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Question

The water is electrolysed in a cell , hydrogen is liberated at one electrode and oxygen is simultaneously liberated at the other . in a particular experiment hydrogen and oxygen so produced were collected together and the total volume measured 16.8 mL at NTP . How many coulombs were passed through the cell in the experiment ? 


Solution

$$ W_{H_{2}} = Z_{H_{2}} Q...(1)$$

$$ W_{O2} =Z_{O2} Q...(2)$$

$$(1)\Rightarrow W_{H_2} = \dfrac{gm\;eq\; wt\; of H_{2}}{96500}\times  Q\dfrac{mole \;wt}{\dfrac{2}{96500}\times Q}$$

$$\Rightarrow \dfrac{W_{H2}}{M_{H2}}= \dfrac{Q}{2\times 96500} \Rightarrow n_{H2}=\dfrac{Q}{96500\times 2}$$

$$\Rightarrow VH\times V_{H2}=\dfrac{2}{96500\times 2} \times 22400 ml$$$$ for O_{2}$$$$WO_{2} = \dfrac{gm eg wt O_{2}}{96500}\times  Q= \dfrac{mol wt O_{2}}{\dfrac{4}{96500}}\times Q$$$$ \Rightarrow \dfrac{W_{02}}{MO_{2}} = \dfrac{Q}{96500\times 4} \therefore no_{2} = \dfrac{Q}{96500\times 4}$$$$ \Rightarrow Vo_{2}=no_{2}\times 22400 ml =\dfrac{Q\times 22400}{96500\times 4}$$$$ A/q , VH_{2}+VO_{2} = \times \dfrac{Q\times 22400}{96500\times 2}+\dfrac{Q\times 22400}{96500\times 4}$$$$ 16.8 = \dfrac{Q}{965} (112+56) =\dfrac{Q}{965}\times 168 $$$$Q = \dfrac{16.8\times 965}{16.8} = 96.5 coulombs $$

Chemistry

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