4}\Rightarrow {10}^{-2}$

Moles of ${\mathrm{O}}_2=\frac{\mathrm{Moles}\mathrm{of}{\mathrm{H}}_2}{2}\Rightarrow \frac{{10}^{-2}}{2}$

We know that, volume at STP$=\frac{{\mathrm{O}}_2\mathrm{Volume}}{22.4}\Rightarrow \frac{{\displaystyle \frac{{10}^{-2}}{2}}}{22.4}\Rightarrow 0.112\mathrm{L}$

Therefore, the volume ${\mathrm{O}}_2$ formed at the anode$=0.112\mathrm{L}$