In the electrolysis of water,why is the volume of gas collected over one electrode double that of the gas collected over the other electrode?
(N/A) During the electrolysis of water,the water molecule $(H_{2}O)$ decomposes into hydrogen gas $(H_{2})$ and oxygen gas $(O_{2})$ in a specific molar ratio.
The chemical equation for this process is:
$2H_{2}O(l) \xrightarrow{\text{Electrolysis}} 2H_{2}(g) + O_{2}(g)$
According to the balanced chemical equation,for every $2$ moles of water decomposed,$2$ moles of hydrogen gas are produced at the cathode and $1$ mole of oxygen gas is produced at the anode.
Therefore,the volume of hydrogen gas collected is double the volume of oxygen gas collected,resulting in a ratio of $2:1$.
The chemical equation for this process is:
$2H_{2}O(l) \xrightarrow{\text{Electrolysis}} 2H_{2}(g) + O_{2}(g)$
According to the balanced chemical equation,for every $2$ moles of water decomposed,$2$ moles of hydrogen gas are produced at the cathode and $1$ mole of oxygen gas is produced at the anode.
Therefore,the volume of hydrogen gas collected is double the volume of oxygen gas collected,resulting in a ratio of $2:1$.
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