Question

An acidic solution of \[C{u^{ + 2}}\] salt containing $0.4g$ of \[C{u^{ + 2}}\] is electrolyzed until all the copper is deposited. The electrolysis is continued for seven more minutes with the volume of solution kept at $100ml$ and the current at $1.2$ amp. Calculate the gas evolved at NTP during the entire electrolysis.

Answer 1

Hint: When electric current is supplied by an external source into the cell it will lead to various chemical changes during the process; this event is commonly known as electrolysis. In electrochemical cells oxidation occurs at anode and reduction will take place at cathode.

Complete answer:
When the process of electrolysis is carried out the process of oxidation and reduction will simultaneously take place at anode and cathode.
At anode water oxidized to release oxygen gas
$2{H_2}O \to 4{H^ + } + {O_2} + 4{e^ - }$
At cathode \[C{u^{ + 2}}\] ions get reduced to copper metal which will be deposited at the electrode.
$C{u^{ + 2}} + 2{e^ - } \to Cu$
Above reaction shows that the equivalent of oxygen liberated is equal to the equivalent of copper deposited.
Equivalent weight of copper metal is the ratio of its molecular mass and valency.
$Eq = \dfrac{{63.54}}{2}$
$Eq = 31.5g$
Since we have $0.4g$ of \[C{u^{ + 2}}\] ions so equivalent weight become
$0.4g$of \[C{u^{ + 2}}\]$ = \dfrac{{0.4}}{{31.5}}g$ equivalent
After solving we finally get the equivalent weight of $0.4g$\[C{u^{ + 2}}\] ions is
$0.4g$of \[C{u^{ + 2}}\]$ = 0.0127g$ equivalent
As we already discussed, the equivalent weight of copper and oxygen releases is equal. therefore, the mass of oxygen deposited at anode will be
Total mass of oxygen deposited at anode $ = 0.0127 \times \dfrac{8}{{32}}$
Total mass of oxygen deposited at anode$ = 0.0031$ moles
After complete deposition of copper ion at cathode further passage of electricity will results in liberation of hydrogen gas at cathode
$2{H_2}O + 2{e^ - } \to {H_2} + 2O{H^ - }$
Now equivalent of oxygen liberated is equal to equivalent of hydrogen liberated
As mentioned in the question
Current $C = 1.2$ amp
Time $t = 7 \times 60$
$t = 420$ seconds
Total amount of charge pass is $q = C \times t$
$q = 1.2 \times 420$
Hence, total amount of charge passed during process is
$q = 504$ coulombs
Amount of oxygen liberated is
$Eq = \dfrac{1}{{96500}} \times 504 \times \dfrac{8}{{32}}$
After solving above equation, we get
$Eq = 0.0013$ moles
Equivalent of hydrogen gas evolved at cathode will be
$Eq = \dfrac{1}{{96500}} \times 504 \times \dfrac{1}{2}$
After solving this we get
$Eq = 0.0026$ moles
Total moles of oxygen evolved during electrolysis is
$M = 0.0031 + 0.0013$
$M = 0.0044$ moles of oxygen
Hence, volume of oxygen evolved during electrolysis at NTP will be
$V = 0.0044 \times 22400$
$V = 98.56ml$ of oxygen gas
Similarly, total moles of hydrogen gas evolved is
$M = 0.0026$ moles of hydrogen
Hence, volume of hydrogen evolved during electrolysis at NTP will be
$V = 0.0026 \times 22400$
$V = 58.24ml$ of hydrogen gas
Hence, during the electrolysis $\left( {98.56ml} \right)$ of oxygen and $\left( {58.24ml} \right)$ of hydrogen gas is evolved at NTP.

Note:
NTP is the normal temperature and pressure of the chemical reaction which have specific values of temperature and pressure. During the NTP volume of one mole of gas will always be equal to $22.4$ litres or $22400$ milli-litres.