Question

Electricity passed through $ CuS{O_4} $ solution results in the deposition of $ 0.4g $ $ Cu $ it cathode what is the volume of gas evolved at anode at STP?

Answer 1

Hint: See the reaction happening on the anode and the cathode for the $ CuS{O_4} $ solution. From that, we will see that oxygen gas will evolve at the anode. Hence, the equivalent of oxygen formed is equal to the equivalent of copper. After that, find the volume of oxygen evolved at anode at STP.
Equivalent of $ Cu = \dfrac{{mass\;of\;Cu\;deposited \times number\;of\;electron \;transfer}}{{molecular\;weight\;of\;Cu}} $
 $ Molecular\;weight\;of\;Cu = 63.5u $ .

Complete answer:
Given that through $ CuS{O_4} $ solution electricity is passed
Therefore, reaction at the anode
 $ \Rightarrow 2{H_2}O \to 4{H^{4 + }} + {O_2} + 4{e^ - } $
Reaction at cathode
 $ \Rightarrow C{u^{2 + }} + 2{e^ - } \to Cu $
From this, we can conclude
the equivalent of oxygen formed $ = $ to the equivalent of copper.
the equivalent of oxygen formed $ = \dfrac{{mass\;of\;Cu \times number\;of\;electron \;transfer}}{{molecular\;weight\;of\;Cu}} $
mass of $ Cu $ deposited is given $ 0.4g $
molecular weight of $ Cu = 63.5u $
 $ \Rightarrow \dfrac{{0.4 \times 2}}{{63.5}} $
the equivalent of oxygen formed $ \Rightarrow 12.58 \times {10^{ - 3}}g $ equivalent
It is already known that $ 1 $ mole of the gas (or $ 32g $ of $ {O_2} $ ) is equivalent to $ 22.4 $ Litres of the oxygen gas. Since the oxygen molecules are four we divide it by $ 4 $ for finding the equivalent of one $ {O_2} $
Hence, the equivalent of one $ {O_2} $ molecule at STP $ \Rightarrow \dfrac{{22.4 \times 12.58 \times {{10}^{ - 3}}}}{4} $
 $ \Rightarrow 0.0704L $
 $ \Rightarrow 7.04ml $
Hence, the volume of gas evolved at the anode at STP is $ 7.04ml.$

Note:
When acidulated water is electrolyzed, oxygen gas is produced at the anode. Because $ O{H^ - } $ is negatively charged, it gravitates towards the anode (positively charged electrode).
 $ SO_4^{2 - } $ , $ O{H^ - } $ , and $ {H^ + } $ are among the ions found in the solution.
The positive charge travels towards the cathode, whereas the negative charge travels towards the anode. As a result, when acidulated water is electrolyzed, hydrogen is collected at the cathode, and oxygen is collected at the anode.