Question

During the electrolysis of CU (2) tetraoxosulphate (6) solution using platinum electrode,0.816g of Cu was deposited on the cathode.calculate the volume of oxygen gas liberated at S.T.P

Answer 1

This process can be expressed with the following equation:

$2CuSO_4+2H_2O \rightarrow 2Cu \downarrow+2H_2SO_4+O_2 \uparrow;$

Now, we have to found the amount of copper metal formed on the cathode in moles:

$n(Cu)=\frac{m(Cu)}{M(Cu)}=\frac{0.816g}{63.5g/mol}=0.0129mol;$

According to the equation, the amount of oxygen formed in moles is half of that of the copper metal:

$n(O_2)=\frac{1}{2}*n(Cu)=\frac{1}{2}*0.0129mol=6.45*10^{-3}mol;$

Finally, the volume of the released oxygen under STP is:

$V(O_2)=n(O_2)*V_m=6.45*10^{-3}mol*22.4L=0.144L$ or 144 mL;

Thus, it was 144 mL of oxygen released in the electrolysis.