Question

A current of 0.75A is passed through an acidic solution of $$Cu{ SO }_{ 4 }$$ for 10 minutes. The volume of oxygen liberated at anode (at STP) will be:

Solution

Given:-Time $$\left( t \right) = 10 \; min = 10 \times 60 = 600 \; s$$Current passed $$\left( i \right) = 0.75 \; A$$From Faraday's law of electrolysis,$$Q = nF$$Whereas, $$n$$ is the no. of moles of electrons used$$\Rightarrow i \times t = nF \; \left( \because q = I \times t \right)$$$$\Rightarrow n = \cfrac{i \times t}{F}$$$$\Rightarrow n = \cfrac{0.75 \times 600}{96500} = \cfrac{4.5}{965} \text{ mol}$$Now,$${H}_{2}O \longrightarrow 4 {H}^{+} + {O}_{2} + 4 {e}^{-}$$Amount of $${O}_{2}$$ released at STP when $$4$$ mole of electrons are used $$= 22400 \; mL$$Amount of $${O}_{2}$$ released at STP when $$\cfrac{4.5}{965}$$ mole of electrons are used $$= \cfrac{22400}{4} \times \left( \cfrac{4.5}{965} \right) = 26.11 \; mL$$Hence te volume of oxygen liberated at anode (at STP) will be $$26.11 \; mL$$.Chemistry

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